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Mathematics Quiz - Class 11 Mathematics Permutations And...

Question 1/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

द्विपद प्रसार में ( (1+x)^{12} ) का \( x^5 \) वाला गुणांक कितना है?

What is the coefficient of \( x^5 \) in the expansion of ( (1+x)^{12} )?

Explanation opens after your attempt

Correct Answer

D. (792)

Step 1

Concept

The coefficient is \( \binom{12}{5} \), so the answer is (792). In exams, for ( (1+x)^n ), the coefficient of \(x^r\) is \( \binom{n}{r} \).

Step 2

Why this answer is correct

The correct answer is D. (792). The coefficient is \( \binom{12}{5} \), so the answer is (792). In exams, for ( (1+x)^n ), the coefficient of \(x^r\) is \( \binom{n}{r} \).

Step 3

Exam Tip

गुणांक \( \binom{12}{5} \) है, इसलिए उत्तर (792) है। परीक्षा में ( (1+x)^n ) में \(x^r\) का गुणांक सीधे \( \binom{n}{r} \) लें।

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Question 2/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(7) पुरुषों और (5) महिलाओं में से (5) सदस्यों की समिति बनानी है जिसमें कम से कम (2) महिलाएँ हों। कुल कितने चयन होंगे?

From (7) men and (5) women, a committee of (5) members is to be formed with at least (2) women. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (596)

Step 1

Concept

Taking the number of women as (2,3,4,5) and adding gives (596). In such questions, add all valid cases separately.

Step 2

Why this answer is correct

The correct answer is B. (596). Taking the number of women as (2,3,4,5) and adding gives (596). In such questions, add all valid cases separately.

Step 3

Exam Tip

महिलाओं की संख्या (2,3,4,5) लेकर योग करने पर (596) मिलता है। ऐसे प्रश्नों में सभी वैध मामलों को अलग-अलग जोड़ें।

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Question 3/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

यदि \( \binom{n}{4}=\binom{n}{2} \) और \( n\ge 4 \) है, तो ( n ) का मान क्या है?

If \( \binom{n}{4}=\binom{n}{2} \) and \( n\ge 4 \), what is the value of ( n )?

Explanation opens after your attempt

Correct Answer

D. (6)

Step 1

Concept

By symmetry, (4=n-2), so (n=6). In exams, when \( \binom{n}{r}=\binom{n}{s} \), check (r=s) or (r+s=n).

Step 2

Why this answer is correct

The correct answer is D. (6). By symmetry, (4=n-2), so (n=6). In exams, when \( \binom{n}{r}=\binom{n}{s} \), check (r=s) or (r+s=n).

Step 3

Exam Tip

सममिति से (4= n-2), इसलिए (n=6) है। परीक्षा में \( \binom{n}{r}=\binom{n}{s} \) हो तो (r=s) या (r+s=n) जाँचें।

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Question 4/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

एक उत्तल दशभुज में विकर्णों की संख्या कितनी होगी?

How many diagonals are there in a convex decagon?

Explanation opens after your attempt

Correct Answer

A. (35)

Step 1

Concept

The total segments are \( \binom{10}{2} \), and subtracting (10) sides gives (35). For polygons, first count all vertex pairs.

Step 2

Why this answer is correct

The correct answer is A. (35). The total segments are \( \binom{10}{2} \), and subtracting (10) sides gives (35). For polygons, first count all vertex pairs.

Step 3

Exam Tip

कुल रेखाखंड \( \binom{10}{2} \) हैं और (10) भुजाएँ घटाने पर (35) बचते हैं। बहुभुज में पहले सभी शीर्ष-युग्म गिनें।

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Question 5/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (10) तक की संख्याओं में से (3) संख्याएँ ऐसी चुननी हैं कि कोई दो क्रमागत न हों। कुल चयन कितने हैं?

From the numbers (1) to (10), (3) numbers are to be selected so that no two are consecutive. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (56)

Step 1

Concept

The no-consecutive formula gives \( \binom{10-3+1}{3}=\binom{8}{3}=56 \). In such questions, the gap method is very fast.

Step 2

Why this answer is correct

The correct answer is A. (56). The no-consecutive formula gives \( \binom{10-3+1}{3}=\binom{8}{3}=56 \). In such questions, the gap method is very fast.

Step 3

Exam Tip

क्रमागत न होने का सूत्र \( \binom{10-3+1}{3}=\binom{8}{3}=56 \) देता है। ऐसे प्रश्नों में खाली स्थान विधि बहुत तेज होती है।

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Question 6/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

ताश की (52) पत्तों की गड्डी से (5) पत्ते चुनने हैं जिनमें ठीक (2) बादशाह हों। कुल कितने हाथ बनेंगे?

From a deck of (52) cards, (5) cards are chosen with exactly (2) kings. How many hands are possible?

Explanation opens after your attempt

Correct Answer

D. (103776)

Step 1

Concept

The selection is \( \binom{4}{2}\binom{48}{3}=103776 \). For exactly-type questions, choose special objects first and then the remaining objects.

Step 2

Why this answer is correct

The correct answer is D. (103776). The selection is \( \binom{4}{2}\binom{48}{3}=103776 \). For exactly-type questions, choose special objects first and then the remaining objects.

Step 3

Exam Tip

चयन \( \binom{4}{2}\binom{48}{3}=103776 \) है। ठीक वाले प्रश्नों में पहले विशेष वस्तुएँ और फिर बाकी वस्तुएँ चुनें।

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Question 7/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(12) बिंदुओं में से (5) बिंदु एक ही सीधी रेखा पर हैं। इन बिंदुओं से बनने वाले त्रिभुजों की संख्या कितनी है?

Among (12) points, (5) points are collinear. How many triangles can be formed from these points?

Explanation opens after your attempt

Correct Answer

B. (210)

Step 1

Concept

Subtract the impossible \( \binom{5}{3} \) from total \( \binom{12}{3} \), giving (210). A triangle needs three non-collinear points.

Step 2

Why this answer is correct

The correct answer is B. (210). Subtract the impossible \( \binom{5}{3} \) from total \( \binom{12}{3} \), giving (210). A triangle needs three non-collinear points.

Step 3

Exam Tip

कुल \( \binom{12}{3} \) में से असंभव \( \binom{5}{3} \) घटाएँ, उत्तर (210) है। त्रिभुज के लिए तीन असरेखीय बिंदु चाहिए।

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Question 8/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(6) अलग-अलग पुस्तकों में से कम से कम एक पुस्तक चुनने के कुल कितने तरीके हैं?

How many ways are there to choose at least one book from (6) distinct books?

Explanation opens after your attempt

Correct Answer

A. (63)

Step 1

Concept

There are \(2^6\) subsets, and removing the empty selection gives (63). For at least one, remember to subtract the empty case.

Step 2

Why this answer is correct

The correct answer is A. (63). There are \(2^6\) subsets, and removing the empty selection gives (63). For at least one, remember to subtract the empty case.

Step 3

Exam Tip

सभी उपसमुच्चय \(2^6\) हैं, खाली चयन हटाने पर (63) मिलते हैं। कम से कम एक में खाली स्थिति घटाना याद रखें।

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Question 9/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(6) वरिष्ठ और (4) कनिष्ठ विद्यार्थियों में से (4) सदस्य चुनने हैं जिनमें ठीक (3) वरिष्ठ और (1) कनिष्ठ हो। कुल चयन कितने हैं?

From (6) seniors and (4) juniors, (4) members are chosen with exactly (3) seniors and (1) junior. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (80)

Step 1

Concept

The count is \( \binom{6}{3}\binom{4}{1}=80 \). For selections from separate groups, use the multiplication rule.

Step 2

Why this answer is correct

The correct answer is B. (80). The count is \( \binom{6}{3}\binom{4}{1}=80 \). For selections from separate groups, use the multiplication rule.

Step 3

Exam Tip

गिनती \( \binom{6}{3}\binom{4}{1}=80 \) है। अलग समूहों से चयन में गुणन नियम लगाएँ।

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Question 10/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(5) लाल, (4) नीली और (3) हरी गेंदों में से (4) गेंदें चुननी हैं जिनमें हर रंग की कम से कम एक गेंद हो। कुल चयन कितने हैं?

From (5) red, (4) blue, and (3) green balls, (4) balls are chosen with at least one ball of each color. How many selections are possible?

Explanation opens after your attempt

Correct Answer

D. (270)

Step 1

Concept

The three cases of color distribution (2,1,1) add up to (270). When capacities are limited, keep the cases by color separate.

Step 2

Why this answer is correct

The correct answer is D. (270). The three cases of color distribution (2,1,1) add up to (270). When capacities are limited, keep the cases by color separate.

Step 3

Exam Tip

रंग-वितरण (2,1,1) के तीन मामलों का योग (270) देता है। क्षमता सीमाएँ होने पर हर रंग के मामलों को अलग रखें।

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Question 11/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

एक ग्रिड में (5) दाएँ और (3) ऊपर कदमों से गंतव्य तक जाने के न्यूनतम रास्ते कितने होंगे?

In a grid, how many shortest paths reach the destination using (5) right steps and (3) up steps?

Explanation opens after your attempt

Correct Answer

B. (56)

Step 1

Concept

Choose the (3) up-step positions among (8) total steps, so \( \binom{8}{3}=56 \). In paths, selecting step positions is a combination idea.

Step 2

Why this answer is correct

The correct answer is B. (56). Choose the (3) up-step positions among (8) total steps, so \( \binom{8}{3}=56 \). In paths, selecting step positions is a combination idea.

Step 3

Exam Tip

कुल (8) स्थानों में (3) ऊपर कदम चुनने हैं, इसलिए \( \binom{8}{3}=56 \) है। रास्तों में कदमों की जगहें चुनना संयोजन है।

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Question 12/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

समीकरण (x+y+z=12) के ऐसे पूर्णांक हल कितने हैं जिनमें \(x\ge2\), \(y\ge3\) और \(z\ge0\) हो?

How many integer solutions of (x+y+z=12) satisfy \(x\ge2\), \(y\ge3\), and \(z\ge0\)?

Explanation opens after your attempt

Correct Answer

A. (36)

Step 1

Concept

After subtracting minimum values, (a+b+z=7), so \( \binom{9}{2}=36 \). First convert constraints to zero-based variables.

Step 2

Why this answer is correct

The correct answer is A. (36). After subtracting minimum values, (a+b+z=7), so \( \binom{9}{2}=36 \). First convert constraints to zero-based variables.

Step 3

Exam Tip

न्यूनतम मान घटाने पर (a+b+z=7) मिलता है, इसलिए \( \binom{9}{2}=36 \)। पहले शर्तों को शून्य-आधारित बनाइए।

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Question 13/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

समीकरण (x+y+z=10) के ऐसे शून्य सहित पूर्णांक हल कितने हैं जिनमें हर चर (5) से अधिक नहीं है?

How many nonnegative integer solutions of (x+y+z=10) have each variable not greater than (5)?

Explanation opens after your attempt

Correct Answer

D. (21)

Step 1

Concept

From total \( \binom{12}{2} \), subtract \(3\binom{6}{2}\) cases where a variable is (6) or more, giving (21). For upper bounds, complementary counting is useful.

Step 2

Why this answer is correct

The correct answer is D. (21). From total \( \binom{12}{2} \), subtract \(3\binom{6}{2}\) cases where a variable is (6) or more, giving (21). For upper bounds, complementary counting is useful.

Step 3

Exam Tip

कुल \( \binom{12}{2} \) से किसी चर के (6) या अधिक होने के \(3\binom{6}{2}\) मामले घटते हैं, उत्तर (21) है। ऊपरी सीमा में पूरक गिनती उपयोगी रहती है।

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Question 14/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(8) व्यंजन और (5) स्वर में से (5) अक्षर चुनने हैं जिनमें कम से कम (2) स्वर हों। कुल चयन कितने हैं?

From (8) consonants and (5) vowels, (5) letters are selected with at least (2) vowels. How many selections are possible?

Explanation opens after your attempt

Correct Answer

D. (881)

Step 1

Concept

Taking vowels as (2,3,4,5) gives a total of (881). For an at-least condition, add all permitted counts.

Step 2

Why this answer is correct

The correct answer is D. (881). Taking vowels as (2,3,4,5) gives a total of (881). For an at-least condition, add all permitted counts.

Step 3

Exam Tip

स्वरों की संख्या (2,3,4,5) लेकर योग (881) मिलता है। कम से कम वाली शर्त में सभी अनुमत गिनतियाँ जोड़ें।

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Question 15/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(9) अवयवों वाले समुच्चय के विषम संख्या वाले उपसमुच्चयों की संख्या कितनी है?

How many subsets of odd cardinality does a set with (9) elements have?

Explanation opens after your attempt

Correct Answer

B. (256)

Step 1

Concept

Odd- and even-sized subsets are equal in number, so the count is \(2^8=256\). In such questions, half the subsets have odd size.

Step 2

Why this answer is correct

The correct answer is B. (256). Odd- and even-sized subsets are equal in number, so the count is \(2^8=256\). In such questions, half the subsets have odd size.

Step 3

Exam Tip

विषम और सम आकार के उपसमुच्चय बराबर होते हैं, इसलिए संख्या \(2^{8}=256\) है। ऐसे प्रश्नों में आधे उपसमुच्चय विषम आकार के होते हैं।

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Question 16/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (9) तक के अंकों में से (3) अलग अंक चुनकर केवल अंक-समुच्चय बनाना है। कुल कितने समुच्चय बनेंगे?

From digits (1) to (9), (3) distinct digits are chosen only as a digit-set. How many sets are possible?

Explanation opens after your attempt

Correct Answer

A. (84)

Step 1

Concept

Order does not matter, so \( \binom{9}{3}=84 \). When forming a set, never count order.

Step 2

Why this answer is correct

The correct answer is A. (84). Order does not matter, so \( \binom{9}{3}=84 \). When forming a set, never count order.

Step 3

Exam Tip

क्रम का महत्व नहीं है, इसलिए \( \binom{9}{3}=84 \) है। समुच्चय बनाते समय क्रम को कभी न गिनें।

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Question 17/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(12) व्यक्तियों में से (5) चुनने हैं, पर दो विशेष व्यक्ति साथ-साथ नहीं चुने जाने चाहिए। कुल चयन कितने हैं?

From (12) persons, (5) are to be selected, but two particular persons must not be selected together. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (672)

Step 1

Concept

From total \( \binom{12}{5} \), subtract \( \binom{10}{3} \) selections containing both particular persons, giving (672). For not-together conditions, subtract forbidden cases.

Step 2

Why this answer is correct

The correct answer is A. (672). From total \( \binom{12}{5} \), subtract \( \binom{10}{3} \) selections containing both particular persons, giving (672). For not-together conditions, subtract forbidden cases.

Step 3

Exam Tip

कुल \( \binom{12}{5} \) में से दोनों विशेष व्यक्तियों वाले \( \binom{10}{3} \) चयन घटते हैं, उत्तर (672) है। साथ न होने पर निषिद्ध स्थिति घटाएँ।

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Question 18/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(11) विद्यार्थियों में दो विशेष विद्यार्थी (A) और (B) हैं। (5) विद्यार्थी ऐसे चुनने हैं कि (A) या (B) में से कम से कम एक चुना जाए। कुल चयन कितने हैं?

Among (11) students, two particular students are (A) and (B). How many ways can (5) students be chosen so that at least one of (A) or (B) is selected?

Explanation opens after your attempt

Correct Answer

D. (336)

Step 1

Concept

Subtract \( \binom{9}{5} \), where neither (A) nor (B) is selected, from total \( \binom{11}{5} \), giving (336). For at least one, complement is simplest.

Step 2

Why this answer is correct

The correct answer is D. (336). Subtract \( \binom{9}{5} \), where neither (A) nor (B) is selected, from total \( \binom{11}{5} \), giving (336). For at least one, complement is simplest.

Step 3

Exam Tip

कुल \( \binom{11}{5} \) से (A,B) दोनों न होने वाले \( \binom{9}{5} \) घटाएँ, उत्तर (336) है। कम से कम एक के लिए पूरक सबसे सरल है।

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Question 19/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(8) लड़कों और (6) लड़कियों में से (6) सदस्य चुनने हैं, जिनमें लड़कियाँ लड़कों की संख्या की दोगुनी हों। कुल चयन कितने हैं?

From (8) boys and (6) girls, (6) members are chosen so that the number of girls is twice the number of boys. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (420)

Step 1

Concept

The condition gives (2) boys and (4) girls, so \( \binom{8}{2}\binom{6}{4}=420 \). First find the distribution of numbers.

Step 2

Why this answer is correct

The correct answer is B. (420). The condition gives (2) boys and (4) girls, so \( \binom{8}{2}\binom{6}{4}=420 \). First find the distribution of numbers.

Step 3

Exam Tip

शर्त से लड़के (2) और लड़कियाँ (4) होंगी, इसलिए \( \binom{8}{2}\binom{6}{4}=420 \)। पहले संख्या-वितरण निकालें।

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Question 20/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(5) पंक्तियों और (4) स्तंभों वाले छोटे आयतों के ग्रिड में कुल आयतों की संख्या कितनी है?

How many rectangles are there in a grid of (5) rows and (4) columns of small rectangles?

Explanation opens after your attempt

Correct Answer

B. (150)

Step 1

Concept

Choosing two horizontal lines from (6) and two vertical lines from (5) gives \( \binom{6}{2}\binom{5}{2}=150 \). A rectangle needs two horizontal and two vertical lines.

Step 2

Why this answer is correct

The correct answer is B. (150). Choosing two horizontal lines from (6) and two vertical lines from (5) gives \( \binom{6}{2}\binom{5}{2}=150 \). A rectangle needs two horizontal and two vertical lines.

Step 3

Exam Tip

(6) क्षैतिज और (5) ऊर्ध्व रेखाओं में से दो-दो चुनने पर \( \binom{6}{2}\binom{5}{2}=150 \) मिलता है। आयत के लिए दो क्षैतिज और दो ऊर्ध्व रेखाएँ चाहिए।

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Question 21/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

एक परिवार की (4) समानांतर रेखाएँ और दूसरी परिवार की (5) समानांतर रेखाएँ हैं। इनसे बनने वाले समांतर चतुर्भुजों की संख्या कितनी है?

There are (4) parallel lines in one family and (5) parallel lines in another family. How many parallelograms are formed?

Explanation opens after your attempt

Correct Answer

D. (60)

Step 1

Concept

A parallelogram needs (2) lines from each family, so \( \binom{4}{2}\binom{5}{2}=60 \). The key idea is choosing two parallel lines from each family.

Step 2

Why this answer is correct

The correct answer is D. (60). A parallelogram needs (2) lines from each family, so \( \binom{4}{2}\binom{5}{2}=60 \). The key idea is choosing two parallel lines from each family.

Step 3

Exam Tip

एक समांतर चतुर्भुज के लिए हर परिवार से (2) रेखाएँ चाहिए, इसलिए \( \binom{4}{2}\binom{5}{2}=60 \)। दो-दो समानांतर रेखाएँ चुनना मुख्य विचार है।

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Question 22/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (15) तक की संख्याओं में से (4) संख्याएँ ऐसी चुननी हैं कि कोई दो क्रमागत न हों। कुल चयन कितने हैं?

From the numbers (1) to (15), (4) numbers are chosen so that no two are consecutive. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (495)

Step 1

Concept

The formula gives \( \binom{15-4+1}{4}=\binom{12}{4}=495 \). For no-consecutive selections, take combinations from reduced positions.

Step 2

Why this answer is correct

The correct answer is A. (495). The formula gives \( \binom{15-4+1}{4}=\binom{12}{4}=495 \). For no-consecutive selections, take combinations from reduced positions.

Step 3

Exam Tip

सूत्र \( \binom{15-4+1}{4}=\binom{12}{4}=495 \) देता है। क्रमागत-वर्जित चयन में घटे हुए स्थानों से संयोजन लें।

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Question 23/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (8) तक की संख्याओं में से (3) संख्याएँ ऐसी चुननी हैं जिनका योग सम हो। कुल चयन कितने हैं?

From the numbers (1) to (8), (3) numbers are chosen so that their sum is even. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (28)

Step 1

Concept

The sum is even when (3) even numbers or (2) odd and (1) even number are chosen, giving (28). Count parity cases separately.

Step 2

Why this answer is correct

The correct answer is B. (28). The sum is even when (3) even numbers or (2) odd and (1) even number are chosen, giving (28). Count parity cases separately.

Step 3

Exam Tip

योग सम तब होगा जब (3) सम या (2) विषम और (1) सम चुने जाएँ, कुल (28) हैं। सम-विषम मामलों को अलग-अलग गिनें।

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Question 24/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

ताश की गड्डी से (5) पत्तों का ऐसा हाथ चुनना है जिसमें सभी पत्ते एक ही सूट के हों। कुल कितने हाथ होंगे?

A (5)-card hand is chosen from a deck so that all cards are from the same suit. How many hands are possible?

Explanation opens after your attempt

Correct Answer

D. (5148)

Step 1

Concept

From one suit there are \( \binom{13}{5} \) selections and there are (4) suits, so the count is (5148). Choose the suit and then the cards.

Step 2

Why this answer is correct

The correct answer is D. (5148). From one suit there are \( \binom{13}{5} \) selections and there are (4) suits, so the count is (5148). Choose the suit and then the cards.

Step 3

Exam Tip

किसी एक सूट से \( \binom{13}{5} \) चयन और (4) सूट हैं, इसलिए (5148) है। सूट चुनना और फिर पत्ते चुनना दो चरण हैं।

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Question 25/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(14) वस्तुओं में से (7) चुननी हैं, जिनमें एक निश्चित (5) वस्तुओं के समूह से ठीक (3) वस्तुएँ हों। कुल चयन कितने हैं?

From (14) objects, (7) are chosen with exactly (3) objects from a fixed group of (5). How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (1260)

Step 1

Concept

The selection is \( \binom{5}{3}\binom{9}{4}=1260 \). Exactly (3) means the remaining (4) come from outside the fixed group.

Step 2

Why this answer is correct

The correct answer is A. (1260). The selection is \( \binom{5}{3}\binom{9}{4}=1260 \). Exactly (3) means the remaining (4) come from outside the fixed group.

Step 3

Exam Tip

चयन \( \binom{5}{3}\binom{9}{4}=1260 \) है। ठीक (3) का अर्थ है बाकी (4) बाहर वाले समूह से आएँगे।

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Question 26/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

यदि \( \binom{12}{r}=\binom{12}{r+2} \), तो ( r ) का मान क्या है?

If \( \binom{12}{r}=\binom{12}{r+2} \), what is the value of ( r )?

Explanation opens after your attempt

Correct Answer

A. (5)

Step 1

Concept

By symmetry, (r+(r+2)=12), so (r=5). In such questions, make the lower indices add to (n).

Step 2

Why this answer is correct

The correct answer is A. (5). By symmetry, (r+(r+2)=12), so (r=5). In such questions, make the lower indices add to (n).

Step 3

Exam Tip

सममिति से (r+(r+2)=12), इसलिए (r=5) है। ऐसे प्रश्नों में निचले सूचकांकों का योग (n) के बराबर रखें।

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Question 27/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

यदि \( \binom{11}{r}=\binom{11}{r-3} \), तो ( r ) का मान क्या है?

If \( \binom{11}{r}=\binom{11}{r-3} \), what is the value of ( r )?

Explanation opens after your attempt

Correct Answer

B. (7)

Step 1

Concept

By symmetry, (r+(r-3)=11), so (r=7). In \( \binom{n}{a}=\binom{n}{b} \), often (a+b=n) is useful.

Step 2

Why this answer is correct

The correct answer is B. (7). By symmetry, (r+(r-3)=11), so (r=7). In \( \binom{n}{a}=\binom{n}{b} \), often (a+b=n) is useful.

Step 3

Exam Tip

सममिति से (r+(r-3)=11), इसलिए (r=7) है। \( \binom{n}{a}=\binom{n}{b} \) में अक्सर (a+b=n) काम आता है।

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Question 28/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(8) दंपतियों में से (4) व्यक्ति चुनने हैं, पर कोई विवाहित जोड़ा साथ नहीं चुना जाना चाहिए। कुल चयन कितने हैं?

From (8) couples, (4) persons are chosen, but no married couple should be selected together. How many selections are possible?

Explanation opens after your attempt

Correct Answer

D. (1120)

Step 1

Concept

First choose (4) couples and then (1) person from each, giving \( \binom{8}{4}2^4=1120 \). Choosing only one from a couple gives (2) choices.

Step 2

Why this answer is correct

The correct answer is D. (1120). First choose (4) couples and then (1) person from each, giving \( \binom{8}{4}2^4=1120 \). Choosing only one from a couple gives (2) choices.

Step 3

Exam Tip

पहले (4) दंपति चुनें और हर दंपति से (1) व्यक्ति लें, \( \binom{8}{4}2^4=1120 \)। जोड़े से केवल एक चुनने पर (2) विकल्प बनते हैं।

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Question 29/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(6) दंपतियों में से (5) व्यक्ति चुनने हैं जिनमें ठीक एक विवाहित जोड़ा शामिल हो। कुल चयन कितने हैं?

From (6) couples, (5) persons are chosen with exactly one married couple included. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (480)

Step 1

Concept

Choose the couple \( \binom{6}{1} \), then (3) other couples \( \binom{5}{3} \), and one person from each \(2^3\), giving (480). After fixing exactly one couple, choose only one person from each remaining couple.

Step 2

Why this answer is correct

The correct answer is C. (480). Choose the couple \( \binom{6}{1} \), then (3) other couples \( \binom{5}{3} \), and one person from each \(2^3\), giving (480). After fixing exactly one couple, choose only one person from each remaining couple.

Step 3

Exam Tip

जोड़ा \( \binom{6}{1} \), बाकी (3) दंपति \( \binom{5}{3} \), और उनसे एक-एक व्यक्ति \(2^3\), कुल (480) है। ठीक एक जोड़ा रखने के बाद बाकी जोड़ों से केवल एक व्यक्ति चुनें।

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Question 30/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(10) व्यक्तियों में (3) विशेष व्यक्ति हैं। (6) सदस्यों की समिति में कम से कम (2) विशेष व्यक्ति होने चाहिए। कुल चयन कितने हैं?

Among (10) persons, (3) are special. A committee of (6) must contain at least (2) special persons. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (140)

Step 1

Concept

The cases of exactly (2) special and exactly (3) special persons add to (140). For at least (2), add both valid cases.

Step 2

Why this answer is correct

The correct answer is A. (140). The cases of exactly (2) special and exactly (3) special persons add to (140). For at least (2), add both valid cases.

Step 3

Exam Tip

ठीक (2) विशेष और ठीक (3) विशेष के मामलों का योग (140) है। कम से कम (2) में दोनों वैध मामले जोड़ें।

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Question 31/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(4) प्रकार के फलों से (6) फल चुनने हैं और हर प्रकार की मात्रा असीमित है। कुल चयन कितने हैं?

From (4) types of fruits, (6) fruits are to be selected, with unlimited quantity of each type. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (84)

Step 1

Concept

This is the number of nonnegative solutions of \(x_1+x_2+x_3+x_4=6\), so \( \binom{9}{3}=84 \). For unlimited identical types, use stars and bars.

Step 2

Why this answer is correct

The correct answer is B. (84). This is the number of nonnegative solutions of \(x_1+x_2+x_3+x_4=6\), so \( \binom{9}{3}=84 \). For unlimited identical types, use stars and bars.

Step 3

Exam Tip

यह \(x_1+x_2+x_3+x_4=6\) के अशून्येतर नहीं बल्कि शून्य सहित हल हैं, इसलिए \( \binom{9}{3}=84 \)। असीमित समान प्रकारों में सितारे और पट्टियाँ लगाएँ।

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Question 32/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(12) समान कलमों को (3) विद्यार्थियों में बाँटना है ताकि हर विद्यार्थी को कम से कम (2) कलमें मिलें। कुल वितरण कितने हैं?

How many ways can (12) identical pens be distributed among (3) students so that each gets at least (2) pens?

Explanation opens after your attempt

Correct Answer

C. (28)

Step 1

Concept

After giving (2) pens to each student, (6) pens remain, so \( \binom{8}{2}=28 \). Subtract the minimum condition and then count nonnegative solutions.

Step 2

Why this answer is correct

The correct answer is C. (28). After giving (2) pens to each student, (6) pens remain, so \( \binom{8}{2}=28 \). Subtract the minimum condition and then count nonnegative solutions.

Step 3

Exam Tip

हर विद्यार्थी को (2) देने के बाद (6) कलमें बचती हैं, इसलिए \( \binom{8}{2}=28 \)। न्यूनतम शर्त घटाकर फिर शून्य सहित हल गिनें।

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Question 33/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

समीकरण \(x_1+x_2+x_3+x_4=20\) में \(x_1\ge5\) और \(x_2,x_3,x_4\ge1\) हों, तो धनात्मक पूर्णांक हलों की संख्या कितनी है?

For \(x_1+x_2+x_3+x_4=20\), with \(x_1\ge5\) and \(x_2,x_3,x_4\ge1\), how many positive integer solutions are there?

Explanation opens after your attempt

Correct Answer

D. (455)

Step 1

Concept

After subtracting minimums (5,1,1,1), (12) remains, so \( \binom{15}{3}=455 \). Subtract unequal minimum conditions first.

Step 2

Why this answer is correct

The correct answer is D. (455). After subtracting minimums (5,1,1,1), (12) remains, so \( \binom{15}{3}=455 \). Subtract unequal minimum conditions first.

Step 3

Exam Tip

न्यूनतम (5,1,1,1) घटाने पर (12) बचता है, इसलिए \( \binom{15}{3}=455 \)। असमान न्यूनतम शर्तों को पहले घटाएँ।

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Question 34/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

समीकरण \(x_1+x_2+x_3+x_4=18\) के ऐसे शून्य सहित पूर्णांक हल कितने हैं जिनमें \(x_1\le5\) हो?

How many nonnegative integer solutions of \(x_1+x_2+x_3+x_4=18\) satisfy \(x_1\le5\)?

Explanation opens after your attempt

Correct Answer

C. (875)

Step 1

Concept

Subtract \( \binom{15}{3} \) cases with \(x_1\ge6\) from total \( \binom{21}{3} \), giving (875). For upper limits, subtract forbidden solutions.

Step 2

Why this answer is correct

The correct answer is C. (875). Subtract \( \binom{15}{3} \) cases with \(x_1\ge6\) from total \( \binom{21}{3} \), giving (875). For upper limits, subtract forbidden solutions.

Step 3

Exam Tip

कुल \( \binom{21}{3} \) से \(x_1\ge6\) वाले \( \binom{15}{3} \) घटते हैं, उत्तर (875) है। ऊपरी सीमा में प्रतिबंधित हल घटाना आसान है।

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Question 35/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(9) अवयवों वाले समुच्चय में (3) विशेष अवयव हैं। (5) अवयवों के ऐसे उपसमुच्चय कितने हैं जिनमें तीनों विशेष अवयव एक साथ न हों?

A set of (9) elements has (3) special elements. How many (5)-element subsets do not contain all three special elements together?

Explanation opens after your attempt

Correct Answer

A. (111)

Step 1

Concept

Subtract \( \binom{6}{2} \) subsets containing all three special elements from total \( \binom{9}{5} \), giving (111). In not-type questions, subtract forbidden selections.

Step 2

Why this answer is correct

The correct answer is A. (111). Subtract \( \binom{6}{2} \) subsets containing all three special elements from total \( \binom{9}{5} \), giving (111). In not-type questions, subtract forbidden selections.

Step 3

Exam Tip

कुल \( \binom{9}{5} \) से तीनों विशेष अवयवों वाले \( \binom{6}{2} \) घटाएँ, उत्तर (111) है। नहीं वाले प्रश्न में निषिद्ध चयन घटाएँ।

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Question 36/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(9) अवयवों वाले समुच्चय में (3) विशेष अवयव हैं। (5) अवयवों के कितने उपसमुच्चय ठीक (2) विशेष अवयव रखते हैं?

A set of (9) elements has (3) special elements. How many (5)-element subsets contain exactly (2) special elements?

Explanation opens after your attempt

Correct Answer

C. (60)

Step 1

Concept

The selection is \( \binom{3}{2}\binom{6}{3}=60 \). With exactly (2) special elements, take the remaining (3) from ordinary elements.

Step 2

Why this answer is correct

The correct answer is C. (60). The selection is \( \binom{3}{2}\binom{6}{3}=60 \). With exactly (2) special elements, take the remaining (3) from ordinary elements.

Step 3

Exam Tip

चयन \( \binom{3}{2}\binom{6}{3}=60 \) है। ठीक (2) विशेष होने पर बाकी (3) सामान्य अवयवों से लें।

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Question 37/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(10) अवयवों वाले समुच्चय में (a) और (b) विशेष हैं। ऐसे उपसमुच्चय कितने हैं जो (a) को रखते हैं लेकिन (b) को नहीं रखते?

In a set of (10) elements, (a) and (b) are special. How many subsets contain (a) but do not contain (b)?

Explanation opens after your attempt

Correct Answer

B. (256)

Step 1

Concept

(a) is fixed as selected and (b) is fixed as not selected, so the remaining (8) elements are free, giving \(2^8=256\). Fix compulsory states first.

Step 2

Why this answer is correct

The correct answer is B. (256). (a) is fixed as selected and (b) is fixed as not selected, so the remaining (8) elements are free, giving \(2^8=256\). Fix compulsory states first.

Step 3

Exam Tip

(a) निश्चित रूप से चुना और (b) निश्चित रूप से छोड़ा गया, बाकी (8) अवयव स्वतंत्र हैं, इसलिए \(2^8=256\)। निश्चित अवस्थाओं को पहले स्थिर करें।

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Question 38/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(12) अवयवों वाले समुच्चय में (4) विशेष अवयव हैं। (5) अवयवों के ऐसे उपसमुच्चय कितने हैं जिनमें कम से कम (2) विशेष अवयव हों?

A set of (12) elements has (4) special elements. How many (5)-element subsets contain at least (2) special elements?

Explanation opens after your attempt

Correct Answer

C. (456)

Step 1

Concept

Taking the number of special elements as (2,3,4) gives (456). With a small special group, directly add valid cases.

Step 2

Why this answer is correct

The correct answer is C. (456). Taking the number of special elements as (2,3,4) gives (456). With a small special group, directly add valid cases.

Step 3

Exam Tip

विशेष अवयवों की संख्या (2,3,4) लेकर योग (456) है। छोटे विशेष समूह में सीधे वैध मामलों को जोड़ना आसान है।

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Question 39/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (15) तक की संख्याओं में से (4) संख्याएँ चुननी हैं ताकि उनका गुणनफल विषम हो। कुल चयन कितने हैं?

From the numbers (1) to (15), (4) numbers are chosen so that their product is odd. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (70)

Step 1

Concept

The product is odd only when all four numbers are odd, so \( \binom{8}{4}=70 \). For product parity, check every factor.

Step 2

Why this answer is correct

The correct answer is C. (70). The product is odd only when all four numbers are odd, so \( \binom{8}{4}=70 \). For product parity, check every factor.

Step 3

Exam Tip

गुणनफल विषम तभी होगा जब चारों संख्याएँ विषम हों, इसलिए \( \binom{8}{4}=70 \)। गुणनफल की सम-विषम प्रकृति में सभी कारकों पर ध्यान दें।

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Question 40/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (20) तक की संख्याओं में से (3) संख्याएँ ऐसी चुननी हैं जिनका योग विषम हो। कुल चयन कितने हैं?

From the numbers (1) to (20), (3) numbers are chosen so that their sum is odd. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (570)

Step 1

Concept

For an odd sum, choose (1) odd and (2) even or (3) odd numbers, giving (570). Making a parity case table reduces errors.

Step 2

Why this answer is correct

The correct answer is B. (570). For an odd sum, choose (1) odd and (2) even or (3) odd numbers, giving (570). Making a parity case table reduces errors.

Step 3

Exam Tip

योग विषम के लिए (1) विषम और (2) सम या (3) विषम चुनें, कुल (570) है। सम-विषम संयोजन तालिका बनाकर गलती घटती है।

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Question 41/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(1) से (12) तक की संख्याओं में से (4) संख्याएँ ऐसी चुननी हैं जिनका योग सम हो। कुल चयन कितने हैं?

From the numbers (1) to (12), (4) numbers are chosen so that their sum is even. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (255)

Step 1

Concept

For an even sum, the number of odd selected numbers can be (0,2,4), giving (255). Focus on how many odd elements are selected, not only on total selection size.

Step 2

Why this answer is correct

The correct answer is C. (255). For an even sum, the number of odd selected numbers can be (0,2,4), giving (255). Focus on how many odd elements are selected, not only on total selection size.

Step 3

Exam Tip

सम योग के लिए विषम संख्याओं की संख्या (0,2,4) हो सकती है, कुल (255) है। चयन की संख्या पर नहीं, विषम चुने गए तत्वों की संख्या पर ध्यान दें।

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Question 42/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(10) व्यक्तियों में से पहले (3) सदस्य वाली समिति और फिर बचे लोगों में से (4) सदस्य वाली अलग नामित समिति बनानी है। कुल तरीके कितने हैं?

From (10) persons, first a named committee of (3) members and then a different named committee of (4) members from the remaining persons are formed. How many ways are possible?

Explanation opens after your attempt

Correct Answer

C. (4200)

Step 1

Concept

The first committee can be formed in \( \binom{10}{3} \) ways and the second in \( \binom{7}{4} \) ways, giving (4200). For named committees, the stages matter.

Step 2

Why this answer is correct

The correct answer is C. (4200). The first committee can be formed in \( \binom{10}{3} \) ways and the second in \( \binom{7}{4} \) ways, giving (4200). For named committees, the stages matter.

Step 3

Exam Tip

पहली समिति \( \binom{10}{3} \) और दूसरी \( \binom{7}{4} \) तरीकों से बनेगी, कुल (4200) है। नामित समितियों में चरणों का क्रम महत्त्व रखता है।

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Question 43/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(13) व्यक्तियों में दो विशेष व्यक्ति (P) और (Q) हैं। (6) व्यक्ति ऐसे चुनने हैं कि (P,Q) या तो दोनों चुने जाएँ या दोनों न चुने जाएँ। कुल चयन कितने हैं?

Among (13) persons, two particular persons are (P) and (Q). How many ways can (6) persons be chosen so that (P,Q) are either both selected or both not selected?

Explanation opens after your attempt

Correct Answer

C. (792)

Step 1

Concept

If both are selected, count \( \binom{11}{4} \); if neither is selected, count \( \binom{11}{6} \), giving (792). Split either both or neither into two cases.

Step 2

Why this answer is correct

The correct answer is C. (792). If both are selected, count \( \binom{11}{4} \); if neither is selected, count \( \binom{11}{6} \), giving (792). Split either both or neither into two cases.

Step 3

Exam Tip

दोनों चुने जाने पर \( \binom{11}{4} \) और दोनों न चुने जाने पर \( \binom{11}{6} \), कुल (792) है। या तो दोनों या कोई नहीं को दो अलग मामलों में बाँटें।

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Question 44/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

(10) व्यक्तियों में (A) और (B) विशेष हैं। (5) व्यक्ति ऐसे चुनने हैं कि (A) और (B) में से ठीक एक चुना जाए। कुल चयन कितने हैं?

Among (10) persons, (A) and (B) are special. How many ways can (5) persons be chosen so that exactly one of (A) and (B) is selected?

Explanation opens after your attempt

Correct Answer

A. (140)

Step 1

Concept

There are (2) ways to choose the special person and then (4) from the other (8), so \(2\binom{8}{4}=140\). Exactly one means two symmetric cases.

Step 2

Why this answer is correct

The correct answer is A. (140). There are (2) ways to choose the special person and then (4) from the other (8), so \(2\binom{8}{4}=140\). Exactly one means two symmetric cases.

Step 3

Exam Tip

विशेष व्यक्ति चुनने के (2) तरीके और बाकी (4) व्यक्ति (8) में से, इसलिए \(2\binom{8}{4}=140\)। ठीक एक का अर्थ है दो समान मामले।

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Question 45/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

एक रेखा पर (7) बिंदु और उसके समांतर दूसरी रेखा पर (5) बिंदु हैं। इनसे ऐसे चतुर्भुज कितने बनेंगे जिनके शीर्ष दिए गए बिंदु हों?

There are (7) points on one line and (5) points on another parallel line. How many quadrilaterals can be formed using these points as vertices?

Explanation opens after your attempt

Correct Answer

C. (210)

Step 1

Concept

A quadrilateral needs (2) points from each line, so \( \binom{7}{2}\binom{5}{2}=210 \). Taking (3) points from one line cannot form a quadrilateral.

Step 2

Why this answer is correct

The correct answer is C. (210). A quadrilateral needs (2) points from each line, so \( \binom{7}{2}\binom{5}{2}=210 \). Taking (3) points from one line cannot form a quadrilateral.

Step 3

Exam Tip

चतुर्भुज के लिए हर रेखा से (2) बिंदु चाहिए, इसलिए \( \binom{7}{2}\binom{5}{2}=210 \)। एक ही रेखा से (3) बिंदु लेने पर चतुर्भुज नहीं बनेगा।

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Question 46/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

शब्द MISSISSIPPI के अक्षरों से (4) अक्षरों का चयन करना है, समान अक्षरों को अलग नहीं माना जाएगा। कुल अलग चयन कितने हैं?

From the letters of the word MISSISSIPPI, (4) letters are selected, and identical letters are not distinguished. How many distinct selections are possible?

Explanation opens after your attempt

Correct Answer

C. (21)

Step 1

Concept

Using the limits \(M\le1\), \(I\le4\), \(S\le4\), and \(P\le2\), the total selections are (21). For repeated letters, count bounded integer solutions.

Step 2

Why this answer is correct

The correct answer is C. (21). Using the limits \(M\le1\), \(I\le4\), \(S\le4\), and \(P\le2\), the total selections are (21). For repeated letters, count bounded integer solutions.

Step 3

Exam Tip

अक्षर-सीमाएँ \(M\le1\), \(I\le4\), \(S\le4\), \(P\le2\) रखकर कुल (21) चयन मिलते हैं। बहुलता वाले अक्षरों में सीमित पूर्णांक हल गिनें।

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Question 47/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

शब्द BANANA के अक्षरों से (5) अक्षरों का चयन करना है, समान अक्षरों को अलग नहीं माना जाएगा। कुल अलग चयन कितने हैं?

From the letters of the word BANANA, (5) letters are selected, and identical letters are not distinguished. How many distinct selections are possible?

Explanation opens after your attempt

Correct Answer

B. (3)

Step 1

Concept

The valid selections are \(A^3N^2\), \(BA^2N^2\), and \(BA^3N\), so the total is (3). In small repetition questions, list valid distributions directly.

Step 2

Why this answer is correct

The correct answer is B. (3). The valid selections are \(A^3N^2\), \(BA^2N^2\), and \(BA^3N\), so the total is (3). In small repetition questions, list valid distributions directly.

Step 3

Exam Tip

वैध चयन \(A^3N^2\), \(BA^2N^2\), और \(BA^3N\) हैं, इसलिए कुल (3) हैं। छोटे बहुलता प्रश्नों में वैध वितरण सीधे लिखें।

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Question 48/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

( (1+x)⁵(1+x)⁷ ) में \(x^8\) का गुणांक कितना है?

What is the coefficient of \(x^8\) in ( (1+x)⁵(1+x)⁷ )?

Explanation opens after your attempt

Correct Answer

D. (495)

Step 1

Concept

The product is ( (1+x)^{12} ), so the coefficient is \( \binom{12}{8}=495 \). In binomial products, combine powers first.

Step 2

Why this answer is correct

The correct answer is D. (495). The product is ( (1+x)^{12} ), so the coefficient is \( \binom{12}{8}=495 \). In binomial products, combine powers first.

Step 3

Exam Tip

गुणनफल ( (1+x)^{12} ) है, इसलिए गुणांक \( \binom{12}{8}=495 \)। द्विपद गुणन में घातें जोड़कर सरल करें।

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Question 49/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

योग \( \sum_{r=0}^{4}\binom{6}{r}\binom{8}{4-r} \) का मान क्या है?

What is the value of \( \sum_{r=0}^{4}\binom{6}{r}\binom{8}{4-r} \)?

Explanation opens after your attempt

Correct Answer

C. (1001)

Step 1

Concept

By Vandermonde's identity, the sum is \( \binom{14}{4}=1001 \). In such sums, view it as total selection after adding upper counts.

Step 2

Why this answer is correct

The correct answer is C. (1001). By Vandermonde's identity, the sum is \( \binom{14}{4}=1001 \). In such sums, view it as total selection after adding upper counts.

Step 3

Exam Tip

वैंडरमोंड सर्वसमिका से योग \( \binom{14}{4}=1001 \) है। ऐसे योग में ऊपर की संख्याएँ जोड़कर कुल चयन देखें।

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Question 50/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 61

यदि \( \binom{n}{2}=45 \) है, तो \( \binom{n}{4} \) का मान क्या होगा?

If \( \binom{n}{2}=45 \), what is the value of \( \binom{n}{4} \)?

Explanation opens after your attempt

Correct Answer

C. (210)

Step 1

Concept

Since \( \binom{10}{2}=45 \), (n=10), so \( \binom{10}{4}=210 \). First find (n), then evaluate the required combination.

Step 2

Why this answer is correct

The correct answer is C. (210). Since \( \binom{10}{2}=45 \), (n=10), so \( \binom{10}{4}=210 \). First find (n), then evaluate the required combination.

Step 3

Exam Tip

\( \binom{10}{2}=45 \) से (n=10), इसलिए \( \binom{10}{4}=210 \)। पहले (n) निकालें, फिर माँगा गया संयोजन रखें।

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Question 51/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{3}=4,^{n}C_{2}\) है, तो (n) का मान क्या है?

If \(^{n}C_{3}=4,^{n}C_{2}\), what is the value of (n)?

Explanation opens after your attempt

Correct Answer

B. (14)

Step 1

Concept

The ratio is \(\frac{^{n}C_{3}}{^{n}C_{2}}=\frac{n-2}{3}\), so (n=14). In exams write the ratio first.

Step 2

Why this answer is correct

The correct answer is B. (14). The ratio is \(\frac{^{n}C_{3}}{^{n}C_{2}}=\frac{n-2}{3}\), so (n=14). In exams write the ratio first.

Step 3

Exam Tip

अनुपात \(\frac{^{n}C_{3}}{^{n}C_{2}}=\frac{n-2}{3}\) होगा, इसलिए (n=14)। परीक्षा में पहले अनुपात लिखें।

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Question 52/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{4}=^{n}C_{6}\) है, तो (n) का मान क्या है?

If \(^{n}C_{4}=^{n}C_{6}\), what is the value of (n)?

Explanation opens after your attempt

Correct Answer

C. (10)

Step 1

Concept

For same (n), \(^{n}C_{r}=^{n}C_{s}\) implies (r=s) or (r+s=n). Here (n=10).

Step 2

Why this answer is correct

The correct answer is C. (10). For same (n), \(^{n}C_{r}=^{n}C_{s}\) implies (r=s) or (r+s=n). Here (n=10).

Step 3

Exam Tip

समान (n) के लिए \(^{n}C_{r}=^{n}C_{s}\) होने पर (r=s) या (r+s=n)। यहां (n=10) है।

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Question 53/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{12}C_{r}=^{12}C_{r+4}\) है, तो (r) का मान क्या है?

If \(^{12}C_{r}=^{12}C_{r+4}\), what is the value of (r)?

Explanation opens after your attempt

Correct Answer

B. (4)

Step 1

Concept

In the equality, (r+(r+4)=12). Hence (r=4).

Step 2

Why this answer is correct

The correct answer is B. (4). In the equality, (r+(r+4)=12). Hence (r=4).

Step 3

Exam Tip

समानता में (r+(r+4)=12) होगा। इससे (r=4) मिलता है।

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Question 54/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{2}=66\) है, तो \(^{n}C_{4}\) का मान क्या होगा?

If \(^{n}C_{2}=66\), what is the value of \(^{n}C_{4}\)?

Explanation opens after your attempt

Correct Answer

A. (495)

Step 1

Concept

From \(^{n}C_{2}=66\), (n=12). Therefore \(^{12}C_{4}=495\).

Step 2

Why this answer is correct

The correct answer is A. (495). From \(^{n}C_{2}=66\), (n=12). Therefore \(^{12}C_{4}=495\).

Step 3

Exam Tip

\(^{n}C_{2}=66\) से (n=12) मिलता है। इसलिए \(^{12}C_{4}=495\)।

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Question 55/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{5}:^{n}C_{4}=7:2\) है, तो (n) का मान क्या है?

If \(^{n}C_{5}:^{n}C_{4}=7:2\), what is the value of (n)?

Explanation opens after your attempt

Correct Answer

D. (21)

Step 1

Concept

\(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}=\frac{7}{2}\) gives no integer (n). Always check validity of the ratio.

Step 2

Why this answer is correct

The correct answer is D. (21). \(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}=\frac{7}{2}\) gives no integer (n). Always check validity of the ratio.

Step 3

Exam Tip

\(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}=\frac{7}{2}\) से (n=21.5) नहीं आता, इसलिए सही अनुपात जांचना जरूरी है। यहां कोई पूर्णांक (n) संभव नहीं है।

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Question 56/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(12) विद्यार्थियों में से (5) की समिति बनानी है, लेकिन दो विशेष विद्यार्थी साथ-साथ ही चुने जा सकते हैं। कितनी समितियां बनेंगी?

A committee of (5) is to be formed from (12) students, but two particular students can be selected only together. How many committees are possible?

Explanation opens after your attempt

Correct Answer

C. (372)

Step 1

Concept

If both special students are selected, count \(^{10}C_{3}\); if both are not selected, count \(^{10}C_{5}\). Total (120+252=372).

Step 2

Why this answer is correct

The correct answer is C. (372). If both special students are selected, count \(^{10}C_{3}\); if both are not selected, count \(^{10}C_{5}\). Total (120+252=372).

Step 3

Exam Tip

दोनों विशेष चुने जाएं तो \(^{10}C_{3}\), और दोनों न चुने जाएं तो \(^{10}C_{5}\)। कुल (120+252=372)।

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Question 57/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(8) पुरुष और (6) महिलाओं में से (5) सदस्यों की समिति बनानी है जिसमें कम से कम (3) महिलाएं हों। कितने चयन संभव हैं?

From (8) men and (6) women, a committee of (5) is to be formed with at least (3) women. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (560)

Step 1

Concept

Cases are (3W2M), (4W1M), and (5W0M). The value is \(^{6}C_{3}{}^{8}C_{2}+^{6}C_{4}{}^{8}C_{1}+^{6}C_{5}=560\).

Step 2

Why this answer is correct

The correct answer is C. (560). Cases are (3W2M), (4W1M), and (5W0M). The value is \(^{6}C_{3}{}^{8}C_{2}+^{6}C_{4}{}^{8}C_{1}+^{6}C_{5}=560\).

Step 3

Exam Tip

मामले (3W2M), (4W1M), (5W0M) हैं। मान \(^{6}C_{3}{}^{8}C_{2}+^{6}C_{4}{}^{8}C_{1}+^{6}C_{5}=560\)।

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Question 58/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(10) खिलाड़ियों में से (4) का समूह चुनना है, लेकिन कप्तान और उपकप्तान दोनों साथ नहीं आ सकते। कितने समूह बनेंगे?

A group of (4) is to be chosen from (10) players, but the captain and vice-captain cannot both be included. How many groups are possible?

Explanation opens after your attempt

Correct Answer

C. (182)

Step 1

Concept

Total selections are \(^{10}C_{4}=210\). Remove \(^{8}C_{2}=28\) selections containing both special players, giving (182).

Step 2

Why this answer is correct

The correct answer is C. (182). Total selections are \(^{10}C_{4}=210\). Remove \(^{8}C_{2}=28\) selections containing both special players, giving (182).

Step 3

Exam Tip

कुल \(^{10}C_{4}=210\) हैं। दोनों विशेष साथ होने वाले \(^{8}C_{2}=28\) हटाएं, उत्तर (182)।

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Question 59/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(7) अलग पुस्तकों में से (4) पुस्तकें चुननी हैं, पर दो निर्धारित पुस्तकें साथ-साथ नहीं चुनी जा सकतीं। कितने चयन संभव हैं?

From (7) distinct books, (4) books are to be selected, but two specified books cannot be selected together. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (25)

Step 1

Concept

Total selections are \(^{7}C_{4}=35\). If both specified books are together, \(^{5}C_{2}=10\) selections are removed.

Step 2

Why this answer is correct

The correct answer is B. (25). Total selections are \(^{7}C_{4}=35\). If both specified books are together, \(^{5}C_{2}=10\) selections are removed.

Step 3

Exam Tip

कुल \(^{7}C_{4}=35\) हैं। दोनों निर्धारित पुस्तकें साथ होने पर \(^{5}C_{2}=10\) चयन हटेंगे।

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Question 60/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(9) बिंदुओं में से कोई (3) सरल रेखा पर नहीं हैं। कितने त्रिभुज बनाए जा सकते हैं?

There are (9) points with no (3) collinear. How many triangles can be formed?

Explanation opens after your attempt

Correct Answer

C. (84)

Step 1

Concept

A triangle needs (3) points. Hence the number is \(^{9}C_{3}=84\).

Step 2

Why this answer is correct

The correct answer is C. (84). A triangle needs (3) points. Hence the number is \(^{9}C_{3}=84\).

Step 3

Exam Tip

त्रिभुज के लिए (3) बिंदु चाहिए। इसलिए संख्या \(^{9}C_{3}=84\) है।

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Question 61/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(11) बिंदुओं में से (5) बिंदु एक ही रेखा पर हैं और बाकी में कोई (3) सरल रेखा पर नहीं हैं। कितने त्रिभुज बनेंगे?

Among (11) points, (5) are collinear and no other (3) points are collinear. How many triangles can be formed?

Explanation opens after your attempt

Correct Answer

A. (155)

Step 1

Concept

Subtract collinear choices \(^{5}C_{3}=10\) from total \(^{11}C_{3}=165\). The answer is (155).

Step 2

Why this answer is correct

The correct answer is A. (155). Subtract collinear choices \(^{5}C_{3}=10\) from total \(^{11}C_{3}=165\). The answer is (155).

Step 3

Exam Tip

कुल \(^{11}C_{3}=165\) से कोलिनियर \(^{5}C_{3}=10\) घटाएं। उत्तर (155) है।

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Question 62/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(10) भुजाओं वाले उत्तल बहुभुज में विकर्णों की संख्या कितनी है?

How many diagonals are there in a convex polygon with (10) sides?

Explanation opens after your attempt

Correct Answer

B. (35)

Step 1

Concept

Total ways to choose two vertices are \(^{10}C_{2}\). Subtract (10) sides to get (35) diagonals.

Step 2

Why this answer is correct

The correct answer is B. (35). Total ways to choose two vertices are \(^{10}C_{2}\). Subtract (10) sides to get (35) diagonals.

Step 3

Exam Tip

दो शीर्ष चुनने के कुल तरीके \(^{10}C_{2}\) हैं। भुजाएं (10) घटाने पर विकर्ण (35) मिलते हैं।

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Question 63/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(12) भुजाओं वाले उत्तल बहुभुज में कितने चतुर्भुज बनाए जा सकते हैं?

How many quadrilaterals can be formed from a convex polygon with (12) sides?

Explanation opens after your attempt

Correct Answer

C. (495)

Step 1

Concept

Any (4) vertices form one quadrilateral. Thus the number is \(^{12}C_{4}=495\).

Step 2

Why this answer is correct

The correct answer is C. (495). Any (4) vertices form one quadrilateral. Thus the number is \(^{12}C_{4}=495\).

Step 3

Exam Tip

किसी भी (4) शीर्षों से एक चतुर्भुज बनता है। इसलिए संख्या \(^{12}C_{4}=495\) है।

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Question 64/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

एक क्रिकेट दल में (7) बल्लेबाज और (6) गेंदबाज हैं। (6) खिलाड़ियों का समूह चुनना है जिसमें कम से कम (2) गेंदबाज हों। कितने तरीके हैं?

A cricket squad has (7) batters and (6) bowlers. A group of (6) players is to be chosen with at least (2) bowlers. How many ways are possible?

Explanation opens after your attempt

Correct Answer

A. (1589)

Step 1

Concept

Subtract cases with (0) and (1) bowler from \(^{13}C_{6}\). (1716-7-120=1589).

Step 2

Why this answer is correct

The correct answer is A. (1589). Subtract cases with (0) and (1) bowler from \(^{13}C_{6}\). (1716-7-120=1589).

Step 3

Exam Tip

कुल \(^{13}C_{6}\) से (0) और (1) गेंदबाज वाले मामले घटाएं। (1716-7-120=1589)।

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Question 65/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(6) भारतीय और (5) विदेशी लेखकों में से (4) लेखकों का पैनल बनाना है जिसमें दोनों प्रकार के लेखक हों। कितने चयन संभव हैं?

From (6) Indian and (5) foreign authors, a panel of (4) authors is to be formed with both types present. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (310)

Step 1

Concept

Subtract all-Indian \(^{6}C_{4}=15\) and all-foreign \(^{5}C_{4}=5\) from total \(^{11}C_{4}=330\). Answer (310).

Step 2

Why this answer is correct

The correct answer is C. (310). Subtract all-Indian \(^{6}C_{4}=15\) and all-foreign \(^{5}C_{4}=5\) from total \(^{11}C_{4}=330\). Answer (310).

Step 3

Exam Tip

कुल \(^{11}C_{4}=330\) से केवल भारतीय \(^{6}C_{4}=15\) और केवल विदेशी \(^{5}C_{4}=5\) हटाएं। उत्तर (310)।

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Question 66/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(15) प्रश्नों में से (10) प्रश्न हल करने हैं। पहले (5) प्रश्नों में से कम से कम (3) हल करने अनिवार्य हैं। चयन कितने होंगे?

From (15) questions, (10) are to be attempted. At least (3) of the first (5) questions must be attempted. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (875)

Step 1

Concept

Choose (3,4,5) from the first (5) and complete from the remaining (10). \(^{5}C_{3}{}^{10}C_{7}+^{5}C_{4}{}^{10}C_{6}+^{5}C_{5}{}^{10}C_{5}=875\).

Step 2

Why this answer is correct

The correct answer is A. (875). Choose (3,4,5) from the first (5) and complete from the remaining (10). \(^{5}C_{3}{}^{10}C_{7}+^{5}C_{4}{}^{10}C_{6}+^{5}C_{5}{}^{10}C_{5}=875\).

Step 3

Exam Tip

पहले (5) में से (3,4,5) चुनकर बाकी (10) से पूर्ति करें। \(^{5}C_{3}{}^{10}C_{7}+^{5}C_{4}{}^{10}C_{6}+^{5}C_{5}{}^{10}C_{5}=875\)।

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Question 67/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

एक परीक्षा में खंड (A) में (6) और खंड (B) में (7) प्रश्न हैं। कुल (8) प्रश्न चुनने हैं और प्रत्येक खंड से कम से कम (3) प्रश्न होने चाहिए। कितने चयन संभव हैं?

An exam has (6) questions in section (A) and (7) in section (B). A total of (8) questions must be chosen with at least (3) from each section. How many selections are possible?

Explanation opens after your attempt

Correct Answer

D. (945)

Step 1

Concept

Cases are (3,5), (4,4), and (5,3). Sum \(^{6}C_{3}{}^{7}C_{5}+^{6}C_{4}{}^{7}C_{4}+^{6}C_{5}{}^{7}C_{3}=945\).

Step 2

Why this answer is correct

The correct answer is D. (945). Cases are (3,5), (4,4), and (5,3). Sum \(^{6}C_{3}{}^{7}C_{5}+^{6}C_{4}{}^{7}C_{4}+^{6}C_{5}{}^{7}C_{3}=945\).

Step 3

Exam Tip

मामले (3,5), (4,4), और (5,3) हैं। योग \(^{6}C_{3}{}^{7}C_{5}+^{6}C_{4}{}^{7}C_{4}+^{6}C_{5}{}^{7}C_{3}=945\)।

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Question 68/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(8) अभ्यर्थियों में से (3) का चयन करना है, पर एक विशेष अभ्यर्थी अवश्य चुना जाए। कितने तरीके होंगे?

From (8) candidates, (3) are to be selected, but one particular candidate must be included. How many ways are possible?

Explanation opens after your attempt

Correct Answer

B. (21)

Step 1

Concept

Fix the particular candidate and choose remaining (2) from (7). The number is \(^{7}C_{2}=21\).

Step 2

Why this answer is correct

The correct answer is B. (21). Fix the particular candidate and choose remaining (2) from (7). The number is \(^{7}C_{2}=21\).

Step 3

Exam Tip

विशेष अभ्यर्थी को निश्चित मानें और बाकी (2) को (7) में से चुनें। संख्या \(^{7}C_{2}=21\)।

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Question 69/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(9) वैज्ञानिकों में से (4) का दल बनाना है, पर दो विरोधी वैज्ञानिक साथ नहीं हो सकते। कितने दल बनेंगे?

A team of (4) is to be formed from (9) scientists, but two rival scientists cannot be together. How many teams are possible?

Explanation opens after your attempt

Correct Answer

A. (105)

Step 1

Concept

Total selections are \(^{9}C_{4}=126\). Remove \(^{7}C_{2}=21\) selections containing both rivals, answer (105).

Step 2

Why this answer is correct

The correct answer is A. (105). Total selections are \(^{9}C_{4}=126\). Remove \(^{7}C_{2}=21\) selections containing both rivals, answer (105).

Step 3

Exam Tip

कुल \(^{9}C_{4}=126\) हैं। दोनों विरोधी साथ होने पर \(^{7}C_{2}=21\) हटाएं, उत्तर (105)।

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Question 70/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(5) गणित और (4) भौतिकी पुस्तकों में से (5) पुस्तकें चुननी हैं, जिनमें ठीक (2) भौतिकी पुस्तकें हों। कितने चयन हैं?

From (5) mathematics and (4) physics books, (5) books are to be selected with exactly (2) physics books. How many selections are there?

Explanation opens after your attempt

Correct Answer

B. (60)

Step 1

Concept

For exactly (2) physics books use \(^{4}C_{2}\), and for (3) mathematics books use \(^{5}C_{3}\). Product is (60).

Step 2

Why this answer is correct

The correct answer is B. (60). For exactly (2) physics books use \(^{4}C_{2}\), and for (3) mathematics books use \(^{5}C_{3}\). Product is (60).

Step 3

Exam Tip

ठीक (2) भौतिकी के लिए \(^{4}C_{2}\) और (3) गणित के लिए \(^{5}C_{3}\)। गुणन से (60) मिलता है।

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Question 71/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(6) लाल, (5) नीली और (4) हरी गेंदों में से (5) गेंदें चुननी हैं, जिनमें हर रंग की कम से कम एक गेंद हो। कितने चयन संभव हैं?

From (6) red, (5) blue, and (4) green balls, (5) balls are to be selected with at least one ball of each color. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (1150)

Step 1

Concept

Use all ordered color distributions of (3,1,1) and (2,2,1). The total is (1150).

Step 2

Why this answer is correct

The correct answer is C. (1150). Use all ordered color distributions of (3,1,1) and (2,2,1). The total is (1150).

Step 3

Exam Tip

रंग-वितरण (3,1,1) और (2,2,1) के सभी क्रम लें। योग (1150) आता है।

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Question 72/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(18) सदस्यों में से (6) की समिति बनानी है। अध्यक्ष और सचिव समिति के सामान्य सदस्य के रूप में साथ नहीं रह सकते। कितनी समितियां बनेंगी?

From (18) members, a committee of (6) is to be formed. The president and secretary cannot both be in the committee as ordinary members. How many committees are possible?

Explanation opens after your attempt

Correct Answer

B. (16128)

Step 1

Concept

Total is \(^{18}C_{6}=18564\), and selections with both special members are \(^{16}C_{4}=1820\), giving (16744). None of the options is correct.

Step 2

Why this answer is correct

The correct answer is B. (16128). Total is \(^{18}C_{6}=18564\), and selections with both special members are \(^{16}C_{4}=1820\), giving (16744). None of the options is correct.

Step 3

Exam Tip

कुल \(^{18}C_{6}=18564\) से दोनों विशेष शामिल वाले \(^{16}C_{4}=1820\) नहीं, सही हटाना यही है और उत्तर (16744) होगा। दिए विकल्पों में सही उत्तर नहीं है।

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Question 73/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(20) विद्यार्थियों में से (6) चुनने हैं। दो विशेष विद्यार्थी में से ठीक एक चुना जाना चाहिए। कितने चयन होंगे?

From (20) students, (6) are to be chosen. Exactly one of two special students must be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (18144)

Step 1

Concept

Choose (1) of the two special students and (5) from the remaining (18). The number is \(^{2}C_{1}{}^{18}C_{5}=17136\).

Step 2

Why this answer is correct

The correct answer is C. (18144). Choose (1) of the two special students and (5) from the remaining (18). The number is \(^{2}C_{1}{}^{18}C_{5}=17136\).

Step 3

Exam Tip

दो विशेष में से (1) चुनें और बाकी (5) को (18) में से चुनें। संख्या \(^{2}C_{1}{}^{18}C_{5}=17136\) होती है।

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Question 74/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि (^{2n}C_{3}=2,^{n}C_{3}+n^{2}(n-1)) है, तो कथन किसके लिए सही है?

If (^{2n}C_{3}=2,^{n}C_{3}+n^{2}(n-1)), for which condition is the statement true?

Explanation opens after your attempt

Correct Answer

A. सभी \(n\geq 3\)all \(n\geq 3\)

Step 1

Concept

While choosing (3) from two groups of size (n), either all (3) come from one group or the split is (2,1). This proves the identity.

Step 2

Why this answer is correct

The correct answer is A. सभी \(n\geq 3\) / all \(n\geq 3\). While choosing (3) from two groups of size (n), either all (3) come from one group or the split is (2,1). This proves the identity.

Step 3

Exam Tip

दो (n) आकार के समूहों से (3) चुनने में सभी (3) एक समूह से या (2,1) विभाजन से आते हैं। यही पहचान है।

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Question 75/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{r}=^{n}C_{r-2}\) और (r>2) है, तो (n) किसके बराबर है?

If \(^{n}C_{r}=^{n}C_{r-2}\) and (r>2), then (n) is equal to what?

Explanation opens after your attempt

Correct Answer

A. (2r-2)

Step 1

Concept

For same (n), the complementary indices add to (n). Hence (r+(r-2)=n).

Step 2

Why this answer is correct

The correct answer is A. (2r-2). For same (n), the complementary indices add to (n). Hence (r+(r-2)=n).

Step 3

Exam Tip

समान (n) में सूचकांकों का योग (n) होता है। इसलिए (r+(r-2)=n)।

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Question 76/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(14) बिंदुओं में से (6) बिंदु एक रेखा पर और (4) दूसरे रेखा पर हैं। दोनों रेखाएं अलग हैं और अन्य कोई (3) कोलिनियर नहीं। कितने त्रिभुज बनेंगे?

Among (14) points, (6) lie on one line and (4) on another line. The lines are distinct and no other (3) points are collinear. How many triangles can be formed?

Explanation opens after your attempt

Correct Answer

B. (324)

Step 1

Concept

Total is \(^{14}C_{3}=364\). Invalid selections are \(^{6}C_{3}+^{4}C_{3}=24\), so the answer is (340).

Step 2

Why this answer is correct

The correct answer is B. (324). Total is \(^{14}C_{3}=364\). Invalid selections are \(^{6}C_{3}+^{4}C_{3}=24\), so the answer is (340).

Step 3

Exam Tip

कुल \(^{14}C_{3}=364\) है। अमान्य चयन \(^{6}C_{3}+^{4}C_{3}=24\), इसलिए उत्तर (340) नहीं बल्कि (340) है।

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Question 77/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(10) अलग-अलग बिंदुओं में से (4) बिंदु एक रेखा पर हैं। कितनी सीधी रेखाएं निर्धारित होंगी?

Among (10) distinct points, (4) points are collinear. How many straight lines are determined?

Explanation opens after your attempt

Correct Answer

D. (40)

Step 1

Concept

Total pairs are \(^{10}C_{2}=45\). The \(^{4}C_{2}=6\) pairs among collinear points give only (1) line, so (45-6+1=40).

Step 2

Why this answer is correct

The correct answer is D. (40). Total pairs are \(^{10}C_{2}=45\). The \(^{4}C_{2}=6\) pairs among collinear points give only (1) line, so (45-6+1=40).

Step 3

Exam Tip

कुल युग्म \(^{10}C_{2}=45\) हैं। (4) कोलिनियर बिंदुओं के \(^{4}C_{2}=6\) युग्मों की जगह (1) रेखा होगी, इसलिए (45-6+1=40)।

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Question 78/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

एक \(5\times 4\) ग्रिड में बाएं-नीचे से दाएं-ऊपर जाने के लिए केवल दाएं और ऊपर चालें चलनी हैं। कुल कितने रास्ते हैं?

In a \(5\times 4\) grid, one moves from bottom-left to top-right using only right and up moves. How many paths are possible?

Explanation opens after your attempt

Correct Answer

A. (126)

Step 1

Concept

There are (9) moves with (5) right and (4) up. Paths are \(^{9}C_{5}=126\).

Step 2

Why this answer is correct

The correct answer is A. (126). There are (9) moves with (5) right and (4) up. Paths are \(^{9}C_{5}=126\).

Step 3

Exam Tip

कुल चालें (9) हैं जिनमें (5) दाएं और (4) ऊपर हैं। रास्ते \(^{9}C_{5}=126\) हैं।

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Question 79/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

\(6\times 5\) ग्रिड में रास्ते में एक निर्धारित बिंदु तक जरूर जाना है, जिसके लिए पहले (2) दाएं और (3) ऊपर चालें चाहिए। कुल रास्ते कितने हैं?

In a \(6\times 5\) grid, a path must pass through a fixed point that requires (2) right and (3) up moves first. How many paths are possible?

Explanation opens after your attempt

Correct Answer

B. (700)

Step 1

Concept

First part has \(^{5}C_{2}=10\) paths and the second part has \(^{6}C_{4}=15\) paths. Total is (150).

Step 2

Why this answer is correct

The correct answer is B. (700). First part has \(^{5}C_{2}=10\) paths and the second part has \(^{6}C_{4}=15\) paths. Total is (150).

Step 3

Exam Tip

पहले भाग के रास्ते \(^{5}C_{2}=10\) और दूसरे भाग के रास्ते \(^{6}C_{4}=15\) नहीं, सही दूसरा \(^{6}C_{4}=15\) है। कुल (150) होता है।

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Question 80/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(1) से (20) तक की संख्याओं में से (4) संख्याएं चुननी हैं ताकि कोई दो लगातार न हों। कितने चयन संभव हैं?

Choose (4) numbers from (1) to (20) so that no two are consecutive. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (3876)

Step 1

Concept

If no two are consecutive, use \(^{n-r+1}C_{r}\). Here \(^{17}C_{4}=2380\), so the correct answer is (2380).

Step 2

Why this answer is correct

The correct answer is C. (3876). If no two are consecutive, use \(^{n-r+1}C_{r}\). Here \(^{17}C_{4}=2380\), so the correct answer is (2380).

Step 3

Exam Tip

कोई दो लगातार न हों तो सूत्र \(^{n-r+1}C_{r}\) है। यहां \(^{17}C_{4}=2380\), इसलिए सही विकल्प (2380) है।

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Question 81/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(1) से (15) तक की संख्याओं में से (5) संख्याएं चुननी हैं, जिनमें कोई दो लगातार न हों। कितने तरीके हैं?

Choose (5) numbers from (1) to (15) such that no two are consecutive. How many ways are possible?

Explanation opens after your attempt

Correct Answer

A. (462)

Step 1

Concept

By the gap method, the count is \(^{15-5+1}C_{5}=^{11}C_{5}=462\). In such problems identify the gap condition first.

Step 2

Why this answer is correct

The correct answer is A. (462). By the gap method, the count is \(^{15-5+1}C_{5}=^{11}C_{5}=462\). In such problems identify the gap condition first.

Step 3

Exam Tip

अलगाव विधि से संख्या \(^{15-5+1}C_{5}=^{11}C_{5}=462\)। ऐसी समस्याओं में पहले अंतर की शर्त पहचानें।

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Question 82/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(12) लोगों में से (5) चुनने हैं, पर तीन विशेष लोगों में से कम से कम (2) चुने जाएं। कितने चयन होंगे?

From (12) people, (5) are to be chosen, but at least (2) of (3) special people must be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (225)

Step 1

Concept

Cases are (2) special and (3) others, or (3) special and (2) others. The value is \(^{3}C_{2}{}^{9}C_{3}+^{3}C_{3}{}^{9}C_{2}=288\).

Step 2

Why this answer is correct

The correct answer is A. (225). Cases are (2) special and (3) others, or (3) special and (2) others. The value is \(^{3}C_{2}{}^{9}C_{3}+^{3}C_{3}{}^{9}C_{2}=288\).

Step 3

Exam Tip

मामले (2) विशेष और (3) अन्य, या (3) विशेष और (2) अन्य हैं। \(^{3}C_{2}{}^{9}C_{3}+^{3}C_{3}{}^{9}C_{2}=288\) होता है।

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Question 83/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(9) प्रश्नों में से (5) चुनने हैं, लेकिन प्रश्न (1) और प्रश्न (2) में से कम से कम एक चुना जाना चाहिए। कितने चयन संभव हैं?

From (9) questions, (5) are to be chosen, but at least one of question (1) and question (2) must be chosen. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (91)

Step 1

Concept

Subtract cases where neither is chosen, \(^{7}C_{5}=21\), from total \(^{9}C_{5}=126\). The answer is (105).

Step 2

Why this answer is correct

The correct answer is A. (91). Subtract cases where neither is chosen, \(^{7}C_{5}=21\), from total \(^{9}C_{5}=126\). The answer is (105).

Step 3

Exam Tip

कुल \(^{9}C_{5}=126\) से दोनों न चुने जाने वाले \(^{7}C_{5}=21\) घटते हैं। उत्तर (105) है।

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Question 84/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(7) शिक्षकों और (8) छात्रों में से (6) सदस्यीय दल बनाना है, जिसमें शिक्षकों की संख्या छात्रों से अधिक हो। कितने दल बनेंगे?

From (7) teachers and (8) students, a (6)-member team is to be formed in which teachers are more than students. How many teams are possible?

Explanation opens after your attempt

Correct Answer

A. (917)

Step 1

Concept

Cases are (4T2S), (5T1S), and (6T0S). Sum is \(^{7}C_{4}{}^{8}C_{2}+^{7}C_{5}{}^{8}C_{1}+^{7}C_{6}=1155\).

Step 2

Why this answer is correct

The correct answer is A. (917). Cases are (4T2S), (5T1S), and (6T0S). Sum is \(^{7}C_{4}{}^{8}C_{2}+^{7}C_{5}{}^{8}C_{1}+^{7}C_{6}=1155\).

Step 3

Exam Tip

शिक्षक अधिक होने के मामले (4T2S), (5T1S), (6T0S) हैं। योग \(^{7}C_{4}{}^{8}C_{2}+^{7}C_{5}{}^{8}C_{1}+^{7}C_{6}=1155\) होता है।

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Question 85/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(16) वस्तुओं में (10) अलग और (6) एक जैसी हैं। (4) वस्तुएं चुनने के कितने तरीके हैं?

Among (16) objects, (10) are distinct and (6) are identical. How many ways are there to select (4) objects?

Explanation opens after your attempt

Correct Answer

D. (385)

Step 1

Concept

We can take (0) to (4) identical objects. The total is \(^{10}C_{4}+^{10}C_{3}+^{10}C_{2}+^{10}C_{1}+1=386\).

Step 2

Why this answer is correct

The correct answer is D. (385). We can take (0) to (4) identical objects. The total is \(^{10}C_{4}+^{10}C_{3}+^{10}C_{2}+^{10}C_{1}+1=386\).

Step 3

Exam Tip

एक जैसी वस्तुओं में से (0) से (4) तक ली जा सकती हैं। योग \(\sum_{k=0}^{4}{}^{10}C_{4-k}=386\) नहीं, सही \(^{10}C_{4}+^{10}C_{3}+^{10}C_{2}+^{10}C_{1}+1=386\) है।

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Question 86/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(7) अलग-अलग फलों और (5) एक जैसी टॉफियों में से कुल (4) वस्तुएं चुननी हैं। कितने चयन संभव हैं?

From (7) distinct fruits and (5) identical candies, (4) objects are to be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (99)

Step 1

Concept

The number of identical candies can be (0) to (4). Sum \(^{7}C_{4}+^{7}C_{3}+^{7}C_{2}+^{7}C_{1}+1=99\).

Step 2

Why this answer is correct

The correct answer is B. (99). The number of identical candies can be (0) to (4). Sum \(^{7}C_{4}+^{7}C_{3}+^{7}C_{2}+^{7}C_{1}+1=99\).

Step 3

Exam Tip

एक जैसी टॉफियों की संख्या (0) से (4) तक हो सकती है। योग \(^{7}C_{4}+^{7}C_{3}+^{7}C_{2}+^{7}C_{1}+1=99\)।

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Question 87/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

कितने तरीकों से (5) पुरस्कार (12) विद्यार्थियों में बांटे जा सकते हैं यदि किसी विद्यार्थी को अधिकतम एक पुरस्कार मिले और पुरस्कार समान हों?

In how many ways can (5) identical prizes be distributed among (12) students if each student gets at most one prize?

Explanation opens after your attempt

Correct Answer

A. (792)

Step 1

Concept

Identical prizes with at most one per student means choosing (5) students. The number is \(^{12}C_{5}=792\).

Step 2

Why this answer is correct

The correct answer is A. (792). Identical prizes with at most one per student means choosing (5) students. The number is \(^{12}C_{5}=792\).

Step 3

Exam Tip

समान पुरस्कार और अधिकतम एक पुरस्कार का अर्थ है केवल (5) विद्यार्थियों का चयन। संख्या \(^{12}C_{5}=792\) है।

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Question 88/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(8) लोगों की सभा में कितने हाथ मिलाने होंगे यदि हर जोड़ा ठीक एक बार हाथ मिलाए?

In a meeting of (8) people, how many handshakes occur if every pair shakes hands exactly once?

Explanation opens after your attempt

Correct Answer

B. (28)

Step 1

Concept

Each handshake corresponds to choosing (2) people. Hence the number is \(^{8}C_{2}=28\).

Step 2

Why this answer is correct

The correct answer is B. (28). Each handshake corresponds to choosing (2) people. Hence the number is \(^{8}C_{2}=28\).

Step 3

Exam Tip

हर हाथ मिलाना (2) लोगों के चयन के बराबर है। इसलिए संख्या \(^{8}C_{2}=28\)।

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Question 89/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(30) लोगों की सभा में (435) हाथ मिलाए गए। यदि हर जोड़ा एक बार हाथ मिलाता है, तो कितने लोग अनुपस्थित थे जब कुल आमंत्रित (35) थे?

In a gathering, (435) handshakes occurred. If every pair shook hands once and (35) people were invited, how many were absent?

Explanation opens after your attempt

Correct Answer

C. (5)

Step 1

Concept

From \(^{n}C_{2}=435\), (n=30) people were present. Therefore absent people are (35-30=5).

Step 2

Why this answer is correct

The correct answer is C. (5). From \(^{n}C_{2}=435\), (n=30) people were present. Therefore absent people are (35-30=5).

Step 3

Exam Tip

\(^{n}C_{2}=435\) से (n=30) उपस्थित हैं। अतः अनुपस्थित (35-30=5)।

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Question 90/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{0}+^{n}C_{1}+^{n}C_{2}=46\) है, तो (n) का मान क्या है?

If \(^{n}C_{0}+^{n}C_{1}+^{n}C_{2}=46\), what is the value of (n)?

Explanation opens after your attempt

Correct Answer

B. (9)

Step 1

Concept

The equation is (1+n+\frac{n(n-1)}{2}=46). Solving gives (n=9).

Step 2

Why this answer is correct

The correct answer is B. (9). The equation is (1+n+\frac{n(n-1)}{2}=46). Solving gives (n=9).

Step 3

Exam Tip

समीकरण (1+n+\frac{n(n-1)}{2}=46) है। इससे (n=9) मिलता है।

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Question 91/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{2}+^{n}C_{3}=^{n+1}C_{3}\) है, तो यह पहचान किस नियम पर आधारित है?

If \(^{n}C_{2}+^{n}C_{3}=^{n+1}C_{3}\), this identity is based on which rule?

Explanation opens after your attempt

Correct Answer

B. पास्कल नियमPascal rule

Step 1

Concept

This is the form \(^{n}C_{r-1}+^{n}C_{r}=^{n+1}C_{r}\). It is called Pascal's rule.

Step 2

Why this answer is correct

The correct answer is B. पास्कल नियम / Pascal rule. This is the form \(^{n}C_{r-1}+^{n}C_{r}=^{n+1}C_{r}\). It is called Pascal's rule.

Step 3

Exam Tip

यह \(^{n}C_{r-1}+^{n}C_{r}=^{n+1}C_{r}\) का रूप है। इसे पास्कल नियम कहते हैं।

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Question 92/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

\(^{15}C_{7}+^{15}C_{8}\) का सरल मान किसके बराबर है?

The simplified value of \(^{15}C_{7}+^{15}C_{8}\) is equal to what?

Explanation opens after your attempt

Correct Answer

D. \(2,^{15}C_{7}\)

Step 1

Concept

Since \(^{15}C_{8}=^{15}C_{7}\), the sum is \(2,^{15}C_{7}\). By Pascal's rule it is also \(^{16}C_{8}\).

Step 2

Why this answer is correct

The correct answer is D. \(2,^{15}C_{7}\). Since \(^{15}C_{8}=^{15}C_{7}\), the sum is \(2,^{15}C_{7}\). By Pascal's rule it is also \(^{16}C_{8}\).

Step 3

Exam Tip

क्योंकि \(^{15}C_{8}=^{15}C_{7}\), योग \(2,^{15}C_{7}\) है। पास्कल से यह \(^{16}C_{8}\) भी है।

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Question 93/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(10) अलग-अलग वस्तुओं में से (3) या (4) वस्तुएं चुनने के कुल तरीके कितने हैं?

How many total ways are there to choose (3) or (4) objects from (10) distinct objects?

Explanation opens after your attempt

Correct Answer

A. (330)

Step 1

Concept

The selection count is \(^{10}C_{3}+^{10}C_{4}=120+210=330\). Here OR means addition.

Step 2

Why this answer is correct

The correct answer is A. (330). The selection count is \(^{10}C_{3}+^{10}C_{4}=120+210=330\). Here OR means addition.

Step 3

Exam Tip

चयन \(^{10}C_{3}+^{10}C_{4}=120+210=330\) है। या का अर्थ यहां जोड़ना है।

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Question 94/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(13) छात्रों में से (5) चुनने हैं, पर एक निश्चित छात्र न चुना जाए। कितने तरीके हैं?

From (13) students, (5) are to be selected, but one fixed student must not be selected. How many ways are possible?

Explanation opens after your attempt

Correct Answer

A. (792)

Step 1

Concept

Remove the forbidden student and choose (5) from (12). The number is \(^{12}C_{5}=792\).

Step 2

Why this answer is correct

The correct answer is A. (792). Remove the forbidden student and choose (5) from (12). The number is \(^{12}C_{5}=792\).

Step 3

Exam Tip

न चुने जाने वाले छात्र को हटाकर (12) में से (5) चुनें। संख्या \(^{12}C_{5}=792\)।

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Question 95/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

एक बैग में (5) अलग लाल और (7) अलग काली गेंदें हैं। (4) गेंदें चुननी हैं जिनमें रंग समान न हो। कितने चयन होंगे?

A bag has (5) distinct red and (7) distinct black balls. (4) balls are to be selected such that not all are of the same color. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (455)

Step 1

Concept

Subtract all-red \(^{5}C_{4}=5\) and all-black \(^{7}C_{4}=35\) from total \(^{12}C_{4}=495\). Answer (455).

Step 2

Why this answer is correct

The correct answer is C. (455). Subtract all-red \(^{5}C_{4}=5\) and all-black \(^{7}C_{4}=35\) from total \(^{12}C_{4}=495\). Answer (455).

Step 3

Exam Tip

कुल \(^{12}C_{4}=495\) से सभी लाल \(^{5}C_{4}=5\) और सभी काली \(^{7}C_{4}=35\) हटाएं। उत्तर (455)।

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Question 96/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(9) अंकों \(1,2,\ldots,9\) में से (4) अंक चुनने हैं जिनका योग सम हो। कितने चयन हैं?

Choose (4) digits from \(1,2,\ldots,9\) such that their sum is even. How many selections are there?

Explanation opens after your attempt

Correct Answer

C. (64)

Step 1

Concept

For an even sum, the number of odd digits must be (0,2,4). Count \(^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}=66\).

Step 2

Why this answer is correct

The correct answer is C. (64). For an even sum, the number of odd digits must be (0,2,4). Count \(^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}=66\).

Step 3

Exam Tip

सम योग के लिए विषम अंकों की संख्या (0,2,4) होनी चाहिए। गणना \(^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}=66\) है।

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Question 97/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(1) से (12) तक की संख्याओं में से (3) संख्याएं चुननी हैं जिनका योग विषम हो। कितने चयन संभव हैं?

Choose (3) numbers from (1) to (12) such that their sum is odd. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (110)

Step 1

Concept

There are (6) odd and (6) even numbers. For odd sum choose (1) or (3) odd numbers, giving \(^{6}C_{1}{}^{6}C_{2}+^{6}C_{3}=110\).

Step 2

Why this answer is correct

The correct answer is B. (110). There are (6) odd and (6) even numbers. For odd sum choose (1) or (3) odd numbers, giving \(^{6}C_{1}{}^{6}C_{2}+^{6}C_{3}=110\).

Step 3

Exam Tip

(6) विषम और (6) सम संख्याएं हैं। विषम योग के लिए (1) या (3) विषम चुनें, संख्या \(^{6}C_{1}{}^{6}C_{2}+^{6}C_{3}=110\)।

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Question 98/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(6) अलग-अलग रेखाएं और (5) अलग-अलग वृत्त हैं। दो आकृतियां चुननी हैं जिनमें कम से कम एक रेखा हो। कितने चयन हैं?

There are (6) distinct lines and (5) distinct circles. Two figures are to be selected with at least one line. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (55)

Step 1

Concept

Subtract circle-only selections \(^{5}C_{2}=10\) from total \(^{11}C_{2}=55\). The correct count is (45).

Step 2

Why this answer is correct

The correct answer is C. (55). Subtract circle-only selections \(^{5}C_{2}=10\) from total \(^{11}C_{2}=55\). The correct count is (45).

Step 3

Exam Tip

कुल \(^{11}C_{2}=55\) से केवल वृत्त \(^{5}C_{2}=10\) घटाने चाहिए। सही संख्या (45) है।

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Question 99/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

(11) खिलाड़ियों में से (7) चुनने हैं, और (4) विशेष खिलाड़ियों में से अधिकतम (2) चुने जा सकते हैं। कितने चयन होंगे?

From (11) players, (7) are to be selected, and at most (2) of (4) special players can be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (126)

Step 1

Concept

The number of special players can be (0,1,2), with (7) ordinary players available. Sum \(^{4}C_{0}{}^{7}C_{7}+^{4}C_{1}{}^{7}C_{6}+^{4}C_{2}{}^{7}C_{5}=155\).

Step 2

Why this answer is correct

The correct answer is A. (126). The number of special players can be (0,1,2), with (7) ordinary players available. Sum \(^{4}C_{0}{}^{7}C_{7}+^{4}C_{1}{}^{7}C_{6}+^{4}C_{2}{}^{7}C_{5}=155\).

Step 3

Exam Tip

विशेष खिलाड़ियों की संख्या (0,1,2) हो सकती है, पर (7) अन्य ही हैं। योग \(^{4}C_{0}{}^{7}C_{7}+^{4}C_{1}{}^{7}C_{6}+^{4}C_{2}{}^{7}C_{5}=155\) है।

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Question 100/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 62

यदि \(^{n}C_{r}=^{n}C_{r+1}\) है, तो (n) और (r) के बीच संबंध क्या है?

If \(^{n}C_{r}=^{n}C_{r+1}\), what is the relation between (n) and (r)?

Explanation opens after your attempt

Correct Answer

B. (n=2r+1)

Step 1

Concept

For equality, complementary indices add to (n). Thus (r+(r+1)=n), so (n=2r+1).

Step 2

Why this answer is correct

The correct answer is B. (n=2r+1). For equality, complementary indices add to (n). Thus (r+(r+1)=n), so (n=2r+1).

Step 3

Exam Tip

समानता में पूरक सूचकांक का योग (n) है। इसलिए (r+(r+1)=n), अतः (n=2r+1)।

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Question 101/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

यदि \(^{n}C_{2}=105\) है, तो \(^{n}C_{3}\) का मान क्या होगा?

If \(^{n}C_{2}=105\), what is the value of \(^{n}C_{3}\)?

Explanation opens after your attempt

Correct Answer

C. (455)

Step 1

Concept

From \(^{n}C_{2}=105\), we get (n=15). Therefore \(^{15}C_{3}=455\).

Step 2

Why this answer is correct

The correct answer is C. (455). From \(^{n}C_{2}=105\), we get (n=15). Therefore \(^{15}C_{3}=455\).

Step 3

Exam Tip

\(^{n}C_{2}=105\) से (n=15) मिलता है। इसलिए \(^{15}C_{3}=455\)।

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Question 102/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

यदि \(^{n}C_{5}:^{n}C_{4}=11:5\) है, तो (n) का मान क्या है?

If \(^{n}C_{5}:^{n}C_{4}=11:5\), what is the value of (n)?

Explanation opens after your attempt

Correct Answer

B. (15)

Step 1

Concept

\(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}\). Equating it to \(\frac{11}{5}\) gives (n=15).

Step 2

Why this answer is correct

The correct answer is B. (15). \(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}\). Equating it to \(\frac{11}{5}\) gives (n=15).

Step 3

Exam Tip

\(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}\) होता है। इसे \(\frac{11}{5}\) के बराबर रखने पर (n=15)।

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Question 103/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

यदि \(^{16}C_{r}=^{16}C_{2r+4}\) है, तो (r) का मान क्या है?

If \(^{16}C_{r}=^{16}C_{2r+4}\), what is the value of (r)?

Explanation opens after your attempt

Correct Answer

A. (4)

Step 1

Concept

The complementary indices must add to (16). Thus (r+2r+4=16) gives (r=4).

Step 2

Why this answer is correct

The correct answer is A. (4). The complementary indices must add to (16). Thus (r+2r+4=16) gives (r=4).

Step 3

Exam Tip

पूरक सूचकांकों का योग (16) होगा। इसलिए (r+2r+4=16) से (r=4)।

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Question 104/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

\(^{13}C_{6}+^{13}C_{7}\) का मान क्या है?

What is the value of \(^{13}C_{6}+^{13}C_{7}\)?

Explanation opens after your attempt

Correct Answer

D. (3432)

Step 1

Concept

By Pascal's rule \(^{13}C_{6}+^{13}C_{7}=^{14}C_{7}\). Its value is (3432).

Step 2

Why this answer is correct

The correct answer is D. (3432). By Pascal's rule \(^{13}C_{6}+^{13}C_{7}=^{14}C_{7}\). Its value is (3432).

Step 3

Exam Tip

पास्कल नियम से \(^{13}C_{6}+^{13}C_{7}=^{14}C_{7}\)। इसका मान (3432) है।

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Question 105/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(18) सदस्यों में से (3) सदस्यों की समिति बनानी है जिसमें एक निश्चित सदस्य अवश्य हो। कितनी समितियां बनेंगी?

A committee of (3) members is to be formed from (18) members with one fixed member included. How many committees are possible?

Explanation opens after your attempt

Correct Answer

A. (136)

Step 1

Concept

Fix the required member first and choose the remaining (2) from (17). The number is \(^{17}C_{2}=136\).

Step 2

Why this answer is correct

The correct answer is A. (136). Fix the required member first and choose the remaining (2) from (17). The number is \(^{17}C_{2}=136\).

Step 3

Exam Tip

निश्चित सदस्य को पहले चुन लें और बाकी (2) को (17) में से चुनें। संख्या \(^{17}C_{2}=136\) है।

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Question 106/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(6) लड़कों और (5) लड़कियों में से (5) विद्यार्थियों का दल बनाना है जिसमें कम से कम (2) लड़के और कम से कम (2) लड़कियां हों। कितने दल बनेंगे?

From (6) boys and (5) girls, a team of (5) students is to be formed with at least (2) boys and at least (2) girls. How many teams are possible?

Explanation opens after your attempt

Correct Answer

C. (350)

Step 1

Concept

The possible cases are (2G3B) and (3G2B). The sum is \(^{5}C_{2}\times{}^{6}C_{3}+^{5}C_{3}\times{}^{6}C_{2}=350\).

Step 2

Why this answer is correct

The correct answer is C. (350). The possible cases are (2G3B) and (3G2B). The sum is \(^{5}C_{2}\times{}^{6}C_{3}+^{5}C_{3}\times{}^{6}C_{2}=350\).

Step 3

Exam Tip

संभव मामले (2G3B) और (3G2B) हैं। योग \(^{5}C_{2}\times{}^{6}C_{3}+^{5}C_{3}\times{}^{6}C_{2}=350\)।

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Question 107/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(10) शिक्षकों और (8) छात्रों में से (6) लोगों का पैनल बनाना है जिसमें ठीक (4) शिक्षक हों। कितने पैनल संभव हैं?

From (10) teachers and (8) students, a panel of (6) people is to be formed with exactly (4) teachers. How many panels are possible?

Explanation opens after your attempt

Correct Answer

D. (5880)

Step 1

Concept

Choose (4) teachers using \(^{10}C_{4}\) and (2) students using \(^{8}C_{2}\). Multiplication gives (5880).

Step 2

Why this answer is correct

The correct answer is D. (5880). Choose (4) teachers using \(^{10}C_{4}\) and (2) students using \(^{8}C_{2}\). Multiplication gives (5880).

Step 3

Exam Tip

ठीक (4) शिक्षक के लिए \(^{10}C_{4}\) और (2) छात्र के लिए \(^{8}C_{2}\) लें। गुणन से (5880) मिलता है।

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Question 108/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(12) लोगों में से (5) चुनने हैं। (3) विशेष लोगों में से ठीक (1) चुना जाना चाहिए। कितने चयन होंगे?

From (12) people, (5) are to be chosen. Exactly (1) of (3) special people must be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (378)

Step 1

Concept

Choose (1) special person and the remaining (4) people from the other (9). The number is \(^{3}C_{1}\times{}^{9}C_{4}=378\).

Step 2

Why this answer is correct

The correct answer is B. (378). Choose (1) special person and the remaining (4) people from the other (9). The number is \(^{3}C_{1}\times{}^{9}C_{4}=378\).

Step 3

Exam Tip

विशेष लोगों में से (1) और बाकी (4) लोगों को शेष (9) में से चुनें। संख्या \(^{3}C_{1}\times{}^{9}C_{4}=378\)।

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Question 109/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(14) विद्यार्थियों में से (6) चुनने हैं। (3) नामित विद्यार्थियों में से कम से कम (1) चुना जाए। कितने चयन होंगे?

From (14) students, (6) are to be selected. At least (1) of (3) named students must be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (2541)

Step 1

Concept

Total selections are \(^{14}C_{6}\). Subtract selections with no named student \(^{11}C_{6}\) to get (2541).

Step 2

Why this answer is correct

The correct answer is C. (2541). Total selections are \(^{14}C_{6}\). Subtract selections with no named student \(^{11}C_{6}\) to get (2541).

Step 3

Exam Tip

कुल चयन \(^{14}C_{6}\) हैं। कोई नामित विद्यार्थी न चुने जाने वाले \(^{11}C_{6}\) घटाने पर (2541) मिलता है।

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Question 110/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(16) सदस्यों में (4) वरिष्ठ सदस्य हैं। (7) सदस्यों की समिति में कम से कम (2) वरिष्ठ सदस्य होने चाहिए। कितनी समितियां बनेंगी?

Among (16) members, (4) are senior members. A committee of (7) must contain at least (2) senior members. How many committees are possible?

Explanation opens after your attempt

Correct Answer

A. (6952)

Step 1

Concept

The number of senior members can be (2), (3), or (4). The sum is \(^{4}C_{2}{}^{12}C_{5}+^{4}C_{3}{}^{12}C_{4}+^{4}C_{4}{}^{12}C_{3}=6952\).

Step 2

Why this answer is correct

The correct answer is A. (6952). The number of senior members can be (2), (3), or (4). The sum is \(^{4}C_{2}{}^{12}C_{5}+^{4}C_{3}{}^{12}C_{4}+^{4}C_{4}{}^{12}C_{3}=6952\).

Step 3

Exam Tip

वरिष्ठ सदस्यों की संख्या (2), (3), या (4) हो सकती है। योग \(^{4}C_{2}{}^{12}C_{5}+^{4}C_{3}{}^{12}C_{4}+^{4}C_{4}{}^{12}C_{3}=6952\)।

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Question 111/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(13) बिंदुओं में (5) बिंदु एक रेखा पर और (4) बिंदु दूसरी रेखा पर हैं। ये दोनों समूह अलग हैं और अन्य कोई (3) बिंदु सरल रेखा पर नहीं हैं। कितने त्रिभुज बनेंगे?

Among (13) points, (5) points lie on one line and (4) points lie on another line. These two groups are disjoint and no other (3) points are collinear. How many triangles can be formed?

Explanation opens after your attempt

Correct Answer

B. (272)

Step 1

Concept

Subtract invalid choices \(^{5}C_{3}+^{4}C_{3}\) from total \(^{13}C_{3}\). The answer is (286-14=272).

Step 2

Why this answer is correct

The correct answer is B. (272). Subtract invalid choices \(^{5}C_{3}+^{4}C_{3}\) from total \(^{13}C_{3}\). The answer is (286-14=272).

Step 3

Exam Tip

कुल \(^{13}C_{3}\) में से अमान्य चयन \(^{5}C_{3}+^{4}C_{3}\) घटाएं। उत्तर (286-14=272)।

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Question 112/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(12) बिंदुओं में से (5) बिंदु एक ही रेखा पर हैं और बाकी में कोई (3) बिंदु सरल रेखा पर नहीं हैं। कितनी अलग-अलग रेखाएं बनेंगी?

Among (12) points, (5) points are collinear and no other (3) points are collinear. How many distinct lines are determined?

Explanation opens after your attempt

Correct Answer

D. (57)

Step 1

Concept

Total pairs are \(^{12}C_{2}\). The \(^{5}C_{2}\) pairs from collinear points give only (1) line, so (66-10+1=57).

Step 2

Why this answer is correct

The correct answer is D. (57). Total pairs are \(^{12}C_{2}\). The \(^{5}C_{2}\) pairs from collinear points give only (1) line, so (66-10+1=57).

Step 3

Exam Tip

कुल युग्म \(^{12}C_{2}\) हैं। (5) कोलिनियर बिंदुओं के \(^{5}C_{2}\) युग्मों की जगह (1) रेखा होगी, इसलिए (66-10+1=57)।

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Question 113/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(18) भुजाओं वाले उत्तल बहुभुज में विकर्णों की संख्या कितनी है?

How many diagonals are there in a convex polygon with (18) sides?

Explanation opens after your attempt

Correct Answer

A. (135)

Step 1

Concept

The ways to choose two vertices are \(^{18}C_{2}\). Subtracting (18) sides gives (135) diagonals.

Step 2

Why this answer is correct

The correct answer is A. (135). The ways to choose two vertices are \(^{18}C_{2}\). Subtracting (18) sides gives (135) diagonals.

Step 3

Exam Tip

दो शीर्ष चुनने के तरीके \(^{18}C_{2}\) हैं। इनमें से (18) भुजाएं घटाने पर (135) विकर्ण मिलते हैं।

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Question 114/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

एक वृत्त पर (11) बिंदु हैं। इनसे कितने चतुर्भुज बनाए जा सकते हैं?

There are (11) points on a circle. How many quadrilaterals can be formed from them?

Explanation opens after your attempt

Correct Answer

C. (330)

Step 1

Concept

Any (4) points on the circle form a quadrilateral. Hence the number is \(^{11}C_{4}=330\).

Step 2

Why this answer is correct

The correct answer is C. (330). Any (4) points on the circle form a quadrilateral. Hence the number is \(^{11}C_{4}=330\).

Step 3

Exam Tip

वृत्त पर कोई भी (4) बिंदु एक चतुर्भुज बनाते हैं। इसलिए संख्या \(^{11}C_{4}=330\)।

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Question 115/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

एक उत्तल (10)-भुज में कोई (3) विकर्ण अंदर एक ही बिंदु पर नहीं मिलते। विकर्णों के आंतरिक प्रतिच्छेद बिंदु कितने होंगे?

In a convex (10)-gon, no (3) diagonals meet at the same interior point. How many interior intersection points of diagonals are there?

Explanation opens after your attempt

Correct Answer

B. (210)

Step 1

Concept

Each interior intersection is determined by choosing (4) vertices. Hence the number is \(^{10}C_{4}=210\).

Step 2

Why this answer is correct

The correct answer is B. (210). Each interior intersection is determined by choosing (4) vertices. Hence the number is \(^{10}C_{4}=210\).

Step 3

Exam Tip

हर आंतरिक प्रतिच्छेद (4) शीर्षों के चयन से बनता है। इसलिए संख्या \(^{10}C_{4}=210\) है।

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Question 116/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(12)-भुज के शीर्षों में से (4) शीर्ष चुनने हैं ताकि कोई दो चुने हुए शीर्ष आसन्न न हों। कितने चयन संभव हैं?

Choose (4) vertices of a (12)-gon so that no two chosen vertices are adjacent. How many selections are possible?

Explanation opens after your attempt

Correct Answer

D. (105)

Step 1

Concept

The circular non-adjacent selection formula is \(\frac{n}{n-r},{}^{n-r}C_{r}\). Here \(\frac{12}{8}\times{}^{8}C_{4}=105\).

Step 2

Why this answer is correct

The correct answer is D. (105). The circular non-adjacent selection formula is \(\frac{n}{n-r},{}^{n-r}C_{r}\). Here \(\frac{12}{8}\times{}^{8}C_{4}=105\).

Step 3

Exam Tip

वृत्तीय गैर-आसन्न चयन का सूत्र \(\frac{n}{n-r},{}^{n-r}C_{r}\) है। यहां \(\frac{12}{8}\times{}^{8}C_{4}=105\)।

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Question 117/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

एक बिंदु से दूसरे बिंदु तक जाने के लिए (7) दाईं चालें और (4) ऊपर चालें चाहिए। सबसे छोटे रास्ते कितने होंगे?

To go from one point to another, (7) right moves and (4) up moves are required. How many shortest paths are possible?

Explanation opens after your attempt

Correct Answer

C. (330)

Step 1

Concept

Choose the positions of (7) right moves among (11) moves. The number is \(^{11}C_{7}=330\).

Step 2

Why this answer is correct

The correct answer is C. (330). Choose the positions of (7) right moves among (11) moves. The number is \(^{11}C_{7}=330\).

Step 3

Exam Tip

कुल (11) चालों में से (7) दाईं चालों के स्थान चुनें। संख्या \(^{11}C_{7}=330\) है।

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Question 118/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

एक ग्रिड पथ को पहले (3) दाईं और (2) ऊपर चालों वाले बिंदु से होकर जाना है। कुल यात्रा में (7) दाईं और (4) ऊपर चालें हैं। ऐसे रास्ते कितने हैं?

A grid path must pass through a point reached after (3) right and (2) up moves. The total journey has (7) right and (4) up moves. How many such paths are there?

Explanation opens after your attempt

Correct Answer

A. (150)

Step 1

Concept

The first part has \(^{5}C_{3}\) paths and the second part has \(^{6}C_{4}\) paths. Total \(10\times15=150\).

Step 2

Why this answer is correct

The correct answer is A. (150). The first part has \(^{5}C_{3}\) paths and the second part has \(^{6}C_{4}\) paths. Total \(10\times15=150\).

Step 3

Exam Tip

पहले भाग में \(^{5}C_{3}\) और दूसरे भाग में \(^{6}C_{4}\) रास्ते हैं। कुल \(10\times15=150\)।

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Question 119/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

कुल (6) दाईं और (5) ऊपर चालों वाला सबसे छोटा ग्रिड पथ उस बिंदु से बचना चाहिए जो (2) दाईं और (2) ऊपर चालों के बाद आता है। कितने रास्ते संभव हैं?

A shortest grid path with (6) right and (5) up moves must avoid the point reached after (2) right and (2) up moves. How many paths are possible?

Explanation opens after your attempt

Correct Answer

B. (252)

Step 1

Concept

Total paths are \(^{11}C_{6}=462\). Subtract paths through that point \(^{4}C_{2}\times{}^{7}C_{4}=210\), giving (252).

Step 2

Why this answer is correct

The correct answer is B. (252). Total paths are \(^{11}C_{6}=462\). Subtract paths through that point \(^{4}C_{2}\times{}^{7}C_{4}=210\), giving (252).

Step 3

Exam Tip

कुल रास्ते \(^{11}C_{6}=462\) हैं। उस बिंदु से होकर जाने वाले \(^{4}C_{2}\times{}^{7}C_{4}=210\) घटाएं, उत्तर (252)।

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Question 120/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(1) से (25) तक की संख्याओं में से (5) संख्याएं चुननी हैं ताकि कोई दो लगातार न हों। कितने चयन होंगे?

Choose (5) numbers from (1) to (25) so that no two are consecutive. How many selections are possible?

Explanation opens after your attempt

Correct Answer

D. (20349)

Step 1

Concept

For non-consecutive selection, use \(^{n-r+1}C_{r}\). Here \(^{21}C_{5}=20349\).

Step 2

Why this answer is correct

The correct answer is D. (20349). For non-consecutive selection, use \(^{n-r+1}C_{r}\). Here \(^{21}C_{5}=20349\).

Step 3

Exam Tip

गैर-लगातार चयन के लिए सूत्र \(^{n-r+1}C_{r}\) है। यहां \(^{21}C_{5}=20349\)।

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Question 121/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

वृत्त पर (10) समान दूरी वाले बिंदुओं में से (3) बिंदु चुनने हैं ताकि कोई दो चुने हुए बिंदु आसन्न न हों। कितने चयन संभव हैं?

From (10) equally spaced points on a circle, (3) points are to be chosen so that no two chosen points are adjacent. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (50)

Step 1

Concept

For circular selection, the count is \(\frac{10}{10-3}\times{}^{7}C_{3}\). Its value is (50).

Step 2

Why this answer is correct

The correct answer is C. (50). For circular selection, the count is \(\frac{10}{10-3}\times{}^{7}C_{3}\). Its value is (50).

Step 3

Exam Tip

वृत्तीय चयन में संख्या \(\frac{10}{10-3}\times{}^{7}C_{3}\) है। इसका मान (50) है।

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Question 122/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

समीकरण (x+y+z=15) के अऋण पूर्णांक हल कितने हैं?

How many non-negative integer solutions does (x+y+z=15) have?

Explanation opens after your attempt

Correct Answer

A. (136)

Step 1

Concept

By the stars and bars method, the count is \(^{15+3-1}C_{3-1}=^{17}C_{2}\). Its value is (136).

Step 2

Why this answer is correct

The correct answer is A. (136). By the stars and bars method, the count is \(^{15+3-1}C_{3-1}=^{17}C_{2}\). Its value is (136).

Step 3

Exam Tip

स्टार और बार विधि से संख्या \(^{15+3-1}C_{3-1}=^{17}C_{2}\) है। इसका मान (136) है।

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Question 123/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

समीकरण \(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=20\) के धन पूर्णांक हल कितने हैं?

How many positive integer solutions does \(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=20\) have?

Explanation opens after your attempt

Correct Answer

D. (3876)

Step 1

Concept

For positive integer solutions, give (1) to each variable first. The count is \(^{19}C_{4}=3876\).

Step 2

Why this answer is correct

The correct answer is D. (3876). For positive integer solutions, give (1) to each variable first. The count is \(^{19}C_{4}=3876\).

Step 3

Exam Tip

धन पूर्णांक हलों के लिए पहले हर चर को (1) दें। संख्या \(^{19}C_{4}=3876\) है।

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Question 124/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

समीकरण \(x_{1}+x_{2}+x_{3}+x_{4}=13\) के अऋण पूर्णांक हलों की संख्या कितनी है यदि हर \(x_i\leq5\) हो?

How many non-negative integer solutions does \(x_{1}+x_{2}+x_{3}+x_{4}=13\) have if each \(x_i\leq5\)?

Explanation opens after your attempt

Correct Answer

B. (104)

Step 1

Concept

Subtract cases with \(x_i\geq6\) from total \(^{16}C_{3}\) and add double-overlap cases. (560-480+24=104).

Step 2

Why this answer is correct

The correct answer is B. (104). Subtract cases with \(x_i\geq6\) from total \(^{16}C_{3}\) and add double-overlap cases. (560-480+24=104).

Step 3

Exam Tip

कुल \(^{16}C_{3}\) से \(x_i\geq6\) वाले मामले घटाकर दोहरे मामले जोड़ें। (560-480+24=104)।

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Question 125/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

समीकरण (x+y+z=10) के अऋण पूर्णांक हलों की संख्या कितनी है यदि \(x\leq3\) हो?

How many non-negative integer solutions does (x+y+z=10) have if \(x\leq3\)?

Explanation opens after your attempt

Correct Answer

C. (38)

Step 1

Concept

There are \(^{12}C_{2}=66\) total solutions. Subtract \(^{8}C_{2}=28\) cases with \(x\geq4\) to get (38).

Step 2

Why this answer is correct

The correct answer is C. (38). There are \(^{12}C_{2}=66\) total solutions. Subtract \(^{8}C_{2}=28\) cases with \(x\geq4\) to get (38).

Step 3

Exam Tip

कुल \(^{12}C_{2}=66\) हल हैं। \(x\geq4\) वाले \(^{8}C_{2}=28\) घटाने पर (38) मिलते हैं।

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Question 126/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(52) पत्तों की गड्डी से (5) पत्तों का हाथ चुनना है जिसमें सभी पत्ते एक ही सूट के हों। कितने हाथ संभव हैं?

From a deck of (52) cards, a (5)-card hand is to be chosen with all cards from the same suit. How many hands are possible?

Explanation opens after your attempt

Correct Answer

A. (5148)

Step 1

Concept

First choose one of (4) suits and then choose (5) cards from (13). The number is \(4\times{}^{13}C_{5}=5148\).

Step 2

Why this answer is correct

The correct answer is A. (5148). First choose one of (4) suits and then choose (5) cards from (13). The number is \(4\times{}^{13}C_{5}=5148\).

Step 3

Exam Tip

पहले (4) सूटों में से सूट चुनें और फिर (13) में से (5) पत्ते चुनें। संख्या \(4\times{}^{13}C_{5}=5148\)।

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Question 127/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(52) पत्तों की गड्डी से (5) पत्तों का हाथ चुनना है जिसमें ठीक (2) दिल के पत्ते हों। कितने हाथ संभव हैं?

From a deck of (52) cards, a (5)-card hand is to be chosen with exactly (2) hearts. How many hands are possible?

Explanation opens after your attempt

Correct Answer

C. (712842)

Step 1

Concept

Choose (2) hearts in \(^{13}C_{2}\) ways and the other (3) cards in \(^{39}C_{3}\) ways. Multiplication gives (712842).

Step 2

Why this answer is correct

The correct answer is C. (712842). Choose (2) hearts in \(^{13}C_{2}\) ways and the other (3) cards in \(^{39}C_{3}\) ways. Multiplication gives (712842).

Step 3

Exam Tip

दिल के (2) पत्ते \(^{13}C_{2}\) तरीकों से और बाकी (3) पत्ते \(^{39}C_{3}\) तरीकों से चुने जाएंगे। गुणन से (712842)।

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Question 128/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(52) पत्तों की गड्डी से (5) पत्तों का हाथ चुनना है जिसमें कम से कम (1) इक्का हो। कितने हाथ संभव हैं?

From a deck of (52) cards, a (5)-card hand is to be chosen with at least (1) ace. How many hands are possible?

Explanation opens after your attempt

Correct Answer

B. (886656)

Step 1

Concept

Subtract hands with no ace \(^{48}C_{5}\) from total \(^{52}C_{5}\). The answer is (886656).

Step 2

Why this answer is correct

The correct answer is B. (886656). Subtract hands with no ace \(^{48}C_{5}\) from total \(^{52}C_{5}\). The answer is (886656).

Step 3

Exam Tip

कुल \(^{52}C_{5}\) से बिना इक्के वाले \(^{48}C_{5}\) हाथ घटाएं। उत्तर (886656) है।

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Question 129/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(52) पत्तों की गड्डी से (6) पत्तों का हाथ चुनना है जिसमें ठीक (3) रानियां हों। कितने हाथ संभव हैं?

From a deck of (52) cards, a (6)-card hand is to be chosen with exactly (3) queens. How many hands are possible?

Explanation opens after your attempt

Correct Answer

C. (69184)

Step 1

Concept

Choose (3) queens from (4) and the other (3) cards from (48) non-queen cards. The number is \(^{4}C_{3}\times{}^{48}C_{3}=69184\).

Step 2

Why this answer is correct

The correct answer is C. (69184). Choose (3) queens from (4) and the other (3) cards from (48) non-queen cards. The number is \(^{4}C_{3}\times{}^{48}C_{3}=69184\).

Step 3

Exam Tip

(4) रानियों में से (3) चुनें और बाकी (3) पत्ते (48) गैर-रानी पत्तों से चुनें। संख्या \(^{4}C_{3}\times{}^{48}C_{3}=69184\)।

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Question 130/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

अंकों \(1,2,\ldots,9\) में से (4) अंक चुनने हैं जिनका योग सम हो। कितने चयन संभव हैं?

Choose (4) digits from \(1,2,\ldots,9\) such that their sum is even. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (66)

Step 1

Concept

For an even sum, the number of odd digits must be (0), (2), or (4). The count is \(^{5}C_{0}{}^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}{}^{4}C_{0}=66\).

Step 2

Why this answer is correct

The correct answer is A. (66). For an even sum, the number of odd digits must be (0), (2), or (4). The count is \(^{5}C_{0}{}^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}{}^{4}C_{0}=66\).

Step 3

Exam Tip

सम योग के लिए विषम अंकों की संख्या (0), (2), या (4) होनी चाहिए। गणना \(^{5}C_{0}{}^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}{}^{4}C_{0}=66\)।

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Question 131/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(1) से (15) तक की संख्याओं में से (4) संख्याएं चुननी हैं जिनमें ठीक (2) संख्याएं (3) की गुणज हों। कितने चयन संभव हैं?

Choose (4) numbers from (1) to (15) such that exactly (2) numbers are multiples of (3). How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (450)

Step 1

Concept

There are (5) multiples of (3) and (10) non-multiples. The selection count is \(^{5}C_{2}\times{}^{10}C_{2}=450\).

Step 2

Why this answer is correct

The correct answer is B. (450). There are (5) multiples of (3) and (10) non-multiples. The selection count is \(^{5}C_{2}\times{}^{10}C_{2}=450\).

Step 3

Exam Tip

(3) के (5) गुणज और (10) गैर-गुणज हैं। चयन \(^{5}C_{2}\times{}^{10}C_{2}=450\) होगा।

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Question 132/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(5) पंक्तियों और (6) स्तंभों वाले छोटे वर्गों के ग्रिड में कुल कितने आयत बनेंगे?

How many rectangles are there in a grid of small squares with (5) rows and (6) columns?

Explanation opens after your attempt

Correct Answer

C. (315)

Step 1

Concept

For a rectangle, choose (2) horizontal lines from (6) and (2) vertical lines from (7). The count is \(^{6}C_{2}\times{}^{7}C_{2}=315\).

Step 2

Why this answer is correct

The correct answer is C. (315). For a rectangle, choose (2) horizontal lines from (6) and (2) vertical lines from (7). The count is \(^{6}C_{2}\times{}^{7}C_{2}=315\).

Step 3

Exam Tip

आयत के लिए (6) क्षैतिज रेखाओं में से (2) और (7) ऊर्ध्व रेखाओं में से (2) चुनें। संख्या \(^{6}C_{2}\times{}^{7}C_{2}=315\)।

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Question 133/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(10) विद्यार्थियों को (5) और (5) के दो अनाम समूहों में बांटना है। कितने तरीके होंगे?

(10) students are to be divided into two unnamed groups of (5) and (5). How many ways are possible?

Explanation opens after your attempt

Correct Answer

A. (126)

Step 1

Concept

First choose (5) students in \(^{10}C_{5}\) ways. Since the two groups are unnamed, divide by (2) to get (126).

Step 2

Why this answer is correct

The correct answer is A. (126). First choose (5) students in \(^{10}C_{5}\) ways. Since the two groups are unnamed, divide by (2) to get (126).

Step 3

Exam Tip

पहले (5) विद्यार्थी \(^{10}C_{5}\) तरीकों से चुने जाते हैं। दो समूह अनाम हैं, इसलिए (2) से भाग दें और उत्तर (126) है।

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Question 134/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(8) लोगों में से दो अलग-अलग जोड़े बनाने हैं ताकि कोई व्यक्ति दो जोड़ों में न आए। कितने तरीके होंगे?

From (8) people, two different pairs are to be formed so that no person appears in both pairs. How many ways are possible?

Explanation opens after your attempt

Correct Answer

D. (210)

Step 1

Concept

First choose (4) people in \(^{8}C_{4}\) ways and split them into two pairs in (3) ways. Total \(70\times3=210\).

Step 2

Why this answer is correct

The correct answer is D. (210). First choose (4) people in \(^{8}C_{4}\) ways and split them into two pairs in (3) ways. Total \(70\times3=210\).

Step 3

Exam Tip

पहले (4) लोगों को \(^{8}C_{4}\) तरीकों से चुनें और उन्हें (3) तरीकों से दो जोड़ों में बांटें। कुल \(70\times3=210\)।

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Question 135/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(6) विवाहित जोड़ों में से (4) लोगों को चुनना है ताकि कोई पूरा जोड़ा न चुना जाए। कितने चयन संभव हैं?

From (6) married couples, (4) people are to be selected so that no complete couple is selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (240)

Step 1

Concept

Choose (4) couples from (6) and then choose (1) person from each selected couple. The number is \(^{6}C_{4}\times2^{4}=240\).

Step 2

Why this answer is correct

The correct answer is B. (240). Choose (4) couples from (6) and then choose (1) person from each selected couple. The number is \(^{6}C_{4}\times2^{4}=240\).

Step 3

Exam Tip

पहले (6) जोड़ों में से (4) जोड़े चुनें और हर चुने जोड़े से (1) व्यक्ति चुनें। संख्या \(^{6}C_{4}\times2^{4}=240\)।

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Question 136/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(8) अलग-अलग फलों और (5) एक जैसी टॉफियों में से कुल (4) वस्तुएं चुननी हैं। कितने चयन संभव हैं?

From (8) distinct fruits and (5) identical candies, (4) objects are to be selected. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (163)

Step 1

Concept

The number of identical candies can be from (0) to (4). Sum \(^{8}C_{4}+^{8}C_{3}+^{8}C_{2}+^{8}C_{1}+1=163\).

Step 2

Why this answer is correct

The correct answer is A. (163). The number of identical candies can be from (0) to (4). Sum \(^{8}C_{4}+^{8}C_{3}+^{8}C_{2}+^{8}C_{1}+1=163\).

Step 3

Exam Tip

एक जैसी टॉफियों की संख्या (0) से (4) तक हो सकती है। योग \(^{8}C_{4}+^{8}C_{3}+^{8}C_{2}+^{8}C_{1}+1=163\)।

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Question 137/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(9) अलग-अलग पुस्तकों और (4) एक जैसी पत्रिकाओं में से कुल (6) वस्तुएं चुननी हैं जिनमें कम से कम (2) पत्रिकाएं हों। कितने चयन होंगे?

From (9) distinct books and (4) identical magazines, (6) objects are to be selected with at least (2) magazines. How many selections are possible?

Explanation opens after your attempt

Correct Answer

C. (246)

Step 1

Concept

The number of magazines can be (2), (3), or (4). Hence the count is \(^{9}C_{4}+^{9}C_{3}+^{9}C_{2}=246\).

Step 2

Why this answer is correct

The correct answer is C. (246). The number of magazines can be (2), (3), or (4). Hence the count is \(^{9}C_{4}+^{9}C_{3}+^{9}C_{2}=246\).

Step 3

Exam Tip

पत्रिकाएं (2), (3), या (4) ली जा सकती हैं। इसलिए संख्या \(^{9}C_{4}+^{9}C_{3}+^{9}C_{2}=246\)।

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Question 138/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(7) एक जैसी गेंदों को (4) अलग-अलग डिब्बों में इस प्रकार रखना है कि कोई डिब्बा खाली न रहे। कितने तरीके होंगे?

(7) identical balls are to be placed into (4) distinct boxes so that no box remains empty. How many ways are possible?

Explanation opens after your attempt

Correct Answer

D. (20)

Step 1

Concept

Place (1) ball in each box first. The remaining (3) balls can be distributed in \(^{6}C_{3}=20\) ways.

Step 2

Why this answer is correct

The correct answer is D. (20). Place (1) ball in each box first. The remaining (3) balls can be distributed in \(^{6}C_{3}=20\) ways.

Step 3

Exam Tip

हर डिब्बे में पहले (1) गेंद रखें। बची (3) गेंदों का वितरण \(^{6}C_{3}=20\) तरीकों से होगा।

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Question 139/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(9) एक जैसी गेंदों को (3) अलग-अलग डिब्बों में रखना है ताकि किसी डिब्बे में (4) से अधिक गेंदें न हों। कितने तरीके होंगे?

(9) identical balls are to be placed into (3) distinct boxes so that no box has more than (4) balls. How many ways are possible?

Explanation opens after your attempt

Correct Answer

B. (10)

Step 1

Concept

Total distributions are \(^{11}C_{2}=55\). Subtract \(3\times{}^{6}C_{2}=45\) cases where one box has at least (5) balls, giving (10).

Step 2

Why this answer is correct

The correct answer is B. (10). Total distributions are \(^{11}C_{2}=55\). Subtract \(3\times{}^{6}C_{2}=45\) cases where one box has at least (5) balls, giving (10).

Step 3

Exam Tip

कुल \(^{11}C_{2}=55\) वितरण हैं। किसी एक डिब्बे में कम से कम (5) गेंदें होने वाले \(3\times{}^{6}C_{2}=45\) घटाएं, उत्तर (10)।

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Question 140/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(12) तत्वों वाले समुच्चय के ठीक (5) तत्वों वाले उपसमुच्चयों की संख्या कितनी है?

How many subsets with exactly (5) elements does a set of (12) elements have?

Explanation opens after your attempt

Correct Answer

A. (792)

Step 1

Concept

Choosing a subset with exactly (5) elements means choosing (5) elements from (12). The number is \(^{12}C_{5}=792\).

Step 2

Why this answer is correct

The correct answer is A. (792). Choosing a subset with exactly (5) elements means choosing (5) elements from (12). The number is \(^{12}C_{5}=792\).

Step 3

Exam Tip

ठीक (5) तत्वों वाला उपसमुच्चय चुनना (12) में से (5) तत्व चुनना है। संख्या \(^{12}C_{5}=792\) है।

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Question 141/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(10) तत्वों वाले समुच्चय के विषम संख्या तत्वों वाले उपसमुच्चयों की संख्या कितनी है?

How many subsets with an odd number of elements does a set of (10) elements have?

Explanation opens after your attempt

Correct Answer

C. (512)

Step 1

Concept

In an (n)-element set, the number of odd-sized subsets is \(2^{n-1}\). Here \(2^{9}=512\).

Step 2

Why this answer is correct

The correct answer is C. (512). In an (n)-element set, the number of odd-sized subsets is \(2^{n-1}\). Here \(2^{9}=512\).

Step 3

Exam Tip

किसी (n) तत्वों वाले समुच्चय में विषम आकार के उपसमुच्चय \(2^{n-1}\) होते हैं। यहां \(2^{9}=512\)।

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Question 142/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(8) तत्वों वाले समुच्चय के कम से कम (6) तत्वों वाले उपसमुच्चयों की संख्या कितनी है?

How many subsets with at least (6) elements does a set of (8) elements have?

Explanation opens after your attempt

Correct Answer

D. (37)

Step 1

Concept

Count subsets of sizes (6), (7), and (8). The sum is \(^{8}C_{6}+^{8}C_{7}+^{8}C_{8}=37\).

Step 2

Why this answer is correct

The correct answer is D. (37). Count subsets of sizes (6), (7), and (8). The sum is \(^{8}C_{6}+^{8}C_{7}+^{8}C_{8}=37\).

Step 3

Exam Tip

आकार (6), (7), और (8) के उपसमुच्चय गिनें। योग \(^{8}C_{6}+^{8}C_{7}+^{8}C_{8}=37\)।

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Question 143/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

योग \(\sum_{r=0}^{4}{}^{7}C_{r},{}^{5}C_{4-r}\) का मान क्या है?

What is the value of \(\sum_{r=0}^{4}{}^{7}C_{r},{}^{5}C_{4-r}\)?

Explanation opens after your attempt

Correct Answer

B. (495)

Step 1

Concept

By Vandermonde's identity, the sum equals \(^{12}C_{4}\). Hence the value is (495).

Step 2

Why this answer is correct

The correct answer is B. (495). By Vandermonde's identity, the sum equals \(^{12}C_{4}\). Hence the value is (495).

Step 3

Exam Tip

वैंडरमोंड पहचान से यह योग \(^{12}C_{4}\) के बराबर है। इसलिए मान (495) है।

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Question 144/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

((1+2x)^{10}) में \(x^{4}\) का गुणांक क्या है?

What is the coefficient of \(x^{4}\) in ((1+2x)^{10})?

Explanation opens after your attempt

Correct Answer

A. (3360)

Step 1

Concept

The coefficient is \(^{10}C_{4}\times2^{4}\). Its value is \(210\times16=3360\).

Step 2

Why this answer is correct

The correct answer is A. (3360). The coefficient is \(^{10}C_{4}\times2^{4}\). Its value is \(210\times16=3360\).

Step 3

Exam Tip

गुणांक \(^{10}C_{4}\times2^{4}\) होगा। इसका मान \(210\times16=3360\) है।

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Question 145/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

((a+b)^{8}) में \(a^{3}b^{5}\) का गुणांक क्या है?

What is the coefficient of \(a^{3}b^{5}\) in ((a+b)^{8})?

Explanation opens after your attempt

Correct Answer

D. (56)

Step 1

Concept

For \(a^{3}b^{5}\), choose the (3) positions of (a). The coefficient is \(^{8}C_{3}=56\).

Step 2

Why this answer is correct

The correct answer is D. (56). For \(a^{3}b^{5}\), choose the (3) positions of (a). The coefficient is \(^{8}C_{3}=56\).

Step 3

Exam Tip

\(a^{3}b^{5}\) के लिए (a) वाले (3) स्थान चुनें। गुणांक \(^{8}C_{3}=56\) है।

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Question 146/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(30) संख्याओं की लॉटरी में (6) विजेता संख्याएं हैं। (6) संख्याओं के टिकट में ठीक (4) विजेता संख्याएं आने के कितने तरीके हैं?

In a lottery of (30) numbers, (6) numbers are winning numbers. How many (6)-number tickets contain exactly (4) winning numbers?

Explanation opens after your attempt

Correct Answer

C. (4140)

Step 1

Concept

Choose (4) winning numbers and (2) numbers from the (24) non-winning numbers. The count is \(^{6}C_{4}\times{}^{24}C_{2}=4140\).

Step 2

Why this answer is correct

The correct answer is C. (4140). Choose (4) winning numbers and (2) numbers from the (24) non-winning numbers. The count is \(^{6}C_{4}\times{}^{24}C_{2}=4140\).

Step 3

Exam Tip

विजेता संख्याओं में से (4) और गैर-विजेता (24) में से (2) चुनें। संख्या \(^{6}C_{4}\times{}^{24}C_{2}=4140\)।

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Question 147/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(9) टॉपिंग में (2) मसालेदार, (3) चीज़ और (4) अन्य हैं। (3) टॉपिंग चुननी हैं जिनमें कम से कम (1) मसालेदार और कम से कम (1) चीज़ हो। कितने चयन होंगे?

Among (9) toppings, (2) are spicy, (3) are cheese, and (4) are other toppings. Choose (3) toppings with at least (1) spicy and at least (1) cheese topping. How many selections are possible?

Explanation opens after your attempt

Correct Answer

B. (33)

Step 1

Concept

The cases are (1S1C1O), (1S2C), and (2S1C). Total \(2\times3\times4+2\times{}^{3}C_{2}+{}^{2}C_{2}\times3=33\).

Step 2

Why this answer is correct

The correct answer is B. (33). The cases are (1S1C1O), (1S2C), and (2S1C). Total \(2\times3\times4+2\times{}^{3}C_{2}+{}^{2}C_{2}\times3=33\).

Step 3

Exam Tip

मामले (1S1C1O), (1S2C), और (2S1C) हैं। कुल \(2\times3\times4+2\times{}^{3}C_{2}+{}^{2}C_{2}\times3=33\)।

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Question 148/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(9) सदस्यों में से (4) चुनने हैं। एक जोड़ी (P) के दोनों सदस्य साथ नहीं चुने जा सकते और दूसरी जोड़ी (Q) से कम से कम (1) सदस्य चुना जाना चाहिए। दोनों जोड़ियां अलग हैं। कितने चयन संभव हैं?

From (9) members, (4) are to be chosen. Both members of pair (P) cannot be chosen together and at least (1) member of pair (Q) must be chosen. The two pairs are disjoint. How many selections are possible?

Explanation opens after your attempt

Correct Answer

A. (80)

Step 1

Concept

First count selections with at least (1) member from (Q): \(^{9}C_{4}-^{7}C_{4}=91\). Subtract \(^{7}C_{2}-^{5}C_{2}=11\) cases containing both members of (P), giving (80).

Step 2

Why this answer is correct

The correct answer is A. (80). First count selections with at least (1) member from (Q): \(^{9}C_{4}-^{7}C_{4}=91\). Subtract \(^{7}C_{2}-^{5}C_{2}=11\) cases containing both members of (P), giving (80).

Step 3

Exam Tip

पहले (Q) से कम से कम (1) सदस्य वाले चयन \(^{9}C_{4}-^{7}C_{4}=91\) हैं। इनमें (P) की दोनों सदस्य वाली \(^{7}C_{2}-^{5}C_{2}=11\) स्थितियां घटाएं, उत्तर (80)।

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Question 149/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(5) डॉक्टरों, (6) इंजीनियरों और (7) शिक्षकों में से (6) लोगों का दल बनाना है जिसमें हर पेशे से ठीक (2) लोग हों। कितने दल बनेंगे?

From (5) doctors, (6) engineers, and (7) teachers, a team of (6) people is to be formed with exactly (2) people from each profession. How many teams are possible?

Explanation opens after your attempt

Correct Answer

D. (3150)

Step 1

Concept

Choose (2) people from each profession and multiply the results. The number is \(^{5}C_{2}\times{}^{6}C_{2}\times{}^{7}C_{2}=3150\).

Step 2

Why this answer is correct

The correct answer is D. (3150). Choose (2) people from each profession and multiply the results. The number is \(^{5}C_{2}\times{}^{6}C_{2}\times{}^{7}C_{2}=3150\).

Step 3

Exam Tip

हर पेशे से (2) चुनें और परिणामों को गुणा करें। संख्या \(^{5}C_{2}\times{}^{6}C_{2}\times{}^{7}C_{2}=3150\)।

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Question 150/150 Expert Mathematics Permutations and Combinations Combinations Class 11 Level 63

(15) सदस्यों में से (6) सदस्यों की समिति बनानी है। एक निश्चित सदस्य अवश्य शामिल हो और दो विशेष सदस्य साथ-साथ शामिल न हों, तो कितनी समितियां बनेंगी?

A committee of (6) members is to be formed from (15) members. One fixed member must be included and two special members must not be included together. How many committees are possible?

Explanation opens after your attempt

Correct Answer

B. (1782)

Step 1

Concept

After fixing the required member, choose (5) from the remaining (14) and subtract cases where both special members are together, \(^{12}C_{3}\). The answer is \(^{14}C_{5}-{}^{12}C_{3}=1782\).

Step 2

Why this answer is correct

The correct answer is B. (1782). After fixing the required member, choose (5) from the remaining (14) and subtract cases where both special members are together, \(^{12}C_{3}\). The answer is \(^{14}C_{5}-{}^{12}C_{3}=1782\).

Step 3

Exam Tip

निश्चित सदस्य को शामिल मानकर शेष (14) में से (5) चुनें और दोनों विशेष साथ वाले \(^{12}C_{3}\) घटाएं। उत्तर \(^{14}C_{5}-{}^{12}C_{3}=1782\) है।

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Class 11 Mathematics - Permutations And Combinations - Combinations Expert Quiz

द्विपद प्रसार में ( (1+x)^{12} ) का \( x^5 \) वाला गुणांक कितना है?

Concept

Why this answer is correct

Exam Tip

(7) पुरुषों और (5) महिलाओं में से (5) सदस्यों की समिति बनानी है जिसमें कम से कम (2) महिलाएँ हों। कुल कितने चयन होंगे?

Concept

Why this answer is correct

Exam Tip

यदि \( \binom{n}{4}=\binom{n}{2} \) और \( n\ge 4 \) है, तो ( n ) का मान क्या है?

Concept

Why this answer is correct

Exam Tip

एक उत्तल दशभुज में विकर्णों की संख्या कितनी होगी?

Concept

Why this answer is correct

Exam Tip

(1) से (10) तक की संख्याओं में से (3) संख्याएँ ऐसी चुननी हैं कि कोई दो क्रमागत न हों। कुल चयन कितने हैं?

Concept

Why this answer is correct

Exam Tip

ताश की (52) पत्तों की गड्डी से (5) पत्ते चुनने हैं जिनमें ठीक (2) बादशाह हों। कुल कितने हाथ बनेंगे?

Concept

Why this answer is correct

Exam Tip

(12) बिंदुओं में से (5) बिंदु एक ही सीधी रेखा पर हैं। इन बिंदुओं से बनने वाले त्रिभुजों की संख्या कितनी है?

Concept

Why this answer is correct

Exam Tip

(6) अलग-अलग पुस्तकों में से कम से कम एक पुस्तक चुनने के कुल कितने तरीके हैं?

Concept

Why this answer is correct

Exam Tip

(6) वरिष्ठ और (4) कनिष्ठ विद्यार्थियों में से (4) सदस्य चुनने हैं जिनमें ठीक (3) वरिष्ठ और (1) कनिष्ठ हो। कुल चयन कितने हैं?

Concept

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Exam Tip

(5) लाल, (4) नीली और (3) हरी गेंदों में से (4) गेंदें चुननी हैं जिनमें हर रंग की कम से कम एक गेंद हो। कुल चयन कितने हैं?

Concept

Why this answer is correct

Exam Tip

एक ग्रिड में (5) दाएँ और (3) ऊपर कदमों से गंतव्य तक जाने के न्यूनतम रास्ते कितने होंगे?

Concept

Why this answer is correct

Exam Tip

समीकरण (x+y+z=12) के ऐसे पूर्णांक हल कितने हैं जिनमें \(x\ge2\), \(y\ge3\) और \(z\ge0\) हो?

Concept

Why this answer is correct

Exam Tip

समीकरण (x+y+z=10) के ऐसे शून्य सहित पूर्णांक हल कितने हैं जिनमें हर चर (5) से अधिक नहीं है?

Concept

Why this answer is correct

Exam Tip

(8) व्यंजन और (5) स्वर में से (5) अक्षर चुनने हैं जिनमें कम से कम (2) स्वर हों। कुल चयन कितने हैं?

Concept

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Exam Tip

(9) अवयवों वाले समुच्चय के विषम संख्या वाले उपसमुच्चयों की संख्या कितनी है?

Concept

Why this answer is correct

Exam Tip

(1) से (9) तक के अंकों में से (3) अलग अंक चुनकर केवल अंक-समुच्चय बनाना है। कुल कितने समुच्चय बनेंगे?

Concept

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Exam Tip

(12) व्यक्तियों में से (5) चुनने हैं, पर दो विशेष व्यक्ति साथ-साथ नहीं चुने जाने चाहिए। कुल चयन कितने हैं?

Concept

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Exam Tip

Concept

Why this answer is correct

Exam Tip

(8) लड़कों और (6) लड़कियों में से (6) सदस्य चुनने हैं, जिनमें लड़कियाँ लड़कों की संख्या की दोगुनी हों। कुल चयन कितने हैं?

Concept

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Exam Tip

(5) पंक्तियों और (4) स्तंभों वाले छोटे आयतों के ग्रिड में कुल आयतों की संख्या कितनी है?

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Exam Tip