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Class 11 Mathematics Expert Quiz

Mathematics Expert Class 11 Level 63

Level 63 • 50/50 questions • 25 seconds per question.

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Question 1/50 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 2/50 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 3/50 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 4/50 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 5/50 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 6/50 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 7/50 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 8/50 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 9/50 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 10/50 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 11/50 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 12/50 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 13/50 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 14/50 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 15/50 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 16/50 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 17/50 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 18/50 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 19/50 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 20/50 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 21/50 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 22/50 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 23/50 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 24/50 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 25/50 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 26/50 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 27/50 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 28/50 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 29/50 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 30/50 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 31/50 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 32/50 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 33/50 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 34/50 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 35/50 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 36/50 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 37/50 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 38/50 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 39/50 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 40/50 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 41/50 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 42/50 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 43/50 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 44/50 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 45/50 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 46/50 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 47/50 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 48/50 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 49/50 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 50/50 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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