Concept-wise Practice

class11 MCQ Questions for Class 11

class11 se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

1581 questions tagged with class11.

(14) विद्यार्थियों में से (7) की टीम बनानी है जिसमें (3) विशेष विद्यार्थी या तो सभी शामिल हों या सभी बाहर हों। कितने तरीके हैं?

From (14) students a team of (7) is to be formed in which (3) special students are either all included or all excluded. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (660)

Step 1

Concept

If all special students are included there are \(\binom{11}{4}=330\) ways and if all are excluded there are \(\binom{11}{7}=330\) ways. The total is (660).

Step 2

Why this answer is correct

The correct answer is B. (660). If all special students are included there are \(\binom{11}{4}=330\) ways and if all are excluded there are \(\binom{11}{7}=330\) ways. The total is (660).

Step 3

Exam Tip

सभी विशेष शामिल हों तो \(\binom{11}{4}=330\) और सभी बाहर हों तो \(\binom{11}{7}=330\) तरीके हैं। कुल (660) है।

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(15) कुर्सियों में से (8) कुर्सियां चुननी हैं और (5) खराब कुर्सियों में से कोई न चुनी जाए। कितने तरीके हैं?

From (15) chairs (8) chairs are to be selected and none of (5) broken chairs is selected. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (45)

Step 1

Concept

After removing (5) broken chairs (10) good chairs remain. The ways to choose (8) are \(\binom{10}{8}=45\).

Step 2

Why this answer is correct

The correct answer is A. (45). After removing (5) broken chairs (10) good chairs remain. The ways to choose (8) are \(\binom{10}{8}=45\).

Step 3

Exam Tip

(5) खराब कुर्सियां हटाने पर (10) अच्छी कुर्सियां बचती हैं। (8) चुनने के तरीके \(\binom{10}{8}=45\) हैं।

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(18) बिंदुओं से त्रिभुज बनाने हैं। यदि (10) बिंदु एक सीध में हैं तो केवल उन्हीं (10) बिंदुओं से बनने वाले असफल चयन कितने हैं?

Triangles are to be formed from (18) points. If (10) points are collinear then how many failed selections come only from those (10) points?

Explanation opens after your attempt
Correct Answer

C. \(\binom{10}{3}\)

Step 1

Concept

For a failed triangle (3) points are chosen from the (10) collinear points. So the number is \(\binom{10}{3}\).

Step 2

Why this answer is correct

The correct answer is C. \(\binom{10}{3}\). For a failed triangle (3) points are chosen from the (10) collinear points. So the number is \(\binom{10}{3}\).

Step 3

Exam Tip

असफल त्रिभुज के लिए (10) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{10}{3}\) है।

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(18) संख्याओं में से (8) संख्याएं चुननी हैं जिनमें (3) निश्चित संख्याएं शामिल हों और (2) निश्चित संख्याएं शामिल न हों। कितने तरीके हैं?

From (18) numbers (8) numbers are to be selected with (3) fixed numbers included and (2) fixed numbers excluded. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (1287)

Step 1

Concept

The (3) numbers are fixed and (2) are excluded. The remaining (5) numbers are chosen from (13), so \(\binom{13}{5}=1287\).

Step 2

Why this answer is correct

The correct answer is B. (1287). The (3) numbers are fixed and (2) are excluded. The remaining (5) numbers are chosen from (13), so \(\binom{13}{5}=1287\).

Step 3

Exam Tip

(3) संख्याएं तय हैं और (2) हट गई हैं। बाकी (5) संख्याएं (13) में से चुनी जाएंगी इसलिए \(\binom{13}{5}=1287\) है।

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(11) विषयों में से (5) विषय चुनने हैं। रसायन अवश्य चुना जाए, जीवविज्ञान न चुना जाए और गणित तथा भौतिकी दोनों साथ में न चुने जाएं। कितने तरीके हैं?

From (11) subjects (5) subjects are to be selected. Chemistry must be selected, biology must not be selected, and mathematics and physics must not both be selected together. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (105)

Step 1

Concept

Chemistry is fixed and biology is excluded. Choose the remaining (4) subjects from (9) and subtract \(\binom{7}{2}\) selections containing both mathematics and physics to get (105).

Step 2

Why this answer is correct

The correct answer is A. (105). Chemistry is fixed and biology is excluded. Choose the remaining (4) subjects from (9) and subtract \(\binom{7}{2}\) selections containing both mathematics and physics to get (105).

Step 3

Exam Tip

रसायन तय और जीवविज्ञान हट गया है। बाकी (4) विषय (9) में से चुनें और गणित-भौतिकी दोनों वाले \(\binom{7}{2}\) चयन घटाएं तो (105) मिलते हैं।

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(13) अलग-अलग अक्षरों में से (6) अक्षर चुनने हैं जिनमें (a) हो, (b) न हो और (c,d) में से ठीक एक अक्षर हो। कितने तरीके हैं?

From (13) distinct letters (6) letters are to be selected containing (a), not containing (b), and containing exactly one of (c,d). How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (252)

Step 1

Concept

The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (4) from the remaining (9), so \(\binom{2}{1}\binom{9}{4}=252\).

Step 2

Why this answer is correct

The correct answer is C. (252). The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (4) from the remaining (9), so \(\binom{2}{1}\binom{9}{4}=252\).

Step 3

Exam Tip

(a) तय है और (b) हट गया है। (c,d) में से (1) और शेष (9) में से (4) चुनेंगे इसलिए \(\binom{2}{1}\binom{9}{4}=252\) है।

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(9) लाल, (10) नीली और (11) हरी गेंदों में से (2) लाल, (2) नीली और (2) हरी गेंद चुनने के कितने तरीके हैं?

From (9) red, (10) blue, and (11) green balls, how many ways are there to choose (2) red, (2) blue, and (2) green balls?

Explanation opens after your attempt
Correct Answer

D. (89100)

Step 1

Concept

Each color condition is counted separately. The ways are \(\binom{9}{2}\binom{10}{2}\binom{11}{2}=89100\).

Step 2

Why this answer is correct

The correct answer is D. (89100). Each color condition is counted separately. The ways are \(\binom{9}{2}\binom{10}{2}\binom{11}{2}=89100\).

Step 3

Exam Tip

हर रंग की शर्त अलग गिनी जाएगी। तरीके \(\binom{9}{2}\binom{10}{2}\binom{11}{2}=89100\) हैं।

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(14) फलों में से (6) फल चुनने हैं लेकिन (6) विशेष फलों में से ठीक (3) चुने जाएं। कितने तरीके हैं?

From (14) fruits (6) fruits are to be selected but exactly (3) of (6) special fruits are chosen. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (1120)

Step 1

Concept

Choose (3) from the special fruits and (3) from the remaining (8). The ways are \(\binom{6}{3}\binom{8}{3}=1120\).

Step 2

Why this answer is correct

The correct answer is B. (1120). Choose (3) from the special fruits and (3) from the remaining (8). The ways are \(\binom{6}{3}\binom{8}{3}=1120\).

Step 3

Exam Tip

विशेष फलों में से (3) और बाकी (8) में से (3) चुनेंगे। तरीके \(\binom{6}{3}\binom{8}{3}=1120\) हैं।

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(13) मित्रों में से (6) को यात्रा के लिए चुनना है और (5) विशेष मित्रों में से कम से कम एक जाना चाहिए। कितने तरीके हैं?

From (13) friends (6) are to be selected for a trip and at least one of (5) special friends must go. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (1688)

Step 1

Concept

Total ways are \(\binom{13}{6}=1716\) and if no special friend goes then \(\binom{8}{6}=28\). Hence (1716-28=1688).

Step 2

Why this answer is correct

The correct answer is C. (1688). Total ways are \(\binom{13}{6}=1716\) and if no special friend goes then \(\binom{8}{6}=28\). Hence (1716-28=1688).

Step 3

Exam Tip

कुल \(\binom{13}{6}=1716\) हैं और कोई विशेष मित्र न जाए तो \(\binom{8}{6}=28\) हैं। इसलिए (1716-28=1688) है।

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(16) उम्मीदवारों में से (6) पुरस्कार विजेताओं का चयन करना है और सभी पुरस्कार समान हैं। कितने तरीके हैं?

From (16) candidates (6) prize winners are to be selected and all prizes are identical. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (8008)

Step 1

Concept

The prizes are identical so only selection is needed. The number of ways is \(\binom{16}{6}=8008\).

Step 2

Why this answer is correct

The correct answer is A. (8008). The prizes are identical so only selection is needed. The number of ways is \(\binom{16}{6}=8008\).

Step 3

Exam Tip

पुरस्कार समान हैं इसलिए केवल चयन होगा। तरीकों की संख्या \(\binom{16}{6}=8008\) है।

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(10) सफेद और (11) काली गेंदों में से (7) गेंदें चुननी हैं जिनमें सभी गेंदों का रंग समान न हो। कितने तरीके हैं?

From (10) white and (11) black balls (7) balls are to be selected such that all balls are not of the same color. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (115830)

Step 1

Concept

Total ways are \(\binom{21}{7}=116280\). Removing all-white \(\binom{10}{7}=120\) and all-black \(\binom{11}{7}=330\) gives (115830).

Step 2

Why this answer is correct

The correct answer is B. (115830). Total ways are \(\binom{21}{7}=116280\). Removing all-white \(\binom{10}{7}=120\) and all-black \(\binom{11}{7}=330\) gives (115830).

Step 3

Exam Tip

कुल \(\binom{21}{7}=116280\) हैं। सभी सफेद \(\binom{10}{7}=120\) और सभी काली \(\binom{11}{7}=330\) हटाने पर (115830) मिलते हैं।

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(11) शिक्षकों और (13) छात्रों में से (7) लोगों का समूह बनाना है जिसमें कम से कम (4) शिक्षक हों। कितने तरीके हैं?

From (11) teachers and (13) students a group of (7) people is to be formed with at least (4) teachers. How many ways are there?

Explanation opens after your attempt
Correct Answer

D. (136752)

Step 1

Concept

The number of teachers can be (4), (5), (6), or (7). The sum of all these cases is (136752).

Step 2

Why this answer is correct

The correct answer is D. (136752). The number of teachers can be (4), (5), (6), or (7). The sum of all these cases is (136752).

Step 3

Exam Tip

शिक्षक (4), (5), (6) या (7) हो सकते हैं। इन सभी मामलों का योग (136752) है।

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(22) विद्यार्थियों में से (8) चुनने हैं ताकि (7) विशेष विद्यार्थियों में से कोई भी शामिल न हो। कितने तरीके हैं?

From (22) students (8) are to be selected so that none of (7) special students is included. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (6435)

Step 1

Concept

After removing (7) special students (15) students remain. Hence there are \(\binom{15}{8}=6435\) ways.

Step 2

Why this answer is correct

The correct answer is A. (6435). After removing (7) special students (15) students remain. Hence there are \(\binom{15}{8}=6435\) ways.

Step 3

Exam Tip

(7) विशेष विद्यार्थियों को हटाने पर (15) विद्यार्थी बचते हैं। इसलिए \(\binom{15}{8}=6435\) तरीके हैं।

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(1) से (35) तक की संख्याओं में से (4) विषम और (2) सम संख्या चुनने के कितने तरीके हैं?

From numbers (1) to (35), how many ways are there to choose (4) odd and (2) even numbers?

Explanation opens after your attempt
Correct Answer

C. (416160)

Step 1

Concept

There are (18) odd numbers and (17) even numbers. The ways are \(\binom{18}{4}\binom{17}{2}=416160\).

Step 2

Why this answer is correct

The correct answer is C. (416160). There are (18) odd numbers and (17) even numbers. The ways are \(\binom{18}{4}\binom{17}{2}=416160\).

Step 3

Exam Tip

विषम संख्याएं (18) और सम संख्याएं (17) हैं। तरीके \(\binom{18}{4}\binom{17}{2}=416160\) हैं।

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(1) से (48) तक की संख्याओं में से (4) ऐसी संख्याएं चुननी हैं जो (6) की गुणज हों और (2) ऐसी संख्याएं जो (6) की गुणज न हों। कितने तरीके हैं?

From numbers (1) to (48), (4) numbers that are multiples of (6) and (2) numbers that are not multiples of (6) are to be selected. How many ways are there?

Explanation opens after your attempt
Correct Answer

D. (54600)

Step 1

Concept

There are (8) multiples of (6) and (40) non-multiples. The ways are \(\binom{8}{4}\binom{40}{2}=54600\).

Step 2

Why this answer is correct

The correct answer is D. (54600). There are (8) multiples of (6) and (40) non-multiples. The ways are \(\binom{8}{4}\binom{40}{2}=54600\).

Step 3

Exam Tip

(6) की (8) गुणज और (40) गैर-गुणज संख्याएं हैं। तरीके \(\binom{8}{4}\binom{40}{2}=54600\) हैं।

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\(A=\{1,2,3,4,5,6,7,8,9,10,11,12,13\}\) के कितने (6)-तत्व उपसमुच्चय (1) को शामिल करते हैं लेकिन (5) और (6) दोनों को साथ शामिल नहीं करते?

How many (6)-element subsets of \(A=\{1,2,3,4,5,6,7,8,9,10,11,12,13\}\) contain (1) but do not contain both (5) and (6) together?

Explanation opens after your attempt
Correct Answer

A. (672)

Step 1

Concept

The element (1) is fixed so choose (5) elements from (12). Subtracting \(\binom{10}{3}=120\) cases containing both (5), (6) gives (792-120=672).

Step 2

Why this answer is correct

The correct answer is A. (672). The element (1) is fixed so choose (5) elements from (12). Subtracting \(\binom{10}{3}=120\) cases containing both (5), (6) gives (792-120=672).

Step 3

Exam Tip

(1) तय है इसलिए (5) तत्व (12) में से चुनेंगे। (5), (6) दोनों होने पर \(\binom{10}{3}=120\) घटाने से (792-120=672) मिलता है।

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कितने (7)-तत्व उपसमुच्चय \(A=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14\}\) से बनाए जा सकते हैं जिनमें (1) और (2) शामिल हों लेकिन (3) और (4) शामिल न हों?

How many (7)-element subsets can be formed from \(A=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14\}\) that contain (1) and (2) but not (3) and (4)?

Explanation opens after your attempt
Correct Answer

B. (252)

Step 1

Concept

The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (5) elements are chosen from (10), so \(\binom{10}{5}=252\).

Step 2

Why this answer is correct

The correct answer is B. (252). The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (5) elements are chosen from (10), so \(\binom{10}{5}=252\).

Step 3

Exam Tip

(1) और (2) तय हैं तथा (3), (4) हट गए हैं। बाकी (5) तत्व (10) में से चुने जाएंगे इसलिए \(\binom{10}{5}=252\) है।

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\(\binom{15}{7}-\binom{14}{7}\) का मान क्या है?

What is the value of \(\binom{15}{7}-\binom{14}{7}\)?

Explanation opens after your attempt
Correct Answer

C. (3003)

Step 1

Concept

\(\binom{15}{7}=6435\) and \(\binom{14}{7}=3432\). The difference is (3003).

Step 2

Why this answer is correct

The correct answer is C. (3003). \(\binom{15}{7}=6435\) and \(\binom{14}{7}=3432\). The difference is (3003).

Step 3

Exam Tip

\(\binom{15}{7}=6435\) और \(\binom{14}{7}=3432\) हैं। अंतर (3003) है।

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\(\binom{13}{6}+\binom{13}{7}\) पास्कल पहचान से किसके बराबर है?

Using Pascal's identity \(\binom{13}{6}+\binom{13}{7}\) is equal to which expression?

Explanation opens after your attempt
Correct Answer

A. \(\binom{14}{7}\)

Step 1

Concept

By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{14}{7}\).

Step 2

Why this answer is correct

The correct answer is A. \(\binom{14}{7}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{14}{7}\).

Step 3

Exam Tip

पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{14}{7}\) है।

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यदि \(\binom{n}{1}+\binom{n}{2}=136\) है तो (n) का मान क्या है?

If \(\binom{n}{1}+\binom{n}{2}=136\), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

B. (16)

Step 1

Concept

It gives (\frac{n(n+1)}{2}=136). Putting (n=16) gives (136).

Step 2

Why this answer is correct

The correct answer is B. (16). It gives (\frac{n(n+1)}{2}=136). Putting (n=16) gives (136).

Step 3

Exam Tip

यह (\frac{n(n+1)}{2}=136) देता है। (n=16) रखने पर (136) मिलता है।

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यदि \(\binom{n}{7}=\binom{n}{11}\) है तो (n) का मान क्या होगा?

If \(\binom{n}{7}=\binom{n}{11}\), what will be the value of (n)?

Explanation opens after your attempt
Correct Answer

C. (18)

Step 1

Concept

Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (7+11=18).

Step 2

Why this answer is correct

The correct answer is C. (18). Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (7+11=18).

Step 3

Exam Tip

\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (7+11=18) है।

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यदि \(\binom{n}{2}=190\) है तो (n) का मान क्या है?

If \(\binom{n}{2}=190\), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

D. (20)

Step 1

Concept

\(\binom{20}{2}=190\). Therefore (n=20) is correct.

Step 2

Why this answer is correct

The correct answer is D. (20). \(\binom{20}{2}=190\). Therefore (n=20) is correct.

Step 3

Exam Tip

\(\binom{20}{2}=190\) होता है। इसलिए (n=20) सही है।

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(10) अलग-अलग पेन और (8) अलग-अलग पेंसिलों में से (7) वस्तुएं चुननी हैं जिनमें कम से कम (2) पेन और कम से कम (2) पेंसिल हों। कितने तरीके हैं?

From (10) different pens and (8) different pencils (7) objects are to be selected with at least (2) pens and at least (2) pencils. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (29736)

Step 1

Concept

The number of pens can be (2), (3), (4), or (5). The total is \(\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736\).

Step 2

Why this answer is correct

The correct answer is B. (29736). The number of pens can be (2), (3), (4), or (5). The total is \(\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736\).

Step 3

Exam Tip

पेन (2), (3), (4) या (5) हो सकते हैं। कुल \(\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736\) है।

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(12) पुरुषों और (10) महिलाओं में से (7) व्यक्तियों की समिति बनानी है जिसमें ठीक (4) पुरुष हों। कितने तरीके हैं?

From (12) men and (10) women a committee of (7) persons is to be formed with exactly (4) men. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (59400)

Step 1

Concept

Exactly (4) men and (3) women are needed. The ways are \(\binom{12}{4}\binom{10}{3}=59400\).

Step 2

Why this answer is correct

The correct answer is C. (59400). Exactly (4) men and (3) women are needed. The ways are \(\binom{12}{4}\binom{10}{3}=59400\).

Step 3

Exam Tip

ठीक (4) पुरुष और (3) महिलाएं चाहिए। तरीके \(\binom{12}{4}\binom{10}{3}=59400\) हैं।

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(17) छात्रों में से (8) छात्रों का चयन करना है। (6) विशेष छात्रों में से कम से कम (3) शामिल हों। कितने तरीके हैं?

From (17) students (8) students are to be selected. At least (3) of (6) special students must be included. How many ways are there?

Explanation opens after your attempt
Correct Answer

D. (15235)

Step 1

Concept

The number of special students can be (3), (4), (5), or (6). The total is \(\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235\).

Step 2

Why this answer is correct

The correct answer is D. (15235). The number of special students can be (3), (4), (5), or (6). The total is \(\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235\).

Step 3

Exam Tip

विशेष छात्र (3), (4), (5) या (6) हो सकते हैं। कुल \(\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235\) है।

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(15) विद्यार्थियों में से (7) विद्यार्थियों की टीम बनानी है। (6) विशेष विद्यार्थियों में से ठीक (3) शामिल होने चाहिए। कितने तरीके हैं?

A team of (7) students is to be formed from (15) students. Exactly (3) of (6) special students must be included. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (2520)

Step 1

Concept

Choose (3) from (6) special students and (4) from the remaining (9). The ways are \(\binom{6}{3}\binom{9}{4}=2520\).

Step 2

Why this answer is correct

The correct answer is B. (2520). Choose (3) from (6) special students and (4) from the remaining (9). The ways are \(\binom{6}{3}\binom{9}{4}=2520\).

Step 3

Exam Tip

(6) विशेष में से (3) और बाकी (9) में से (4) चुनेंगे। तरीके \(\binom{6}{3}\binom{9}{4}=2520\) हैं।

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(13) रंगों में से (6) रंग चुनने हैं। एक निश्चित रंग अवश्य हो, दूसरा निश्चित रंग न हो और दो अन्य रंगों में से कम से कम एक रंग हो। कितने तरीके हैं?

From (13) colors (6) colors are to be selected. One fixed color must be included, another fixed color must be excluded, and at least one of two other colors must be included. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (336)

Step 1

Concept

One color is fixed and one is excluded. Choose the remaining (5) colors from (11) and subtract \(\binom{9}{5}\) selections missing both other colors to get (336).

Step 2

Why this answer is correct

The correct answer is A. (336). One color is fixed and one is excluded. Choose the remaining (5) colors from (11) and subtract \(\binom{9}{5}\) selections missing both other colors to get (336).

Step 3

Exam Tip

एक रंग तय और एक हट गया है। बाकी (5) रंग (11) में से चुनें और दोनों अन्य रंग न आने वाले \(\binom{9}{5}\) चयन घटाएं तो (336) मिलेगा।

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(15) अलग-अलग उपहारों में से कम से कम (4) और अधिकतम (6) उपहार चुनने के कितने तरीके हैं?

In how many ways can at least (4) and at most (6) gifts be selected from (15) different gifts?

Explanation opens after your attempt
Correct Answer

C. (9373)

Step 1

Concept

The selection can be of (4), (5), or (6) gifts. The total is \(\binom{15}{4}+\binom{15}{5}+\binom{15}{6}=9373\).

Step 2

Why this answer is correct

The correct answer is C. (9373). The selection can be of (4), (5), or (6) gifts. The total is \(\binom{15}{4}+\binom{15}{5}+\binom{15}{6}=9373\).

Step 3

Exam Tip

चयन (4), (5) या (6) उपहारों का होगा। कुल \(\binom{15}{4}+\binom{15}{5}+\binom{15}{6}=9373\) है।

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(11) डॉक्टरों और (8) नर्सों में से (7) लोगों की टीम बनानी है जिसमें डॉक्टरों की संख्या नर्सों से अधिक हो। कितने तरीके हैं?

From (11) doctors and (8) nurses a team of (7) people is to be formed with more doctors than nurses. How many ways are there?

Explanation opens after your attempt
Correct Answer

D. (35442)

Step 1

Concept

The number of doctors will be (4), (5), (6), or (7). The total is \(\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442\).

Step 2

Why this answer is correct

The correct answer is D. (35442). The number of doctors will be (4), (5), (6), or (7). The total is \(\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442\).

Step 3

Exam Tip

डॉक्टरों की संख्या (4), (5), (6) या (7) होगी। कुल \(\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442\) है।

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(10) अंग्रेजी और (11) हिंदी पुस्तकों में से (7) पुस्तकें चुननी हैं जिनमें हर भाषा से कम से कम (2) पुस्तकें हों। कितने तरीके हैं?

From (10) English and (11) Hindi books (7) books are to be selected with at least (2) books from each language. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (108900)

Step 1

Concept

The number of English books can be (2), (3), (4), or (5). The sum of all valid cases is (108900).

Step 2

Why this answer is correct

The correct answer is B. (108900). The number of English books can be (2), (3), (4), or (5). The sum of all valid cases is (108900).

Step 3

Exam Tip

अंग्रेजी पुस्तकों की संख्या (2), (3), (4) या (5) हो सकती है। सभी वैध मामलों का योग (108900) है।

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