Class 11 Mathematics Hard Quiz

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(20) बिंदुओं में से (8) बिंदु एक सीध में हैं और (5) अन्य बिंदु दूसरी सीध में हैं। कोई अन्य (3) बिंदु समरेखीय नहीं हैं। कितने त्रिभुज बनेंगे?

Among (20) points (8) points are collinear and another (5) points are collinear on a different line. No other (3) points are collinear. How many triangles can be formed?

Explanation opens after your attempt
Correct Answer

B. (1074)

Step 1

Concept

Total triples are \(\binom{20}{3}=1140\) and failed triples are \(\binom{8}{3}+\binom{5}{3}=66\). Hence (1140-66=1074).

Step 2

Why this answer is correct

The correct answer is B. (1074). Total triples are \(\binom{20}{3}=1140\) and failed triples are \(\binom{8}{3}+\binom{5}{3}=66\). Hence (1140-66=1074).

Step 3

Exam Tip

कुल \(\binom{20}{3}=1140\) त्रिक हैं और असफल त्रिक \(\binom{8}{3}+\binom{5}{3}=66\) हैं। इसलिए (1140-66=1074) है।

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(10) पुरुषों और (8) महिलाओं में से (6) व्यक्तियों की समिति बनानी है जिसमें कम से कम (2) पुरुष और कम से कम (2) महिलाएं हों। कितने तरीके हैं?

From (10) men and (8) women a committee of (6) persons is to be formed with at least (2) men and at least (2) women. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (15750)

Step 1

Concept

The number of men can be (2), (3), or (4). The total is \(\binom{10}{2}\binom{8}{4}+\binom{10}{3}\binom{8}{3}+\binom{10}{4}\binom{8}{2}=15750\).

Step 2

Why this answer is correct

The correct answer is C. (15750). The number of men can be (2), (3), or (4). The total is \(\binom{10}{2}\binom{8}{4}+\binom{10}{3}\binom{8}{3}+\binom{10}{4}\binom{8}{2}=15750\).

Step 3

Exam Tip

पुरुषों की संख्या (2), (3) या (4) हो सकती है। कुल \(\binom{10}{2}\binom{8}{4}+\binom{10}{3}\binom{8}{3}+\binom{10}{4}\binom{8}{2}=15750\) है।

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(5) गणित, (4) भौतिकी और (3) रसायन पुस्तकों में से (5) पुस्तकें चुननी हैं जिनमें हर विषय की कम से कम (1) पुस्तक हो। कितने तरीके हैं?

From (5) mathematics, (4) physics, and (3) chemistry books, (5) books are to be selected with at least (1) book from each subject. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (621)

Step 1

Concept

Subtract selections missing at least one subject from \(\binom{12}{5}=792\). The correct count is (621).

Step 2

Why this answer is correct

The correct answer is A. (621). Subtract selections missing at least one subject from \(\binom{12}{5}=792\). The correct count is (621).

Step 3

Exam Tip

कुल \(\binom{12}{5}=792\) में से किसी विषय के न होने वाले चयन घटाएं। सही गिनती (621) है।

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(14) विद्यार्थियों में से (6) चुनने हैं। (4) विशेष विद्यार्थियों में से ठीक (2) या ठीक (3) शामिल हों। कितने तरीके हैं?

From (14) students (6) are to be selected. Exactly (2) or exactly (3) of (4) special students must be included. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (2100)

Step 1

Concept

The cases are (2) special and (3) special students. The total is \(\binom{4}{2}\binom{10}{4}+\binom{4}{3}\binom{10}{3}=2100\).

Step 2

Why this answer is correct

The correct answer is C. (2100). The cases are (2) special and (3) special students. The total is \(\binom{4}{2}\binom{10}{4}+\binom{4}{3}\binom{10}{3}=2100\).

Step 3

Exam Tip

मामले (2) विशेष और (3) विशेष के हैं। कुल \(\binom{4}{2}\binom{10}{4}+\binom{4}{3}\binom{10}{3}=2100\) है।

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(16) वस्तुओं में से (7) वस्तुएं चुननी हैं और (6) विशेष वस्तुओं में से कम से कम (3) चुनी जाएं। कितने तरीके हैं?

From (16) objects (7) objects are to be selected and at least (3) of (6) special objects are chosen. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (6960)

Step 1

Concept

The number of special objects can be (3), (4), (5), or (6). The total is \(\sum_{r=3}^{6}\binom{6}{r}\binom{10}{7-r}=6960\).

Step 2

Why this answer is correct

The correct answer is B. (6960). The number of special objects can be (3), (4), (5), or (6). The total is \(\sum_{r=3}^{6}\binom{6}{r}\binom{10}{7-r}=6960\).

Step 3

Exam Tip

विशेष वस्तुएं (3), (4), (5) या (6) हो सकती हैं। कुल \(\sum_{r=3}^{6}\binom{6}{r}\binom{10}{7-r}=6960\) है।

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(12) अलग-अलग अक्षरों में से (5) अक्षर चुनने हैं जिनमें (a) और (b) में से ठीक एक हो तथा (c) और (d) दोनों न हों। कितने तरीके हैं?

From (12) distinct letters (5) letters are to be selected with exactly one of (a,b) and not both (c,d). How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (300)

Step 1

Concept

Choose (1) from (a,b), then choose the remaining (4) so that (c,d) are not both included. The correct expression is (\binom{2}{1}\left\(\binom{10}{4}-\binom{8}{2}\right\)=364).

Step 2

Why this answer is correct

The correct answer is C. (300). Choose (1) from (a,b), then choose the remaining (4) so that (c,d) are not both included. The correct expression is (\binom{2}{1}\left\(\binom{10}{4}-\binom{8}{2}\right\)=364).

Step 3

Exam Tip

(a,b) में से (1) चुनकर बाकी (4) ऐसे चुनें कि (c,d) दोनों साथ न आएं। तरीके (\binom{2}{1}\left\(\binom{10}{4}-\binom{8}{2}\right\)=364) नहीं, सही (\binom{2}{1}\left\(\binom{10}{4}-\binom{8}{2}\right\)=364) है।

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(1) से (40) तक की संख्याओं में से (5) संख्याएं चुननी हैं जिनमें ठीक (2) संख्याएं (4) की गुणज हों। कितने तरीके हैं?

From numbers (1) to (40), (5) numbers are to be selected with exactly (2) numbers being multiples of (4). How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (122760)

Step 1

Concept

There are (10) multiples of (4) and (30) non-multiples. The ways are \(\binom{10}{2}\binom{30}{3}=122760\).

Step 2

Why this answer is correct

The correct answer is A. (122760). There are (10) multiples of (4) and (30) non-multiples. The ways are \(\binom{10}{2}\binom{30}{3}=122760\).

Step 3

Exam Tip

(4) की (10) गुणज और (30) गैर-गुणज संख्याएं हैं। तरीके \(\binom{10}{2}\binom{30}{3}=122760\) हैं।

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(13) खिलाड़ियों में से (6) खिलाड़ी चुनने हैं। (3) विशेष खिलाड़ियों में से कम से कम (1) और अधिकतम (2) चुने जाएं। कितने तरीके हैं?

From (13) players (6) players are to be selected. At least (1) and at most (2) of (3) special players are selected. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (1470)

Step 1

Concept

The number of special players can be (1) or (2). The total is \(\binom{3}{1}\binom{10}{5}+\binom{3}{2}\binom{10}{4}=1470\).

Step 2

Why this answer is correct

The correct answer is A. (1470). The number of special players can be (1) or (2). The total is \(\binom{3}{1}\binom{10}{5}+\binom{3}{2}\binom{10}{4}=1470\).

Step 3

Exam Tip

विशेष खिलाड़ी (1) या (2) चुने जा सकते हैं। कुल \(\binom{3}{1}\binom{10}{5}+\binom{3}{2}\binom{10}{4}=1470\) है।

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(10) विज्ञान और (6) कला विद्यार्थियों में से (6) विद्यार्थी चुनने हैं जिनमें विज्ञान विद्यार्थियों की संख्या कला विद्यार्थियों की संख्या से ठीक (2) अधिक हो। कितने तरीके हैं?

From (10) science and (6) arts students (6) students are to be selected such that the number of science students is exactly (2) more than the number of arts students. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (3150)

Step 1

Concept

If science students are (x) and arts students are (y), then (x+y=6) and (x=y+2), so (x=4), (y=2). The ways are \(\binom{10}{4}\binom{6}{2}=3150\).

Step 2

Why this answer is correct

The correct answer is B. (3150). If science students are (x) and arts students are (y), then (x+y=6) and (x=y+2), so (x=4), (y=2). The ways are \(\binom{10}{4}\binom{6}{2}=3150\).

Step 3

Exam Tip

यदि विज्ञान (x) और कला (y) हों तो (x+y=6) और (x=y+2), इसलिए (x=4), (y=2)। तरीके \(\binom{10}{4}\binom{6}{2}=3150\) हैं।

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(16) बिंदुओं में से (6) बिंदु एक सीध में हैं और अन्य (4) बिंदु भी एक अलग सीध में हैं। कोई अन्य (3) बिंदु एक सीध में नहीं हैं। कितने त्रिभुज बनेंगे?

Among (16) points (6) points are collinear and another (4) points are collinear on a different line. No other (3) points are collinear. How many triangles can be formed?

Explanation opens after your attempt
Correct Answer

A. (536)

Step 1

Concept

Total triples are \(\binom{16}{3}=560\). Failed triples are \(\binom{6}{3}+\binom{4}{3}=24\), so (560-24=536).

Step 2

Why this answer is correct

The correct answer is A. (536). Total triples are \(\binom{16}{3}=560\). Failed triples are \(\binom{6}{3}+\binom{4}{3}=24\), so (560-24=536).

Step 3

Exam Tip

कुल \(\binom{16}{3}=560\) त्रिक हैं। असफल त्रिक \(\binom{6}{3}+\binom{4}{3}=24\) हैं इसलिए (560-24=536) है।

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एक लीग में (14) टीमें हैं और हर दो टीमों के बीच (3) मैच खेले जाते हैं। कुल मैच कितने होंगे?

A league has (14) teams and (3) matches are played between every pair of teams. How many matches will be played?

Explanation opens after your attempt
Correct Answer

C. (273)

Step 1

Concept

The pairs of teams are \(\binom{14}{2}=91\). With (3) matches for each pair the total is (273).

Step 2

Why this answer is correct

The correct answer is C. (273). The pairs of teams are \(\binom{14}{2}=91\). With (3) matches for each pair the total is (273).

Step 3

Exam Tip

टीमों की जोड़ियां \(\binom{14}{2}=91\) हैं। हर जोड़ी के (3) मैच होने से कुल (273) मैच होंगे।

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(11) व्यक्तियों में से (5) व्यक्तियों की समिति बनानी है और (2) विशेष व्यक्ति साथ में नहीं चुने जा सकते। कितने तरीके होंगे?

A committee of (5) persons is to be formed from (11) persons and (2) special persons cannot be selected together. How many ways are possible?

Explanation opens after your attempt
Correct Answer

B. (378)

Step 1

Concept

Total ways are \(\binom{11}{5}=462\) and ways with both special persons are \(\binom{9}{3}=84\). Hence (462-84=378).

Step 2

Why this answer is correct

The correct answer is B. (378). Total ways are \(\binom{11}{5}=462\) and ways with both special persons are \(\binom{9}{3}=84\). Hence (462-84=378).

Step 3

Exam Tip

कुल \(\binom{11}{5}=462\) हैं और दोनों विशेष साथ हों तो \(\binom{9}{3}=84\) हैं। इसलिए (462-84=378) तरीके हैं।

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(10) गणित और (7) भौतिकी पुस्तकों में से (6) पुस्तकें चुननी हैं जिनमें कम से कम (3) गणित पुस्तकें हों। कितने तरीके हैं?

From (10) mathematics and (7) physics books (6) books are to be selected with at least (3) mathematics books. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (10584)

Step 1

Concept

The cases are (3), (4), (5), and (6) mathematics books. Their correct sum is (10584).

Step 2

Why this answer is correct

The correct answer is C. (10584). The cases are (3), (4), (5), and (6) mathematics books. Their correct sum is (10584).

Step 3

Exam Tip

मामले (3), (4), (5) और (6) गणित पुस्तकों के हैं। इनका सही योग (10584) है।

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(8) अंग्रेजी और (9) हिंदी पुस्तकों में से (5) पुस्तकें चुननी हैं जिनमें दोनों भाषाओं की पुस्तकें हों। कितने तरीके हैं?

From (8) English and (9) Hindi books (5) books are to be selected with books from both languages. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (6006)

Step 1

Concept

Total ways are \(\binom{17}{5}=6188\). Removing only English \(\binom{8}{5}=56\) and only Hindi \(\binom{9}{5}=126\) gives (6006).

Step 2

Why this answer is correct

The correct answer is A. (6006). Total ways are \(\binom{17}{5}=6188\). Removing only English \(\binom{8}{5}=56\) and only Hindi \(\binom{9}{5}=126\) gives (6006).

Step 3

Exam Tip

कुल \(\binom{17}{5}=6188\) हैं। केवल अंग्रेजी \(\binom{8}{5}=56\) और केवल हिंदी \(\binom{9}{5}=126\) हटाने पर (6006) मिलते हैं।

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(9) डॉक्टरों और (7) नर्सों में से (6) लोगों की टीम बनानी है जिसमें डॉक्टरों की संख्या नर्सों से अधिक हो। कितने तरीके हैं?

From (9) doctors and (7) nurses a team of (6) people is to be formed with more doctors than nurses. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (3612)

Step 1

Concept

The number of doctors will be (4), (5), or (6). The total is \(\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612\).

Step 2

Why this answer is correct

The correct answer is C. (3612). The number of doctors will be (4), (5), or (6). The total is \(\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612\).

Step 3

Exam Tip

डॉक्टरों की संख्या (4), (5) या (6) होगी। कुल \(\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612\) है।

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(13) अलग-अलग उपहारों में से कम से कम (3) और अधिकतम (5) उपहार चुनने के कितने तरीके हैं?

In how many ways can at least (3) and at most (5) gifts be selected from (13) different gifts?

Explanation opens after your attempt
Correct Answer

C. (2288)

Step 1

Concept

The selection can be of (3), (4), or (5) gifts. The total is \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}=2288\).

Step 2

Why this answer is correct

The correct answer is C. (2288). The selection can be of (3), (4), or (5) gifts. The total is \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}=2288\).

Step 3

Exam Tip

चयन (3), (4) या (5) उपहारों का होगा। कुल \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}=2288\) है।

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(11) रंगों में से (5) रंग चुनने हैं लेकिन एक निश्चित रंग अवश्य हो और दूसरा निश्चित रंग न हो। कितने तरीके हैं?

From (11) colors (5) colors are to be selected with one fixed color included and another fixed color excluded. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (126)

Step 1

Concept

One color is fixed and one is removed so choose the remaining (4) colors from (9). The ways are \(\binom{9}{4}=126\).

Step 2

Why this answer is correct

The correct answer is B. (126). One color is fixed and one is removed so choose the remaining (4) colors from (9). The ways are \(\binom{9}{4}=126\).

Step 3

Exam Tip

एक रंग तय है और एक हट गया है इसलिए बाकी (4) रंग (9) में से चुनेंगे। तरीके \(\binom{9}{4}=126\) हैं।

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(12) विद्यार्थियों में से (6) विद्यार्थियों की टीम बनानी है। (5) विशेष विद्यार्थियों में से ठीक (3) शामिल होने चाहिए। कितने तरीके हैं?

A team of (6) students is to be formed from (12) students. Exactly (3) of (5) special students must be included. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (350)

Step 1

Concept

Choose (3) from (5) special students and (3) from the remaining (7). The ways are \(\binom{5}{3}\binom{7}{3}=350\).

Step 2

Why this answer is correct

The correct answer is C. (350). Choose (3) from (5) special students and (3) from the remaining (7). The ways are \(\binom{5}{3}\binom{7}{3}=350\).

Step 3

Exam Tip

(5) विशेष में से (3) और बाकी (7) में से (3) चुनेंगे। तरीके \(\binom{5}{3}\binom{7}{3}=350\) हैं।

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(15) छात्रों में से (6) छात्रों का चयन करना है। (5) विशेष छात्रों में से कम से कम (2) शामिल हों। कितने तरीके हैं?

From (15) students (6) students are to be selected. At least (2) of (5) special students must be included. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (3535)

Step 1

Concept

Total ways are \(\binom{15}{6}=5005\). Removing selections with (0) and (1) special student gives \(5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535\).

Step 2

Why this answer is correct

The correct answer is A. (3535). Total ways are \(\binom{15}{6}=5005\). Removing selections with (0) and (1) special student gives \(5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535\).

Step 3

Exam Tip

कुल \(\binom{15}{6}=5005\) हैं। (0) और (1) विशेष वाले चयन हटाने पर \(5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535\) है।

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(10) पुरुषों और (8) महिलाओं में से (6) व्यक्तियों की समिति बनानी है जिसमें ठीक (3) पुरुष हों। कितने तरीके हैं?

From (10) men and (8) women a committee of (6) persons is to be formed with exactly (3) men. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (6720)

Step 1

Concept

Exactly (3) men and (3) women are needed. The ways are \(\binom{10}{3}\binom{8}{3}=6720\).

Step 2

Why this answer is correct

The correct answer is C. (6720). Exactly (3) men and (3) women are needed. The ways are \(\binom{10}{3}\binom{8}{3}=6720\).

Step 3

Exam Tip

ठीक (3) पुरुष और (3) महिलाएं चाहिए। तरीके \(\binom{10}{3}\binom{8}{3}=6720\) हैं।

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(9) अलग-अलग पेन और (8) अलग-अलग पेंसिलों में से (6) वस्तुएं चुननी हैं जिनमें कम से कम (2) पेन और कम से कम (2) पेंसिल हों। कितने तरीके हैं?

From (9) different pens and (8) different pencils (6) 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

C. (10752)

Step 1

Concept

The number of pens can be (2), (3), or (4). The total is \(\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752\).

Step 2

Why this answer is correct

The correct answer is C. (10752). The number of pens can be (2), (3), or (4). The total is \(\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752\).

Step 3

Exam Tip

पेन (2), (3) या (4) हो सकते हैं। कुल \(\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752\) है।

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यदि \(\binom{n}{2}=120\) है तो (n) का मान क्या है?

If \(\binom{n}{2}=120\), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

C. (16)

Step 1

Concept

\(\binom{16}{2}=120\). Therefore (n=16) is correct.

Step 2

Why this answer is correct

The correct answer is C. (16). \(\binom{16}{2}=120\). Therefore (n=16) is correct.

Step 3

Exam Tip

\(\binom{16}{2}=120\) होता है। इसलिए (n=16) सही है।

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यदि \(\binom{n}{5}=\binom{n}{9}\) है तो (n) का मान क्या होगा?

If \(\binom{n}{5}=\binom{n}{9}\), what will be the value of (n)?

Explanation opens after your attempt
Correct Answer

C. (14)

Step 1

Concept

Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (5+9=14).

Step 2

Why this answer is correct

The correct answer is C. (14). Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (5+9=14).

Step 3

Exam Tip

\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (5+9=14) है।

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यदि \(\binom{n}{1}+\binom{n}{2}=78\) है तो (n) का मान क्या है?

If \(\binom{n}{1}+\binom{n}{2}=78\), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

C. (12)

Step 1

Concept

It gives (n+\frac{n(n-1)}{2}=78). Putting (n=12) gives (12+66=78).

Step 2

Why this answer is correct

The correct answer is C. (12). It gives (n+\frac{n(n-1)}{2}=78). Putting (n=12) gives (12+66=78).

Step 3

Exam Tip

यह (n+\frac{n(n-1)}{2}=78) देता है। (n=12) रखने पर (12+66=78) मिलता है।

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\(\binom{11}{4}+\binom{11}{5}\) पास्कल पहचान से किसके बराबर है?

Using Pascal's identity \(\binom{11}{4}+\binom{11}{5}\) is equal to which expression?

Explanation opens after your attempt
Correct Answer

B. \(\binom{12}{5}\)

Step 1

Concept

By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{12}{5}\).

Step 2

Why this answer is correct

The correct answer is B. \(\binom{12}{5}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{12}{5}\).

Step 3

Exam Tip

पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{12}{5}\) है।

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\(\binom{13}{6}-\binom{12}{6}\) का मान क्या है?

What is the value of \(\binom{13}{6}-\binom{12}{6}\)?

Explanation opens after your attempt
Correct Answer

C. (792)

Step 1

Concept

\(\binom{13}{6}=1716\) and \(\binom{12}{6}=924\). The difference is (792).

Step 2

Why this answer is correct

The correct answer is C. (792). \(\binom{13}{6}=1716\) and \(\binom{12}{6}=924\). The difference is (792).

Step 3

Exam Tip

\(\binom{13}{6}=1716\) और \(\binom{12}{6}=924\) हैं। अंतर (792) है।

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

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

Explanation opens after your attempt
Correct Answer

B. (84)

Step 1

Concept

The elements (1) and (2) are fixed and (3) is excluded. The remaining (3) elements are chosen from (9), so \(\binom{9}{3}=84\).

Step 2

Why this answer is correct

The correct answer is B. (84). The elements (1) and (2) are fixed and (3) is excluded. The remaining (3) elements are chosen from (9), so \(\binom{9}{3}=84\).

Step 3

Exam Tip

(1) और (2) तय हैं और (3) हट गया है। बाकी (3) तत्व (9) में से चुने जाएंगे इसलिए \(\binom{9}{3}=84\) है।

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

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

Explanation opens after your attempt
Correct Answer

C. (336)

Step 1

Concept

Total subsets are \(\binom{11}{6}=462\) and those containing both (4), (5) are \(\binom{9}{4}=126\). Hence (462-126=336).

Step 2

Why this answer is correct

The correct answer is C. (336). Total subsets are \(\binom{11}{6}=462\) and those containing both (4), (5) are \(\binom{9}{4}=126\). Hence (462-126=336).

Step 3

Exam Tip

कुल \(\binom{11}{6}=462\) हैं और (4), (5) दोनों हों तो \(\binom{9}{4}=126\) हैं। इसलिए (462-126=336) है।

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(1) से (30) तक की संख्याओं में से (3) ऐसी संख्याएं चुननी हैं जो (3) की गुणज हों और (2) ऐसी संख्याएं जो (3) की गुणज न हों। कितने तरीके हैं?

From numbers (1) to (30), (3) numbers that are multiples of (3) and (2) numbers that are not multiples of (3) are to be selected. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (22800)

Step 1

Concept

There are (10) multiples of (3) and (20) non-multiples. The ways are \(\binom{10}{3}\binom{20}{2}=22800\).

Step 2

Why this answer is correct

The correct answer is A. (22800). There are (10) multiples of (3) and (20) non-multiples. The ways are \(\binom{10}{3}\binom{20}{2}=22800\).

Step 3

Exam Tip

(3) की (10) गुणज और (20) गैर-गुणज संख्याएं हैं। तरीके \(\binom{10}{3}\binom{20}{2}=22800\) हैं।

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(1) से (25) तक की संख्याओं में से (3) विषम और (2) सम संख्या चुनने के कितने तरीके हैं?

From numbers (1) to (25), how many ways are there to choose (3) odd and (2) even numbers?

Explanation opens after your attempt
Correct Answer

C. (18876)

Step 1

Concept

There are (13) odd numbers and (12) even numbers. The ways are \(\binom{13}{3}\binom{12}{2}=18876\).

Step 2

Why this answer is correct

The correct answer is C. (18876). There are (13) odd numbers and (12) even numbers. The ways are \(\binom{13}{3}\binom{12}{2}=18876\).

Step 3

Exam Tip

विषम संख्याएं (13) और सम संख्याएं (12) हैं। तरीके \(\binom{13}{3}\binom{12}{2}=18876\) हैं।

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(18) विद्यार्थियों में से (7) चुनने हैं ताकि (5) विशेष विद्यार्थियों में से कोई भी शामिल न हो। कितने तरीके हैं?

From (18) students (7) are to be selected so that none of (5) special students is included. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (1716)

Step 1

Concept

After removing (5) special students (13) students remain. Hence there are \(\binom{13}{7}=1716\) ways.

Step 2

Why this answer is correct

The correct answer is B. (1716). After removing (5) special students (13) students remain. Hence there are \(\binom{13}{7}=1716\) ways.

Step 3

Exam Tip

(5) विशेष विद्यार्थियों को हटाने पर (13) विद्यार्थी बचते हैं। इसलिए \(\binom{13}{7}=1716\) तरीके हैं।

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(9) शिक्षकों और (11) छात्रों में से (6) लोगों का समूह बनाना है जिसमें कम से कम (3) शिक्षक हों। कितने तरीके हैं?

From (9) teachers and (11) students a group of (6) people is to be formed with at least (3) teachers. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (22260)

Step 1

Concept

The number of teachers can be (3), (4), (5), or (6). The sum of all these cases is (22260).

Step 2

Why this answer is correct

The correct answer is C. (22260). The number of teachers can be (3), (4), (5), or (6). The sum of all these cases is (22260).

Step 3

Exam Tip

शिक्षक (3), (4), (5) या (6) हो सकते हैं। इन सभी मामलों का योग (22260) है।

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(8) सफेद और (9) काली गेंदों में से (6) गेंदें चुननी हैं जिनमें सभी गेंदों का रंग समान न हो। कितने तरीके हैं?

From (8) white and (9) black balls (6) 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. (12264)

Step 1

Concept

Total ways are \(\binom{17}{6}=12376\). Removing all-white \(\binom{8}{6}=28\) and all-black \(\binom{9}{6}=84\) gives (12264).

Step 2

Why this answer is correct

The correct answer is B. (12264). Total ways are \(\binom{17}{6}=12376\). Removing all-white \(\binom{8}{6}=28\) and all-black \(\binom{9}{6}=84\) gives (12264).

Step 3

Exam Tip

कुल \(\binom{17}{6}=12376\) हैं। सभी सफेद \(\binom{8}{6}=28\) और सभी काली \(\binom{9}{6}=84\) हटाने पर (12264) मिलते हैं।

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(13) उम्मीदवारों में से (5) पुरस्कार विजेताओं का चयन करना है और सभी पुरस्कार समान हैं। कितने तरीके हैं?

From (13) candidates (5) prize winners are to be selected and all prizes are identical. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (1287)

Step 1

Concept

The prizes are identical so only selection is needed. The number of ways is \(\binom{13}{5}=1287\).

Step 2

Why this answer is correct

The correct answer is A. (1287). The prizes are identical so only selection is needed. The number of ways is \(\binom{13}{5}=1287\).

Step 3

Exam Tip

पुरस्कार समान हैं इसलिए केवल चयन होगा। तरीकों की संख्या \(\binom{13}{5}=1287\) है।

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(11) मित्रों में से (5) को यात्रा के लिए चुनना है और (3) विशेष मित्रों में से कम से कम एक जाना चाहिए। कितने तरीके हैं?

From (11) friends (5) are to be selected for a trip and at least one of (3) special friends must go. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (406)

Step 1

Concept

Total ways are \(\binom{11}{5}=462\) and if no special friend goes then \(\binom{8}{5}=56\). Hence (462-56=406).

Step 2

Why this answer is correct

The correct answer is B. (406). Total ways are \(\binom{11}{5}=462\) and if no special friend goes then \(\binom{8}{5}=56\). Hence (462-56=406).

Step 3

Exam Tip

कुल \(\binom{11}{5}=462\) हैं और कोई विशेष मित्र न जाए तो \(\binom{8}{5}=56\) हैं। इसलिए (462-56=406) है।

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(12) फलों में से (5) फल चुनने हैं लेकिन (5) विशेष फलों में से ठीक (2) चुने जाएं। कितने तरीके हैं?

From (12) fruits (5) fruits are to be selected but exactly (2) of (5) special fruits are chosen. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (350)

Step 1

Concept

Choose (2) from the special fruits and (3) from the remaining (7). The ways are \(\binom{5}{2}\binom{7}{3}=350\).

Step 2

Why this answer is correct

The correct answer is C. (350). Choose (2) from the special fruits and (3) from the remaining (7). The ways are \(\binom{5}{2}\binom{7}{3}=350\).

Step 3

Exam Tip

विशेष फलों में से (2) और बाकी (7) में से (3) चुनेंगे। तरीके \(\binom{5}{2}\binom{7}{3}=350\) हैं।

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(7) लाल, (8) नीली और (9) हरी गेंदों में से (2) लाल, (2) नीली और (1) हरी गेंद चुनने के कितने तरीके हैं?

From (7) red, (8) blue, and (9) green balls, how many ways are there to choose (2) red, (2) blue, and (1) green ball?

Explanation opens after your attempt
Correct Answer

C. (5292)

Step 1

Concept

Each color condition is counted separately. The ways are \(\binom{7}{2}\binom{8}{2}\binom{9}{1}=5292\).

Step 2

Why this answer is correct

The correct answer is C. (5292). Each color condition is counted separately. The ways are \(\binom{7}{2}\binom{8}{2}\binom{9}{1}=5292\).

Step 3

Exam Tip

हर रंग की शर्त अलग गिनी जाएगी। तरीके \(\binom{7}{2}\binom{8}{2}\binom{9}{1}=5292\) हैं।

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(10) अलग-अलग अक्षरों में से (5) अक्षर चुनने हैं जिनमें (a) हो, (b) न हो और (c,d) में से ठीक एक अक्षर हो। कितने तरीके हैं?

From (10) distinct letters (5) 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

B. (40)

Step 1

Concept

The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (3) from the remaining (6), so \(\binom{2}{1}\binom{6}{3}=40\).

Step 2

Why this answer is correct

The correct answer is B. (40). The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (3) from the remaining (6), so \(\binom{2}{1}\binom{6}{3}=40\).

Step 3

Exam Tip

(a) तय है और (b) हट गया है। (c,d) में से (1) और शेष (6) में से (3) चुनेंगे इसलिए \(\binom{2}{1}\binom{6}{3}=40\) है।

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(9) विषयों में से (4) विषय चुनने हैं। रसायन अवश्य चुना जाए लेकिन गणित और भौतिकी दोनों साथ में न चुने जाएं। कितने तरीके हैं?

From (9) subjects (4) subjects are to be selected. Chemistry must be selected but mathematics and physics must not both be selected together. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (50)

Step 1

Concept

Chemistry is fixed so choose the remaining (3) subjects from (8). Removing (6) selections containing both mathematics and physics leaves (50) ways.

Step 2

Why this answer is correct

The correct answer is B. (50). Chemistry is fixed so choose the remaining (3) subjects from (8). Removing (6) selections containing both mathematics and physics leaves (50) ways.

Step 3

Exam Tip

रसायन तय है इसलिए बाकी (3) विषय (8) में से चुनेंगे। गणित और भौतिकी दोनों साथ वाले (6) चयन हटाने पर (50) तरीके बचते हैं।

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(15) संख्याओं में से (6) संख्याएं चुननी हैं जिनमें (2) निश्चित संख्याएं शामिल हों और (2) निश्चित संख्याएं शामिल न हों। कितने तरीके हैं?

From (15) numbers (6) numbers are to be selected with (2) fixed numbers included and (2) fixed numbers excluded. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (330)

Step 1

Concept

The (2) numbers are fixed and (2) are excluded. The remaining (4) numbers are chosen from (11) in \(\binom{11}{4}=330\) ways.

Step 2

Why this answer is correct

The correct answer is B. (330). The (2) numbers are fixed and (2) are excluded. The remaining (4) numbers are chosen from (11) in \(\binom{11}{4}=330\) ways.

Step 3

Exam Tip

(2) संख्याएं तय हैं और (2) हट गई हैं। बाकी (4) संख्याएं (11) में से \(\binom{11}{4}=330\) तरीकों से चुनी जाएंगी।

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(16) बिंदुओं से त्रिभुज बनाने हैं। यदि (8) बिंदु एक सीध में हैं तो केवल उन्हीं (8) बिंदुओं से बनने वाले असफल चयन कितने हैं?

Triangles are to be formed from (16) points. If (8) points are collinear then how many failed selections come only from those (8) points?

Explanation opens after your attempt
Correct Answer

B. \(\binom{8}{3}\)

Step 1

Concept

For a failed triangle (3) points are chosen from the (8) collinear points. So the number is \(\binom{8}{3}\).

Step 2

Why this answer is correct

The correct answer is B. \(\binom{8}{3}\). For a failed triangle (3) points are chosen from the (8) collinear points. So the number is \(\binom{8}{3}\).

Step 3

Exam Tip

असफल त्रिभुज के लिए (8) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{8}{3}\) है।

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(12) कुर्सियों में से (6) कुर्सियां चुननी हैं और (4) खराब कुर्सियों में से कोई न चुनी जाए। कितने तरीके हैं?

From (12) chairs (6) chairs are to be selected and none of (4) broken chairs is selected. How many ways are there?

Explanation opens after your attempt
Correct Answer

A. (28)

Step 1

Concept

After removing (4) broken chairs (8) good chairs remain. The ways to choose (6) are \(\binom{8}{6}=28\).

Step 2

Why this answer is correct

The correct answer is A. (28). After removing (4) broken chairs (8) good chairs remain. The ways to choose (6) are \(\binom{8}{6}=28\).

Step 3

Exam Tip

(4) खराब कुर्सियां हटाने पर (8) अच्छी कुर्सियां बचती हैं। (6) चुनने के तरीके \(\binom{8}{6}=28\) हैं।

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(12) विद्यार्थियों में से (6) की टीम बनानी है जिसमें (2) विशेष विद्यार्थी दोनों शामिल हों या दोनों बाहर हों। कितने तरीके हैं?

From (12) students a team of (6) is to be formed in which (2) special students are either both included or both excluded. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (420)

Step 1

Concept

If both are included there are \(\binom{10}{4}=210\) ways and if both are excluded there are \(\binom{10}{6}=210\) ways. The total is (420).

Step 2

Why this answer is correct

The correct answer is B. (420). If both are included there are \(\binom{10}{4}=210\) ways and if both are excluded there are \(\binom{10}{6}=210\) ways. The total is (420).

Step 3

Exam Tip

दोनों शामिल हों तो \(\binom{10}{4}=210\) और दोनों बाहर हों तो \(\binom{10}{6}=210\) तरीके हैं। कुल (420) है।

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(12) प्रश्नों में से (7) प्रश्न चुनने हैं और अंतिम (5) प्रश्नों में से कम से कम (3) प्रश्न चुनने हैं। कितने चयन होंगे?

From (12) questions (7) are to be selected and at least (3) of the last (5) questions must be selected. How many selections are there?

Explanation opens after your attempt
Correct Answer

C. (546)

Step 1

Concept

You can choose (3), (4), or (5) from the last (5) questions. The total is \(\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=546\).

Step 2

Why this answer is correct

The correct answer is C. (546). You can choose (3), (4), or (5) from the last (5) questions. The total is \(\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=546\).

Step 3

Exam Tip

अंतिम (5) में से (3), (4) या (5) प्रश्न चुने जा सकते हैं। कुल \(\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=546\) है।

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(8) वरिष्ठ और (10) कनिष्ठ कर्मचारियों में से (6) लोगों की समिति बनानी है जिसमें वरिष्ठों की संख्या कनिष्ठों से कम हो। कितने तरीके हैं?

From (8) senior and (10) junior employees a committee of (6) is to be formed with fewer seniors than juniors. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (8106)

Step 1

Concept

The number of seniors can be (0), (1), or (2). The total is \(\binom{8}{0}\binom{10}{6}+\binom{8}{1}\binom{10}{5}+\binom{8}{2}\binom{10}{4}=8106\).

Step 2

Why this answer is correct

The correct answer is C. (8106). The number of seniors can be (0), (1), or (2). The total is \(\binom{8}{0}\binom{10}{6}+\binom{8}{1}\binom{10}{5}+\binom{8}{2}\binom{10}{4}=8106\).

Step 3

Exam Tip

वरिष्ठों की संख्या (0), (1) या (2) हो सकती है। कुल \(\binom{8}{0}\binom{10}{6}+\binom{8}{1}\binom{10}{5}+\binom{8}{2}\binom{10}{4}=8106\) है।

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(13) अलग-अलग कार्डों में से (5) कार्ड चुनने हैं जिनमें दो निश्चित कार्डों में से ठीक एक कार्ड हो। कितने तरीके हैं?

From (13) distinct cards (5) cards are to be selected containing exactly one of two fixed cards. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (660)

Step 1

Concept

Choose (1) of the two fixed cards and (4) cards from the remaining (11). The ways are \(\binom{2}{1}\binom{11}{4}=660\).

Step 2

Why this answer is correct

The correct answer is B. (660). Choose (1) of the two fixed cards and (4) cards from the remaining (11). The ways are \(\binom{2}{1}\binom{11}{4}=660\).

Step 3

Exam Tip

दो निश्चित कार्डों में से (1) चुनें और बाकी (4) कार्ड (11) में से चुनें। तरीके \(\binom{2}{1}\binom{11}{4}=660\) हैं।

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(12) अलग-अलग सिक्कों में से सम संख्या में सिक्के चुनने के कितने तरीके हैं?

In how many ways can an even number of coins be selected from (12) different coins?

Explanation opens after your attempt
Correct Answer

C. (2048)

Step 1

Concept

The number of even selections is \(2^{12-1}=2048\). Even and odd selections are equal.

Step 2

Why this answer is correct

The correct answer is C. (2048). The number of even selections is \(2^{12-1}=2048\). Even and odd selections are equal.

Step 3

Exam Tip

सम चयन की संख्या \(2^{12-1}=2048\) होती है। सम और विषम चयन बराबर होते हैं।

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(11) अलग-अलग खिलौनों में से विषम संख्या में खिलौने चुनने के कितने तरीके हैं?

In how many ways can an odd number of toys be selected from (11) different toys?

Explanation opens after your attempt
Correct Answer

C. (1024)

Step 1

Concept

The number of odd selections is \(2^{11-1}=1024\). Remember that even and odd selections are equal in such questions.

Step 2

Why this answer is correct

The correct answer is C. (1024). The number of odd selections is \(2^{11-1}=1024\). Remember that even and odd selections are equal in such questions.

Step 3

Exam Tip

विषम चयन की संख्या \(2^{11-1}=1024\) होती है। परीक्षा में सम और विषम चयन की संख्या बराबर याद रखें।

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(11) खिलाड़ियों में से (5) खिलाड़ियों का चयन करना है। एक कप्तान पहले से तय है और उसे चुना नहीं जाना है लेकिन उपकप्तान अवश्य चुना जाना है। कितने तरीके हैं?

From (11) players (5) players are to be selected. One captain is already fixed and must not be selected but the vice-captain must be selected. How many ways are there?

Explanation opens after your attempt
Correct Answer

B. (126)

Step 1

Concept

The captain is excluded and the vice-captain is fixed. The remaining (4) players are chosen from (9), so \(\binom{9}{4}=126\).

Step 2

Why this answer is correct

The correct answer is B. (126). The captain is excluded and the vice-captain is fixed. The remaining (4) players are chosen from (9), so \(\binom{9}{4}=126\).

Step 3

Exam Tip

कप्तान हट गया और उपकप्तान तय है। बाकी (4) खिलाड़ी (9) में से चुने जाएंगे इसलिए \(\binom{9}{4}=126\) है।

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(16) वस्तुओं में से (6) वस्तुएं चुननी हैं और (5) विशेष वस्तुओं में से अधिकतम (2) चुनी जाएं। कितने तरीके हैं?

From (16) objects (6) objects are to be selected and at most (2) of (5) special objects are chosen. How many ways are there?

Explanation opens after your attempt
Correct Answer

C. (6072)

Step 1

Concept

The number of special objects can be (0), (1), or (2). The total is \(\binom{5}{0}\binom{11}{6}+\binom{5}{1}\binom{11}{5}+\binom{5}{2}\binom{11}{4}=6072\).

Step 2

Why this answer is correct

The correct answer is C. (6072). The number of special objects can be (0), (1), or (2). The total is \(\binom{5}{0}\binom{11}{6}+\binom{5}{1}\binom{11}{5}+\binom{5}{2}\binom{11}{4}=6072\).

Step 3

Exam Tip

विशेष वस्तुएं (0), (1) या (2) चुनी जा सकती हैं। कुल \(\binom{5}{0}\binom{11}{6}+\binom{5}{1}\binom{11}{5}+\binom{5}{2}\binom{11}{4}=6072\) है।

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FAQs

Class 11 Mathematics Quiz FAQs

How many questions are in this quiz?

This level is designed for 50 active questions. Currently 50 questions are available for the selected class and difficulty.

Is there a timer in this quiz?

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Can I open each question separately?

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