Class 11 Mathematics Medium Quiz

Mathematics Medium Class 11 Level 59

Level 59 • 50/50 questions • 35 seconds per question.

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Question 1/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(9) अलग-अलग पेनों को एक पंक्ति में कितने तरीकों से रखा जा सकता है?

In how many ways can (9) distinct pens be arranged in a row?

Explanation opens after your attempt
Correct Answer

C. (362880)

Step 1

Concept

The arrangement of (9) distinct objects is (9!). In exams, use factorial directly when all objects are distinct.

Step 2

Why this answer is correct

The correct answer is C. (362880). The arrangement of (9) distinct objects is (9!). In exams, use factorial directly when all objects are distinct.

Step 3

Exam Tip

अलग-अलग (9) वस्तुओं की व्यवस्था (9!) होती है। परीक्षा में सभी वस्तुएं अलग हों तो सीधे factorial लगाएं।

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Question 2/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) विद्यार्थियों में से चार अलग-अलग पदों के लिए चयन कितने तरीकों से होगा?

In how many ways can (4) different posts be filled from (8) students?

Explanation opens after your attempt
Correct Answer

B. (1680)

Step 1

Concept

Order matters for different posts, so \(^{8}P_4=1680\). In exams, do not use combination when posts are different.

Step 2

Why this answer is correct

The correct answer is B. (1680). Order matters for different posts, so \(^{8}P_4=1680\). In exams, do not use combination when posts are different.

Step 3

Exam Tip

अलग पदों में क्रम महत्वपूर्ण है, इसलिए \(^{8}P_4=1680\)। परीक्षा में पद अलग हों तो combination नहीं लगाएं।

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Question 3/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (0,1,2,3,4,5,6) से बिना पुनरावृत्ति (3) अंकों की कितनी संख्याएं बनेंगी?

How many (3)-digit numbers can be formed from (0,1,2,3,4,5,6) without repetition?

Explanation opens after your attempt
Correct Answer

D. (180)

Step 1

Concept

The hundreds place cannot be (0), so \(6\cdot6\cdot5=180\). In exams, always check the zero restriction on the first digit.

Step 2

Why this answer is correct

The correct answer is D. (180). The hundreds place cannot be (0), so \(6\cdot6\cdot5=180\). In exams, always check the zero restriction on the first digit.

Step 3

Exam Tip

सैकड़े के स्थान पर (0) नहीं आएगा, इसलिए \(6\cdot6\cdot5=180\)। परीक्षा में पहले अंक की शून्य-शर्त हमेशा जांचें।

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Question 4/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (1,2,3,4,5,6,7) से बिना पुनरावृत्ति (5) अंकों की कितनी संख्याएं बनेंगी जो (5) पर समाप्त हों?

How many (5)-digit numbers can be formed from (1,2,3,4,5,6,7) without repetition and ending in (5)?

Explanation opens after your attempt
Correct Answer

A. (360)

Step 1

Concept

The last digit is fixed as (5), and the remaining (4) places are filled in \(^{6}P_4\) ways. So the answer is (360).

Step 2

Why this answer is correct

The correct answer is A. (360). The last digit is fixed as (5), and the remaining (4) places are filled in \(^{6}P_4\) ways. So the answer is (360).

Step 3

Exam Tip

अंतिम अंक (5) निश्चित है और बाकी (4) स्थान \(^{6}P_4\) तरीकों से भरेंगे। इसलिए उत्तर (360) है।

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Question 5/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

शब्द व्यवस्था के (8) अक्षरों में व अक्षर (2) बार आता है। अलग व्यवस्थाएं कितनी होंगी?

A word has (8) letters in which one letter is repeated (2) times. How many distinct arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (20160)

Step 1

Concept

One letter is repeated (2) times, so the number is \(\frac{8!}{2!}=20160\). In exams, divide by factorials of repeated letters.

Step 2

Why this answer is correct

The correct answer is B. (20160). One letter is repeated (2) times, so the number is \(\frac{8!}{2!}=20160\). In exams, divide by factorials of repeated letters.

Step 3

Exam Tip

एक अक्षर (2) बार समान है, इसलिए संख्या \(\frac{8!}{2!}=20160\) होगी। परीक्षा में समान अक्षरों से भाग दें।

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Question 6/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) लड़कियों और (3) लड़कों को एक पंक्ति में ऐसे कितने तरीकों से बैठाया जाए कि तीनों लड़के साथ बैठें?

In how many ways can (7) girls and (3) boys be seated in a row so that all three boys sit together?

Explanation opens after your attempt
Correct Answer

C. (241920)

Step 1

Concept

Treat the three boys as one block, then there are (8) units. The arrangement is \(8!\cdot3!=241920\).

Step 2

Why this answer is correct

The correct answer is C. (241920). Treat the three boys as one block, then there are (8) units. The arrangement is \(8!\cdot3!=241920\).

Step 3

Exam Tip

तीनों लड़कों को एक ब्लॉक मानें, तब कुल (8) इकाइयां हैं। व्यवस्था \(8!\cdot3!=241920\) होगी।

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Question 7/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) व्यक्तियों को गोल मेज पर कितने तरीकों से बैठाया जा सकता है?

In how many ways can (8) people be seated around a circular table?

Explanation opens after your attempt
Correct Answer

A. (5040)

Step 1

Concept

Rotations are considered the same around a circular table, so the number is ((8-1)!=5040). In exams, use ((n-1)!) for circular table seating.

Step 2

Why this answer is correct

The correct answer is A. (5040). Rotations are considered the same around a circular table, so the number is ((8-1)!=5040). In exams, use ((n-1)!) for circular table seating.

Step 3

Exam Tip

गोल मेज पर घुमाव समान माने जाते हैं, इसलिए संख्या ((8-1)!=5040) है। परीक्षा में circular table में ((n-1)!) लगाएं।

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Question 8/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग मोतियों की माला कितने तरीकों से बन सकती है यदि पलटना भी समान माना जाए?

In how many ways can a necklace be made using (6) distinct beads if flipping is also considered the same?

Explanation opens after your attempt
Correct Answer

B. (60)

Step 1

Concept

In a necklace, both rotation and reflection are the same, so (\frac{(6-1)!}{2}=60). In exams, distinguish necklace from circular seating.

Step 2

Why this answer is correct

The correct answer is B. (60). In a necklace, both rotation and reflection are the same, so (\frac{(6-1)!}{2}=60). In exams, distinguish necklace from circular seating.

Step 3

Exam Tip

माला में घुमाव और पलटना दोनों समान हैं, इसलिए (\frac{(6-1)!}{2}=60)। परीक्षा में necklace को circular seating से अलग पहचानें।

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Question 9/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(10) धावकों में से प्रथम और द्वितीय स्थान कितने तरीकों से तय हो सकते हैं?

In how many ways can first and second positions be decided among (10) runners?

Explanation opens after your attempt
Correct Answer

C. (90)

Step 1

Concept

Positions are ordered, so \(^{10}P_2=10\cdot9=90\). In exams, rank or position indicates permutation.

Step 2

Why this answer is correct

The correct answer is C. (90). Positions are ordered, so \(^{10}P_2=10\cdot9=90\). In exams, rank or position indicates permutation.

Step 3

Exam Tip

स्थान क्रमित हैं, इसलिए \(^{10}P_2=10\cdot9=90\)। परीक्षा में rank या position आने पर permutation लगाएं।

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Question 10/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (2,3,5,6,7,8) से बिना पुनरावृत्ति (4) अंकों की कितनी सम संख्याएं बनेंगी?

How many (4)-digit even numbers can be formed from (2,3,5,6,7,8) without repetition?

Explanation opens after your attempt
Correct Answer

A. (180)

Step 1

Concept

The units place has (3) choices from (2,6,8), and the rest have \(5\cdot4\cdot3\) ways. The total is (180).

Step 2

Why this answer is correct

The correct answer is A. (180). The units place has (3) choices from (2,6,8), and the rest have \(5\cdot4\cdot3\) ways. The total is (180).

Step 3

Exam Tip

इकाई स्थान पर (2,6,8) में से (3) विकल्प हैं और बाकी \(5\cdot4\cdot3\) तरीके हैं। कुल (180) है।

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Question 11/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (0,2,3,4,7,9) से बिना पुनरावृत्ति (4) अंकों की कितनी विषम संख्याएं बनेंगी?

How many (4)-digit odd numbers can be formed from (0,2,3,4,7,9) without repetition?

Explanation opens after your attempt
Correct Answer

B. (150)

Step 1

Concept

There are (3) odd choices for the units place, then (4) valid choices for the thousands place after excluding (0). The total is \(3\cdot4\cdot4\cdot3=144\).

Step 2

Why this answer is correct

The correct answer is B. (150). There are (3) odd choices for the units place, then (4) valid choices for the thousands place after excluding (0). The total is \(3\cdot4\cdot4\cdot3=144\).

Step 3

Exam Tip

इकाई स्थान पर (3) विषम विकल्प हैं, फिर हजार के स्थान पर (0) छोड़कर (4) विकल्प बचते हैं। कुल \(3\cdot4\cdot4\cdot3=144\) होता है।

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Question 12/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग गणित की और (4) अलग-अलग भौतिकी की पुस्तकों को ऐसे कितने तरीकों से रखें कि हर विषय की पुस्तकें साथ रहें?

In how many ways can (5) distinct mathematics books and (4) distinct physics books be arranged so that books of each subject remain together?

Explanation opens after your attempt
Correct Answer

B. (34560)

Step 1

Concept

Treat the two subjects as two blocks, so the arrangement is \(2!\cdot5!\cdot4!=5760\). In exams, arrange both the blocks and the items inside blocks.

Step 2

Why this answer is correct

The correct answer is B. (34560). Treat the two subjects as two blocks, so the arrangement is \(2!\cdot5!\cdot4!=5760\). In exams, arrange both the blocks and the items inside blocks.

Step 3

Exam Tip

दो विषयों को दो ब्लॉक मानें, इसलिए व्यवस्था \(2!\cdot5!\cdot4!=5760\) है। परीक्षा में ब्लॉक और ब्लॉक के अंदर दोनों की व्यवस्था करें।

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Question 13/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) अलग-अलग अक्षरों से (4) अक्षरों के गुप्त-कोड कितने बनेंगे यदि पुनरावृत्ति नहीं है?

How many (4)-letter codes can be formed from (7) distinct letters if repetition is not allowed?

Explanation opens after your attempt
Correct Answer

A. (840)

Step 1

Concept

Order matters in a code, so \(^{7}P_4=840\). In exams, think of permutation when forming codes.

Step 2

Why this answer is correct

The correct answer is A. (840). Order matters in a code, so \(^{7}P_4=840\). In exams, think of permutation when forming codes.

Step 3

Exam Tip

गुप्त-कोड में क्रम महत्वपूर्ण है, इसलिए \(^{7}P_4=840\)। परीक्षा में code बनने पर permutation सोचें।

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Question 14/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

शब्द परीक्षा के (7) अक्षरों में एक अक्षर (2) बार आता है। अलग व्यवस्थाएं कितनी होंगी?

A word has (7) letters with one letter repeated (2) times. How many distinct arrangements are possible?

Explanation opens after your attempt
Correct Answer

A. (2520)

Step 1

Concept

There are (7) letters and one letter is repeated (2) times, so \(\frac{7!}{2!}=2520\). In exams, count repeated letters carefully.

Step 2

Why this answer is correct

The correct answer is A. (2520). There are (7) letters and one letter is repeated (2) times, so \(\frac{7!}{2!}=2520\). In exams, count repeated letters carefully.

Step 3

Exam Tip

कुल अक्षर (7) हैं और एक अक्षर (2) बार समान है, इसलिए \(\frac{7!}{2!}=2520\)। परीक्षा में repeated letter को ध्यान से गिनें।

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Question 15/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) विद्यार्थियों को (8) अलग-अलग कुर्सियों पर कितने तरीकों से बैठाया जा सकता है?

In how many ways can (6) students be seated on (8) distinct chairs?

Explanation opens after your attempt
Correct Answer

A. (20160)

Step 1

Concept

The chairs are distinct and (6) students are seated, so \(^{8}P_6=20160\). In exams, permutation may apply even when some places remain empty.

Step 2

Why this answer is correct

The correct answer is A. (20160). The chairs are distinct and (6) students are seated, so \(^{8}P_6=20160\). In exams, permutation may apply even when some places remain empty.

Step 3

Exam Tip

कुर्सियां अलग हैं और (6) विद्यार्थी बैठेंगे, इसलिए \(^{8}P_6=20160\)। परीक्षा में खाली स्थानों के बावजूद permutation लग सकता है।

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Question 16/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(9) लोगों की पंक्ति में दो विशेष व्यक्ति हमेशा साथ रहें तो व्यवस्थाएं कितनी होंगी?

In a row of (9) people, how many arrangements are possible if two particular people always remain together?

Explanation opens after your attempt
Correct Answer

A. (80640)

Step 1

Concept

Treat the two particular people as one block, then \(8!\cdot2!=80640\). In exams, use the block method for together conditions.

Step 2

Why this answer is correct

The correct answer is A. (80640). Treat the two particular people as one block, then \(8!\cdot2!=80640\). In exams, use the block method for together conditions.

Step 3

Exam Tip

दो विशेष व्यक्तियों को एक ब्लॉक मानें, तब \(8!\cdot2!=80640\)। परीक्षा में साथ की शर्त में block method लगाएं।

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Question 17/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(9) लोगों की पंक्ति में दो विशेष व्यक्ति साथ न बैठें तो व्यवस्थाएं कितनी होंगी?

In a row of (9) people, how many arrangements are possible if two particular people do not sit together?

Explanation opens after your attempt
Correct Answer

A. (282240)

Step 1

Concept

Subtract the together cases \(8!\cdot2!\) from total (9!). The answer is (362880-80640=282240).

Step 2

Why this answer is correct

The correct answer is A. (282240). Subtract the together cases \(8!\cdot2!\) from total (9!). The answer is (362880-80640=282240).

Step 3

Exam Tip

कुल (9!) व्यवस्थाओं में से साथ बैठने वाली \(8!\cdot2!\) घटाएं। उत्तर (362880-80640=282240) है।

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Question 18/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(4) पुरुष और (4) महिलाएं एक पंक्ति में वैकल्पिक रूप से कितने तरीकों से बैठ सकते हैं?

In how many ways can (4) men and (4) women sit alternately in a row?

Explanation opens after your attempt
Correct Answer

B. (1152)

Step 1

Concept

There are two possible patterns and each has \(4!\cdot4!\) arrangements. The total is \(2\cdot4!\cdot4!=1152\).

Step 2

Why this answer is correct

The correct answer is B. (1152). There are two possible patterns and each has \(4!\cdot4!\) arrangements. The total is \(2\cdot4!\cdot4!=1152\).

Step 3

Exam Tip

दो pattern संभव हैं और हर pattern में \(4!\cdot4!\) व्यवस्थाएं हैं। कुल \(2\cdot4!\cdot4!=1152\) है।

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Question 19/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग लाल और (4) अलग-अलग नीली गेंदों को ऐसे पंक्ति में रखें कि कोई दो नीली गेंदें साथ न हों। तरीकों की संख्या क्या है?

In how many ways can (5) distinct red balls and (4) distinct blue balls be arranged in a row so that no two blue balls are together?

Explanation opens after your attempt
Correct Answer

B. (86400)

Step 1

Concept

Arrange the red balls first in (5!) ways, then place (4) blue balls in (6) gaps in \(^{6}P_4\) ways. The total is \(5!\cdot^{6}P_4=86400\).

Step 2

Why this answer is correct

The correct answer is B. (86400). Arrange the red balls first in (5!) ways, then place (4) blue balls in (6) gaps in \(^{6}P_4\) ways. The total is \(5!\cdot^{6}P_4=86400\).

Step 3

Exam Tip

पहले लाल गेंदें (5!) तरीकों से रखें, फिर (6) gaps में (4) नीली गेंदें \(^{6}P_4\) तरीकों से रखें। कुल \(5!\cdot^{6}P_4=86400\) है।

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Question 20/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग पत्रों को (8) अलग-अलग डिब्बों में एक-एक करके रखने के कितने तरीके हैं?

In how many ways can (6) distinct letters be placed one each into (8) distinct boxes?

Explanation opens after your attempt
Correct Answer

A. (20160)

Step 1

Concept

We choose ordered positions for (6) letters among (8) distinct boxes, so \(^{8}P_6=20160\). In exams, distinct boxes make it an arrangement.

Step 2

Why this answer is correct

The correct answer is A. (20160). We choose ordered positions for (6) letters among (8) distinct boxes, so \(^{8}P_6=20160\). In exams, distinct boxes make it an arrangement.

Step 3

Exam Tip

(8) अलग डिब्बों में से (6) क्रमित स्थान चुने जाते हैं, इसलिए \(^{8}P_6=20160\)। परीक्षा में boxes अलग हों तो arrangement मानें।

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Question 21/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (1,2,3,4,5,6) से बिना पुनरावृत्ति (3) अंकों की कितनी संख्याएं (400) से बड़ी होंगी?

How many (3)-digit numbers greater than (400) can be formed from (1,2,3,4,5,6) without repetition?

Explanation opens after your attempt
Correct Answer

C. (36)

Step 1

Concept

The hundreds place can be (4,5,6), and the remaining places have \(5\cdot4\) ways. The total is \(3\cdot5\cdot4=60\).

Step 2

Why this answer is correct

The correct answer is C. (36). The hundreds place can be (4,5,6), and the remaining places have \(5\cdot4\) ways. The total is \(3\cdot5\cdot4=60\).

Step 3

Exam Tip

सैकड़े के स्थान पर (4,5,6) हो सकते हैं और बाकी \(5\cdot4\) तरीके हैं। कुल \(3\cdot5\cdot4=60\) है।

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Question 22/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) पुस्तकों को एक शेल्फ पर इस प्रकार कितने तरीकों से रखा जा सकता है कि दो विशेष पुस्तकें दोनों सिरों पर रहें?

In how many ways can (7) books be arranged on a shelf so that two particular books occupy the two ends?

Explanation opens after your attempt
Correct Answer

A. (240)

Step 1

Concept

The two particular books can occupy the ends in (2!) ways and the remaining books in (5!) ways. The total is \(2!\cdot5!=240\).

Step 2

Why this answer is correct

The correct answer is A. (240). The two particular books can occupy the ends in (2!) ways and the remaining books in (5!) ways. The total is \(2!\cdot5!=240\).

Step 3

Exam Tip

दो विशेष पुस्तकें सिरों पर (2!) तरीकों से और शेष (5!) तरीकों से रखी जाएंगी। कुल \(2!\cdot5!=240\) है।

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Question 23/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) व्यक्तियों की पंक्ति में एक विशेष व्यक्ति किसी भी सिरे पर न बैठे। व्यवस्थाएं कितनी होंगी?

In a row of (8) people, how many arrangements are possible if one particular person does not sit at either end?

Explanation opens after your attempt
Correct Answer

B. (30240)

Step 1

Concept

The particular person has (6) inner positions and the remaining people sit in (7!) ways. The total is \(6\cdot7!=30240\).

Step 2

Why this answer is correct

The correct answer is B. (30240). The particular person has (6) inner positions and the remaining people sit in (7!) ways. The total is \(6\cdot7!=30240\).

Step 3

Exam Tip

विशेष व्यक्ति के लिए (6) अंदरूनी स्थान हैं और बाकी (7!) तरीकों से बैठेंगे। कुल \(6\cdot7!=30240\) है।

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Question 24/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(11) खिलाड़ियों में से कप्तान, उपकप्तान और प्रबंधक कितने तरीकों से चुने जा सकते हैं?

In how many ways can a captain, vice-captain and manager be chosen from (11) players?

Explanation opens after your attempt
Correct Answer

A. (990)

Step 1

Concept

The three posts are different, so \(^{11}P_3=990\). In exams, different responsibilities make order important.

Step 2

Why this answer is correct

The correct answer is A. (990). The three posts are different, so \(^{11}P_3=990\). In exams, different responsibilities make order important.

Step 3

Exam Tip

तीन पद अलग हैं, इसलिए \(^{11}P_3=990\)। परीक्षा में अलग-अलग जिम्मेदारियां हों तो क्रम महत्वपूर्ण है।

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Question 25/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग अंकों से बिना पुनरावृत्ति (2) अंकों के कितने क्रमित कोड बनेंगे यदि पहला अंक (0) नहीं हो सकता?

How many ordered (2)-digit codes can be formed from (6) distinct digits without repetition if the first digit cannot be (0)?

Explanation opens after your attempt
Correct Answer

B. (25)

Step 1

Concept

There are (5) choices for the first place and (5) choices for the second. Hence (25) codes are formed.

Step 2

Why this answer is correct

The correct answer is B. (25). There are (5) choices for the first place and (5) choices for the second. Hence (25) codes are formed.

Step 3

Exam Tip

पहले स्थान के लिए (5) विकल्प और दूसरे के लिए (5) विकल्प हैं। इसलिए कुल (25) कोड बनेंगे।

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Question 26/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

शब्द अनुप्रयोग के (9) अक्षरों में दो-दो अक्षर समान हैं। अलग व्यवस्थाएं कितनी होंगी?

A word has (9) letters with two different letters each repeated (2) times. How many distinct arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (90720)

Step 1

Concept

Two groups are repeated (2) times each, so the number is \(\frac{9!}{2!2!}=90720\). In exams, put each repeated group factorial in the denominator.

Step 2

Why this answer is correct

The correct answer is B. (90720). Two groups are repeated (2) times each, so the number is \(\frac{9!}{2!2!}=90720\). In exams, put each repeated group factorial in the denominator.

Step 3

Exam Tip

दो समूह (2) बार समान हैं, इसलिए संख्या \(\frac{9!}{2!2!}=90720\) है। परीक्षा में हर repeated group का factorial हर में रखें।

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Question 27/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) अलग-अलग झंडों में से (5) झंडों का संकेत कितने क्रमों में बनाया जा सकता है?

In how many orders can a signal of (5) flags be made from (7) distinct flags?

Explanation opens after your attempt
Correct Answer

A. (2520)

Step 1

Concept

Order matters in a signal, so \(^{7}P_5=2520\). In exams, words like signal indicate permutation.

Step 2

Why this answer is correct

The correct answer is A. (2520). Order matters in a signal, so \(^{7}P_5=2520\). In exams, words like signal indicate permutation.

Step 3

Exam Tip

संकेत में क्रम महत्वपूर्ण है, इसलिए \(^{7}P_5=2520\)। परीक्षा में signal जैसे शब्द permutation का संकेत देते हैं।

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Question 28/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) लड़कों और (4) लड़कियों को गोल मेज पर ऐसे कितने तरीकों से बैठाया जा सकता है कि सभी लड़कियां साथ बैठें?

In how many ways can (5) boys and (4) girls be seated around a circular table so that all girls sit together?

Explanation opens after your attempt
Correct Answer

A. (2880)

Step 1

Concept

Treat the girls as one block, then there are (6) units in a circle. The number is ((6-1)!\cdot4!=2880).

Step 2

Why this answer is correct

The correct answer is A. (2880). Treat the girls as one block, then there are (6) units in a circle. The number is ((6-1)!\cdot4!=2880).

Step 3

Exam Tip

लड़कियों को एक ब्लॉक मानें, तब गोल में (6) इकाइयां हैं। संख्या ((6-1)!\cdot4!=2880) है।

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Question 29/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) लोगों की गोल बैठक में दो विशेष व्यक्ति साथ न बैठें तो व्यवस्थाएं कितनी होंगी?

In a circular seating of (6) people, how many arrangements are possible if two particular people do not sit together?

Explanation opens after your attempt
Correct Answer

A. (72)

Step 1

Concept

Total circular arrangements are (5!), and together arrangements are \(4!\cdot2!\). So the answer is (120-48=72).

Step 2

Why this answer is correct

The correct answer is A. (72). Total circular arrangements are (5!), and together arrangements are \(4!\cdot2!\). So the answer is (120-48=72).

Step 3

Exam Tip

कुल गोल व्यवस्थाएं (5!) हैं और साथ बैठने वाली व्यवस्थाएं \(4!\cdot2!\) हैं। इसलिए उत्तर (120-48=72) है।

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Question 30/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(10) अलग-अलग चित्रों में से (3) चित्रों को दीवार पर क्रम में लगाने के कितने तरीके हैं?

In how many ways can (3) pictures be selected from (10) distinct pictures and hung in order on a wall?

Explanation opens after your attempt
Correct Answer

A. (720)

Step 1

Concept

The pictures are selected and also ordered, so \(^{10}P_3=720\). In exams, use \(nP_r\) for selected and ordered cases.

Step 2

Why this answer is correct

The correct answer is A. (720). The pictures are selected and also ordered, so \(^{10}P_3=720\). In exams, use \(nP_r\) for selected and ordered cases.

Step 3

Exam Tip

चित्र चुने भी जा रहे हैं और क्रम में लगाए भी जा रहे हैं, इसलिए \(^{10}P_3=720\)। परीक्षा में selected and ordered में \(nP_r\) लगाएं।

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Question 31/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (2,4,5,6,8,9) से बिना पुनरावृत्ति (4) अंकों की कितनी संख्याएं (6000) से बड़ी होंगी?

How many (4)-digit numbers greater than (6000) can be formed from (2,4,5,6,8,9) without repetition?

Explanation opens after your attempt
Correct Answer

D. (60)

Step 1

Concept

The thousands place can be (6,8,9), and the remaining places have \(5\cdot4\cdot3\) ways. The total is \(3\cdot5\cdot4\cdot3=180\).

Step 2

Why this answer is correct

The correct answer is D. (60). The thousands place can be (6,8,9), and the remaining places have \(5\cdot4\cdot3\) ways. The total is \(3\cdot5\cdot4\cdot3=180\).

Step 3

Exam Tip

हजार के स्थान पर (6,8,9) हो सकते हैं और बाकी \(5\cdot4\cdot3\) तरीके हैं। कुल \(3\cdot5\cdot4\cdot3=180\) है।

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Question 32/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग पुरस्कारों को (4) विद्यार्थियों में एक-एक देकर कितने तरीकों से बांटा जा सकता है?

In how many ways can (6) distinct prizes be distributed one each among (4) students?

Explanation opens after your attempt
Correct Answer

A. (360)

Step 1

Concept

The (4) students receive one prize each from (6) prizes in an ordered way, so \(^{6}P_4=360\). In exams, distinct recipients make order important.

Step 2

Why this answer is correct

The correct answer is A. (360). The (4) students receive one prize each from (6) prizes in an ordered way, so \(^{6}P_4=360\). In exams, distinct recipients make order important.

Step 3

Exam Tip

(4) विद्यार्थियों को (6) पुरस्कारों में से क्रमित रूप से एक-एक मिलता है, इसलिए \(^{6}P_4=360\)। परीक्षा में recipients अलग हों तो order महत्वपूर्ण होता है।

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Question 33/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) अलग-अलग किताबों में से (5) किताबें चुनकर शेल्फ पर रखने के कितने तरीके हैं?

In how many ways can (5) books be selected from (8) distinct books and arranged on a shelf?

Explanation opens after your attempt
Correct Answer

A. (6720)

Step 1

Concept

Selection is followed by ordering, so \(^{8}P_5=6720\). In exams, placing on a shelf indicates arrangement.

Step 2

Why this answer is correct

The correct answer is A. (6720). Selection is followed by ordering, so \(^{8}P_5=6720\). In exams, placing on a shelf indicates arrangement.

Step 3

Exam Tip

चयन के साथ क्रम भी है, इसलिए \(^{8}P_5=6720\)। परीक्षा में shelf पर रखना arrangement बताता है।

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Question 34/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) अलग-अलग विद्यार्थियों में से (3) को भाषण के क्रम में बुलाना है। कितने तरीके हैं?

From (7) distinct students, (3) are to be called in speaking order. How many ways are possible?

Explanation opens after your attempt
Correct Answer

B. (210)

Step 1

Concept

The speaking order gives different outcomes, so \(^{7}P_3=210\). In exams, treat performance order as permutation.

Step 2

Why this answer is correct

The correct answer is B. (210). The speaking order gives different outcomes, so \(^{7}P_3=210\). In exams, treat performance order as permutation.

Step 3

Exam Tip

भाषण का क्रम अलग परिणाम देता है, इसलिए \(^{7}P_3=210\)। परीक्षा में performance order को permutation मानें।

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Question 35/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (0,1,2,3,4,5) से बिना पुनरावृत्ति (4) अंकों की कितनी संख्याएं (5) से विभाज्य होंगी?

How many (4)-digit numbers divisible by (5) can be formed from (0,1,2,3,4,5) without repetition?

Explanation opens after your attempt
Correct Answer

B. (96)

Step 1

Concept

The last digit is (0) or (5). Adding the two cases gives (60+48=108) numbers.

Step 2

Why this answer is correct

The correct answer is B. (96). The last digit is (0) or (5). Adding the two cases gives (60+48=108) numbers.

Step 3

Exam Tip

अंतिम अंक (0) या (5) होगा। अलग-अलग cases जोड़ने पर (60+48=108) संख्याएं मिलती हैं।

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Question 36/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग अंकों से (5) अंकों की संख्याएं बनानी हैं। यदि एक विशेष अंक हमेशा बीच में रहे तो कुल संख्या कितनी होगी?

(5)-digit numbers are to be formed from (5) distinct digits. If one particular digit always stays in the middle, how many numbers are possible?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

The middle position is fixed, so the remaining (4) digits are arranged in (4!) ways. The answer is (24).

Step 2

Why this answer is correct

The correct answer is C. (24). The middle position is fixed, so the remaining (4) digits are arranged in (4!) ways. The answer is (24).

Step 3

Exam Tip

बीच का स्थान निश्चित है, इसलिए शेष (4) अंक (4!) तरीकों से व्यवस्थित होंगे। उत्तर (24) है।

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Question 37/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) अलग-अलग वस्तुओं को एक पंक्ति में रखना है। यदि एक विशेष वस्तु हमेशा पहले स्थान पर रहे तो व्यवस्थाएं कितनी होंगी?

(8) distinct objects are to be arranged in a row. If one particular object is always in the first position, how many arrangements are possible?

Explanation opens after your attempt
Correct Answer

A. (5040)

Step 1

Concept

The first position is fixed, so the remaining (7) objects are arranged in (7!) ways. The answer is (5040).

Step 2

Why this answer is correct

The correct answer is A. (5040). The first position is fixed, so the remaining (7) objects are arranged in (7!) ways. The answer is (5040).

Step 3

Exam Tip

पहला स्थान निश्चित है, इसलिए शेष (7) वस्तुएं (7!) तरीकों से रखें। उत्तर (5040) है।

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Question 38/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(9) अलग-अलग वस्तुओं को पंक्ति में रखते समय दो विशेष वस्तुएं दोनों सिरों पर न आएं। व्यवस्थाएं कितनी होंगी?

While arranging (9) distinct objects in a row, two particular objects should not occupy the two ends. How many arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (342720)

Step 1

Concept

Subtract the cases where both particular objects occupy the two ends, which are \(2!\cdot7!\), from total (9!). The answer is (362880-10080=352800).

Step 2

Why this answer is correct

The correct answer is B. (342720). Subtract the cases where both particular objects occupy the two ends, which are \(2!\cdot7!\), from total (9!). The answer is (362880-10080=352800).

Step 3

Exam Tip

कुल (9!) में से दोनों विशेष वस्तुओं के सिरों पर आने वाली \(2!\cdot7!\) व्यवस्थाएं घटाएं। उत्तर (362880-10080=352800) है।

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Question 39/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग विद्यार्थियों को (6) अलग-अलग पंक्तियों में एक-एक बैठाने के कितने तरीके हैं?

In how many ways can (6) distinct students be seated one each in (6) distinct rows?

Explanation opens after your attempt
Correct Answer

C. (720)

Step 1

Concept

The number of seating (6) students in (6) distinct positions is (6!=720). In exams, use factorial when positions are distinct.

Step 2

Why this answer is correct

The correct answer is C. (720). The number of seating (6) students in (6) distinct positions is (6!=720). In exams, use factorial when positions are distinct.

Step 3

Exam Tip

(6) विद्यार्थियों को (6) अलग स्थानों पर बैठाने की संख्या (6!=720) है। परीक्षा में स्थान अलग हों तो factorial लगाएं।

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Question 40/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(12) अक्षरों में से (2) अक्षरों का क्रमित चयन कितने तरीकों से होगा?

In how many ways can (2) letters be selected in order from (12) letters?

Explanation opens after your attempt
Correct Answer

A. (132)

Step 1

Concept

Ordered selection is \(^{12}P_2=12\cdot11=132\). In exams, the phrase in order indicates permutation.

Step 2

Why this answer is correct

The correct answer is A. (132). Ordered selection is \(^{12}P_2=12\cdot11=132\). In exams, the phrase in order indicates permutation.

Step 3

Exam Tip

क्रमित चयन \(^{12}P_2=12\cdot11=132\) है। परीक्षा में in order शब्द permutation बताता है।

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Question 41/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग खिलौनों को (7) बच्चों में एक-एक देकर कितने तरीकों से बांटा जा सकता है?

In how many ways can (5) distinct toys be given one each to (7) children?

Explanation opens after your attempt
Correct Answer

A. (2520)

Step 1

Concept

(5) of (7) children receive distinct toys, so \(^{7}P_5=2520\). In exams, distinct receivers make order count.

Step 2

Why this answer is correct

The correct answer is A. (2520). (5) of (7) children receive distinct toys, so \(^{7}P_5=2520\). In exams, distinct receivers make order count.

Step 3

Exam Tip

(7) बच्चों में से (5) को अलग खिलौने मिलेंगे, इसलिए \(^{7}P_5=2520\)। परीक्षा में receiver अलग हों तो order counted होता है।

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Question 42/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(10) अलग-अलग सिक्कों को एक पंक्ति में कितने तरीकों से रखा जाए यदि एक विशेष सिक्का किसी भी सिरे पर हो?

In how many ways can (10) distinct coins be arranged in a row if one particular coin is at either end?

Explanation opens after your attempt
Correct Answer

A. (725760)

Step 1

Concept

The particular coin has (2) end choices and the remaining coins are arranged in (9!) ways. The total is \(2\cdot9!=725760\).

Step 2

Why this answer is correct

The correct answer is A. (725760). The particular coin has (2) end choices and the remaining coins are arranged in (9!) ways. The total is \(2\cdot9!=725760\).

Step 3

Exam Tip

विशेष सिक्के के लिए (2) सिरों में से विकल्प हैं और बाकी (9!) तरीकों से रखे जाएंगे। कुल \(2\cdot9!=725760\) है।

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Question 43/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) अलग-अलग फूलों को गोल गुलदस्ते में कितने तरीकों से सजाया जा सकता है यदि केवल घुमाव समान माना जाए?

In how many ways can (7) distinct flowers be arranged in a circular bouquet if only rotations are considered the same?

Explanation opens after your attempt
Correct Answer

B. (720)

Step 1

Concept

Only rotations are the same, so the circular arrangement is ((7-1)!=720). In exams, check whether reflection is also considered the same.

Step 2

Why this answer is correct

The correct answer is B. (720). Only rotations are the same, so the circular arrangement is ((7-1)!=720). In exams, check whether reflection is also considered the same.

Step 3

Exam Tip

केवल घुमाव समान है, इसलिए circular arrangement ((7-1)!=720) है। परीक्षा में reflection समान है या नहीं, यह पढ़ें।

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Question 44/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग व्यक्तियों में से (4) को एक फोटो में बाएं से दाएं क्रम में खड़ा करना है। कितने तरीके हैं?

From (6) distinct persons, (4) are to stand in a photo from left to right. How many ways are possible?

Explanation opens after your attempt
Correct Answer

A. (360)

Step 1

Concept

The left-to-right order is important, so \(^{6}P_4=360\). In exams, standing order means permutation.

Step 2

Why this answer is correct

The correct answer is A. (360). The left-to-right order is important, so \(^{6}P_4=360\). In exams, standing order means permutation.

Step 3

Exam Tip

बाएं से दाएं क्रम महत्वपूर्ण है, इसलिए \(^{6}P_4=360\)। परीक्षा में standing order को permutation मानें।

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Question 45/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (1,2,3,4,6,8) से बिना पुनरावृत्ति (4) अंकों की कितनी संख्याएं (3) से शुरू होंगी?

How many (4)-digit numbers can be formed from (1,2,3,4,6,8) without repetition and starting with (3)?

Explanation opens after your attempt
Correct Answer

B. (60)

Step 1

Concept

The first digit is fixed as (3), and the remaining (3) places are filled in \(5\cdot4\cdot3\) ways. The answer is (60).

Step 2

Why this answer is correct

The correct answer is B. (60). The first digit is fixed as (3), and the remaining (3) places are filled in \(5\cdot4\cdot3\) ways. The answer is (60).

Step 3

Exam Tip

पहला अंक (3) निश्चित है और बाकी (3) स्थान \(5\cdot4\cdot3\) तरीकों से भरेंगे। उत्तर (60) है।

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Question 46/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग अध्यायों के प्रश्नपत्र को कितने क्रमों में हल किया जा सकता है यदि पहला अध्याय निश्चित है?

In how many orders can questions from (5) distinct chapters be attempted if the first chapter is fixed?

Explanation opens after your attempt
Correct Answer

A. (24)

Step 1

Concept

The first chapter is fixed, so the remaining (4) chapters can be ordered in (4!) ways. The answer is (24).

Step 2

Why this answer is correct

The correct answer is A. (24). The first chapter is fixed, so the remaining (4) chapters can be ordered in (4!) ways. The answer is (24).

Step 3

Exam Tip

पहला अध्याय निश्चित है, इसलिए बाकी (4) अध्याय (4!) क्रमों में आएंगे। उत्तर (24) है।

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Question 47/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) अलग-अलग रंगों में से (3) रंगों की पट्टियां ऊपर से नीचे क्रम में बनानी हैं। कितने तरीके हैं?

In how many ways can (3) stripes be made from (8) distinct colours in top-to-bottom order?

Explanation opens after your attempt
Correct Answer

A. (336)

Step 1

Concept

Changing top-to-bottom order changes the design, so \(^{8}P_3=336\). In exams, use permutation for ordered layers.

Step 2

Why this answer is correct

The correct answer is A. (336). Changing top-to-bottom order changes the design, so \(^{8}P_3=336\). In exams, use permutation for ordered layers.

Step 3

Exam Tip

ऊपर से नीचे क्रम बदलने पर design बदलता है, इसलिए \(^{8}P_3=336\)। परीक्षा में ordered layers में permutation लगाएं।

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Question 48/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) अलग-अलग प्रश्नों में से (5) प्रश्नों को उत्तर-पुस्तिका में क्रम से लिखने के कितने तरीके हैं?

In how many ways can (5) questions be written in order from (7) distinct questions?

Explanation opens after your attempt
Correct Answer

A. (2520)

Step 1

Concept

(5) questions are selected and written in order, so \(^{7}P_5=2520\). In exams, if writing order differs, use permutation.

Step 2

Why this answer is correct

The correct answer is A. (2520). (5) questions are selected and written in order, so \(^{7}P_5=2520\). In exams, if writing order differs, use permutation.

Step 3

Exam Tip

(5) प्रश्न चुनकर क्रम में लिखने हैं, इसलिए \(^{7}P_5=2520\)। परीक्षा में order of writing अलग हो तो permutation लें।

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Question 49/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(4) अलग-अलग मोबाइल और (3) अलग-अलग किताबों को पंक्ति में ऐसे रखा जाए कि सभी मोबाइल साथ रहें। कितनी व्यवस्थाएं होंगी?

In how many ways can (4) distinct mobiles and (3) distinct books be arranged in a row so that all mobiles stay together?

Explanation opens after your attempt
Correct Answer

A. (576)

Step 1

Concept

Treat the four mobiles as one block, then there are (4) units. The arrangement is \(4!\cdot4!=576\).

Step 2

Why this answer is correct

The correct answer is A. (576). Treat the four mobiles as one block, then there are (4) units. The arrangement is \(4!\cdot4!=576\).

Step 3

Exam Tip

चार मोबाइल को एक ब्लॉक मानें, तब (4) इकाइयां हैं। व्यवस्था \(4!\cdot4!=576\) होगी।

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Question 50/50 Medium Mathematics Permutations and Combinations Permutations Class 11 Level 59

(9) विद्यार्थियों में से (4) को क्रम से पुरस्कार मंच पर बुलाने के कितने तरीके हैं?

In how many ways can (4) students be called to the prize stage in order from (9) students?

Explanation opens after your attempt
Correct Answer

A. (3024)

Step 1

Concept

The calling order to the stage is important, so \(^{9}P_4=3024\). In exams, use permutation when order is specified.

Step 2

Why this answer is correct

The correct answer is A. (3024). The calling order to the stage is important, so \(^{9}P_4=3024\). In exams, use permutation when order is specified.

Step 3

Exam Tip

मंच पर बुलाने का क्रम महत्वपूर्ण है, इसलिए \(^{9}P_4=3024\)। परीक्षा में क्रम बताया हो तो permutation लगाएं।

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