Class 11 Mathematics - Permutations And Combinations - Permutations Hard Quiz

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अंकों (1,2,3,4,5) से (4) अंकों की ऐसी संख्याएँ बनानी हैं जिनमें कोई अंक दोहराया न जाए और संख्या सम हो। कुल कितनी संख्याएँ बनेंगी?

Using the digits (1,2,3,4,5), how many (4)-digit numbers can be formed without repetition and divisible by (2)?

Explanation opens after your attempt
Correct Answer

A. (48)

Step 1

Concept

There are (2) choices for the units digit and \(4 \times 3 \times 2\) ways for the remaining places. In exams apply the last digit condition first.

Step 2

Why this answer is correct

The correct answer is A. (48). There are (2) choices for the units digit and \(4 \times 3 \times 2\) ways for the remaining places. In exams apply the last digit condition first.

Step 3

Exam Tip

इकाई स्थान पर (2) या (4) के लिए (2) विकल्प हैं और बाकी स्थानों के लिए \(4 \times 3 \times 2\) तरीके हैं। परीक्षा में अंतिम स्थान की शर्त पहले लगाएँ।

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अंकों (0,1,2,3,4,5) से (5) अंकों की ऐसी संख्याएँ बनानी हैं जिनमें पुनरावृत्ति न हो और संख्या (5) से विभाज्य हो। कुल कितनी संख्याएँ बनेंगी?

Using the digits (0,1,2,3,4,5), how many (5)-digit numbers can be formed without repetition and divisible by (5)?

Explanation opens after your attempt
Correct Answer

A. (216)

Step 1

Concept

If the units digit is (0), there are \(5 \times 4 \times 3 \times 2\) ways; if it is (5), the first digit has (4) choices. Add the two cases carefully.

Step 2

Why this answer is correct

The correct answer is A. (216). If the units digit is (0), there are \(5 \times 4 \times 3 \times 2\) ways; if it is (5), the first digit has (4) choices. Add the two cases carefully.

Step 3

Exam Tip

इकाई स्थान (0) होने पर \(5 \times 4 \times 3 \times 2\) और इकाई (5) होने पर पहले स्थान के लिए (4) विकल्प मिलते हैं। दोनों मामलों को जोड़ना मुख्य कदम है।

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किसी विद्यालय के प्रवेश कूट में पहले (2) अलग अक्षर और फिर (3) अलग अंक लिखे जाते हैं। यदि अक्षर (A,B,C,D,E) से और अंक (1,2,3,4,5,6) से चुने जाएँ तो कुल कितने कूट बनेंगे?

An admission code has first (2) distinct letters followed by (3) distinct digits. If letters are chosen from (A,B,C,D,E) and digits from (1,2,3,4,5,6), how many codes are possible?

Explanation opens after your attempt
Correct Answer

C. (2400)

Step 1

Concept

Letters can be chosen in \(5 \times 4\) ways and digits in \(6 \times 5 \times 4\) ways. Multiply independent stages by the fundamental principle.

Step 2

Why this answer is correct

The correct answer is C. (2400). Letters can be chosen in \(5 \times 4\) ways and digits in \(6 \times 5 \times 4\) ways. Multiply independent stages by the fundamental principle.

Step 3

Exam Tip

अक्षरों के लिए \(5 \times 4\) और अंकों के लिए \(6 \times 5 \times 4\) तरीके हैं। स्वतंत्र चरणों को गुणा करना मौलिक सिद्धांत है।

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शब्द (MATRIX) के सभी अक्षरों से ऐसे क्रम बनाने हैं जिनमें स्वर साथ-साथ न हों। कुल कितने क्रम बनेंगे?

Using all letters of the word (MATRIX), how many arrangements are possible in which the vowels are not together?

Explanation opens after your attempt
Correct Answer

A. (480)

Step 1

Concept

Subtract arrangements where (A,I) are together from total (6!). Treat together letters as one block for such conditions.

Step 2

Why this answer is correct

The correct answer is A. (480). Subtract arrangements where (A,I) are together from total (6!). Treat together letters as one block for such conditions.

Step 3

Exam Tip

कुल (6!) क्रमों से उन क्रमों को घटाएँ जिनमें (A,I) साथ हों। साथ-साथ वाली शर्त में ब्लॉक बनाना उपयोगी है।

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अंकों (1,2,3,4,5,6,7) से (4) अंकों की ऐसी संख्याएँ कितनी बनेंगी जिनमें पुनरावृत्ति न हो और (4) हमेशा (2) के बाएँ आए?

How many (4)-digit numbers can be formed from digits (1,2,3,4,5,6,7) without repetition such that (4) always appears to the left of (2)?

Explanation opens after your attempt
Correct Answer

B. (300)

Step 1

Concept

Choose (4) digits including (2) and (4), and in half of their arrangements (4) is left of (2). Symmetry saves time here.

Step 2

Why this answer is correct

The correct answer is B. (300). Choose (4) digits including (2) and (4), and in half of their arrangements (4) is left of (2). Symmetry saves time here.

Step 3

Exam Tip

पहले (2) और (4) सहित (4) अंक चुनें और उनके आधे क्रमों में (4), (2) के बाएँ होगा। सममिति का उपयोग ऐसे प्रश्नों में समय बचाता है।

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किसी कक्षा में (7) विद्यार्थियों में से अध्यक्ष, सचिव और कोषाध्यक्ष चुनने हैं। एक ही विद्यार्थी एक से अधिक पद नहीं ले सकता। चयन कितने तरीकों से होगा?

From (7) students, a president, secretary, and treasurer are to be chosen. No student can hold more than one post. In how many ways can this be done?

Explanation opens after your attempt
Correct Answer

A. (210)

Step 1

Concept

The posts are distinct, so the number of ways is \(7 \times 6 \times 5\). Use arrangements, not combinations, when positions are different.

Step 2

Why this answer is correct

The correct answer is A. (210). The posts are distinct, so the number of ways is \(7 \times 6 \times 5\). Use arrangements, not combinations, when positions are different.

Step 3

Exam Tip

पद अलग-अलग हैं इसलिए \(7 \times 6 \times 5\) तरीके होंगे। पदों में क्रम महत्वपूर्ण हो तो संयोजन नहीं लगाएँ।

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अंकों (2,3,4,5,6,7,8) से (5) अंकों की कितनी संख्याएँ बनेंगी जिनमें पुनरावृत्ति न हो और संख्या (4) से विभाज्य हो?

Using digits (2,3,4,5,6,7,8), how many (5)-digit numbers can be formed without repetition and divisible by (4)?

Explanation opens after your attempt
Correct Answer

D. (300)

Step 1

Concept

Check divisibility by (4) using the last two digits and fill the remaining (3) places for each valid ending pair. Fix the last two places first.

Step 2

Why this answer is correct

The correct answer is D. (300). Check divisibility by (4) using the last two digits and fill the remaining (3) places for each valid ending pair. Fix the last two places first.

Step 3

Exam Tip

अंतिम दो अंकों से (4) से विभाज्यता जाँचें और हर सही जोड़े के लिए बाकी (3) स्थान भरें। अंत के दो स्थान पहले तय करें।

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(9) व्यक्तियों में से (4) व्यक्तियों की पंक्ति बनानी है। दो निश्चित व्यक्ति एक साथ न आएँ। ऐसी पंक्तियाँ कितनी होंगी?

From (9) persons, a row of (4) persons is to be formed. Two particular persons must not sit together. How many rows are possible?

Explanation opens after your attempt
Correct Answer

B. (2784)

Step 1

Concept

Subtract arrangements where the two particular persons are together from \(^{9}P_{4}\). Treat the pair as a block when they are together.

Step 2

Why this answer is correct

The correct answer is B. (2784). Subtract arrangements where the two particular persons are together from \(^{9}P_{4}\). Treat the pair as a block when they are together.

Step 3

Exam Tip

कुल \(^{9}P_{4}\) से उन व्यवस्थाओं को घटाएँ जिनमें दोनों निश्चित व्यक्ति साथ हों। साथ आने पर उन्हें ब्लॉक मानें।

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एक पुस्तक शेल्फ में (4) अलग गणित, (3) अलग रसायन और (2) अलग जीवविज्ञान की पुस्तकें हैं। यदि हर विषय की पुस्तकें साथ-साथ रखनी हों तो कुल व्यवस्थाएँ कितनी होंगी?

A shelf has (4) distinct mathematics books, (3) distinct chemistry books, and (2) distinct biology books. If books of each subject must be kept together, how many arrangements are possible?

Explanation opens after your attempt
Correct Answer

A. (1728)

Step 1

Concept

The three subject blocks can be arranged in (3!) ways and books inside blocks in (4!3!2!) ways. Use the block method for together conditions.

Step 2

Why this answer is correct

The correct answer is A. (1728). The three subject blocks can be arranged in (3!) ways and books inside blocks in (4!3!2!) ways. Use the block method for together conditions.

Step 3

Exam Tip

तीन विषयों के ब्लॉक (3!) तरीकों से और ब्लॉकों के भीतर पुस्तकें (4!3!2!) तरीकों से सजेंगी। साथ रखने में ब्लॉक विधि लगाएँ।

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शब्द (SCHOOL) के अक्षरों से कितने भिन्न क्रम बनेंगे?

How many distinct arrangements can be made using all letters of the word (SCHOOL)?

Explanation opens after your attempt
Correct Answer

B. (360)

Step 1

Concept

There are (6) letters with (O) repeated twice, so the count is \(\frac{6!}{2!}\). Do not forget to divide by repeated letters.

Step 2

Why this answer is correct

The correct answer is B. (360). There are (6) letters with (O) repeated twice, so the count is \(\frac{6!}{2!}\). Do not forget to divide by repeated letters.

Step 3

Exam Tip

कुल (6) अक्षर हैं और (O) दो बार आया है, इसलिए संख्या \(\frac{6!}{2!}\) होगी। समान अक्षरों पर विभाजन करना न भूलें।

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(4) अलग गणित और (3) अलग अंग्रेजी की पुस्तकों को पंक्ति में सजाना है। कोई भी दो अंग्रेजी पुस्तकें साथ न हों। कुल व्यवस्थाएँ कितनी होंगी?

(4) distinct mathematics books and (3) distinct English books are to be arranged in a row. No two English books should be together. How many arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (1440)

Step 1

Concept

First arrange the mathematics books in (4!) ways, then place the (3) English books in (5) gaps with order. The gap method is useful for separation restrictions.

Step 2

Why this answer is correct

The correct answer is B. (1440). First arrange the mathematics books in (4!) ways, then place the (3) English books in (5) gaps with order. The gap method is useful for separation restrictions.

Step 3

Exam Tip

पहले गणित की पुस्तकें (4!) तरीकों से रखें और बने (5) खाली स्थानों में (3) अंग्रेजी पुस्तकें क्रम सहित रखें। अलग-अलग रखने में खाली स्थान विधि उपयोगी है।

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अक्षरों (A,B,C,D,E,F) से (4) अक्षरों के ऐसे शब्द कितने बनेंगे जिनमें (A) हो पर (B) न हो और अक्षर दोहराए न जाएँ?

Using letters (A,B,C,D,E,F), how many (4)-letter words can be formed without repetition such that (A) is included but (B) is not included?

Explanation opens after your attempt
Correct Answer

B. (96)

Step 1

Concept

Choose (3) letters from (C,D,E,F) along with (A), then arrange the (4) letters in (4!) ways. Apply inclusion and exclusion restrictions first.

Step 2

Why this answer is correct

The correct answer is B. (96). Choose (3) letters from (C,D,E,F) along with (A), then arrange the (4) letters in (4!) ways. Apply inclusion and exclusion restrictions first.

Step 3

Exam Tip

(A) के साथ (C,D,E,F) में से (3) अक्षर चुनें और (4!) क्रम लगाएँ। शामिल और बाहर वाली शर्तों को पहले लागू करें।

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(7) विद्यार्थियों को (3) पुरस्कार देने हैं। कोई विद्यार्थी एक से अधिक पुरस्कार नहीं पा सकता और पुरस्कार अलग-अलग हैं। कुल कितने तरीके होंगे?

(3) distinct prizes are to be awarded to (7) students. No student can receive more than one prize. How many ways are possible?

Explanation opens after your attempt
Correct Answer

A. (210)

Step 1

Concept

The distinct prizes have (7,6,5) choices respectively. Order matters when prizes are different.

Step 2

Why this answer is correct

The correct answer is A. (210). The distinct prizes have (7,6,5) choices respectively. Order matters when prizes are different.

Step 3

Exam Tip

अलग पुरस्कारों के लिए क्रमशः (7,6,5) विकल्प हैं। अलग पुरस्कारों में क्रम का महत्व होता है।

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एक पासवाक्य में (5) अलग अक्षर हैं। इसे ऐसे सजाना है कि (2) निश्चित अक्षरों के बीच ठीक (2) अक्षर आएँ। कुल कितनी व्यवस्थाएँ होंगी?

A password has (5) distinct letters. They are to be arranged so that exactly (2) letters lie between two particular letters. How many arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (36)

Step 1

Concept

The two particular letters can occupy positions ((1,4)) or ((2,5)), and their internal order has (2!) ways. The remaining (3) letters are arranged in (3!) ways.

Step 2

Why this answer is correct

The correct answer is B. (36). The two particular letters can occupy positions ((1,4)) or ((2,5)), and their internal order has (2!) ways. The remaining (3) letters are arranged in (3!) ways.

Step 3

Exam Tip

दो निश्चित अक्षरों के स्थान ((1,4)) या ((2,5)) हो सकते हैं और उनका आपसी क्रम (2!) तरीकों से होगा। शेष (3) अक्षर (3!) तरीकों से सजेंगे।

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