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Mathematics Quiz - Class 11 Mathematics Permutations And...

Question 1/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(8) अलग-अलग ट्रॉफियों को एक शोकेस में सीधी पंक्ति में कितने तरीकों से सजाया जा सकता है?

In how many ways can (8) different trophies be arranged in a straight row in a showcase?

Explanation opens after your attempt

Correct Answer

A. (40320)

Step 1

Concept

The arrangement of (8) distinct objects is (8!=40320). Use factorial for all distinct objects.

Step 2

Why this answer is correct

The correct answer is A. (40320). The arrangement of (8) distinct objects is (8!=40320). Use factorial for all distinct objects.

Step 3

Exam Tip

(8) अलग वस्तुओं की व्यवस्था (8!=40320) होती है। सभी अलग वस्तुओं के लिए factorial लगाएं।

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Question 2/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(12) विद्यार्थियों में से कक्षा-प्रतिनिधि और सह-प्रतिनिधि चुनने के तरीके कितने हैं?

In how many ways can a class representative and assistant representative be chosen from (12) students?

Explanation opens after your attempt

Correct Answer

B. (132)

Step 1

Concept

The two posts are distinct, so \({}^{12}P_{2}=12\times11=132\). Order matters for different posts.

Step 2

Why this answer is correct

The correct answer is B. (132). The two posts are distinct, so \({}^{12}P_{2}=12\times11=132\). Order matters for different posts.

Step 3

Exam Tip

दो पद अलग हैं इसलिए \({}^{12}P_{2}=12\times11=132\)। अलग पदों में क्रम महत्त्वपूर्ण होता है।

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Question 3/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (TRAIN) के सभी अक्षरों से कितने अलग शब्द बनाए जा सकते हैं?

How many distinct words can be formed using all letters of the word (TRAIN)?

Explanation opens after your attempt

Correct Answer

C. (120)

Step 1

Concept

The word (TRAIN) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.

Step 2

Why this answer is correct

The correct answer is C. (120). The word (TRAIN) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.

Step 3

Exam Tip

(TRAIN) में (5) अलग अक्षर हैं इसलिए (5!=120)। सभी अक्षर अलग हों तो denominator नहीं लगता।

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Question 4/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (2,3,5,7) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बनेंगी?

How many (3)-digit numbers can be formed from digits (2,3,5,7) without repetition?

Explanation opens after your attempt

Correct Answer

D. (24)

Step 1

Concept

For three places, there are \(4\times3\times2=24\) ways. Without repetition, each next choice decreases.

Step 2

Why this answer is correct

The correct answer is D. (24). For three places, there are \(4\times3\times2=24\) ways. Without repetition, each next choice decreases.

Step 3

Exam Tip

तीन स्थानों के लिए \(4\times3\times2=24\) तरीके हैं। बिना पुनरावृत्ति में हर अगला विकल्प घटता है।

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Question 5/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{11}P_{2}\) का मान क्या है?

What is the value of \({}^{11}P_{2}\)?

Explanation opens after your attempt

Correct Answer

A. (110)

Step 1

Concept

\({}^{11}P_{2}=11\times10=110\). When (r=2), take two decreasing factors.

Step 2

Why this answer is correct

The correct answer is A. (110). \({}^{11}P_{2}=11\times10=110\). When (r=2), take two decreasing factors.

Step 3

Exam Tip

\({}^{11}P_{2}=11\times10=110\) होता है। (r=2) होने पर दो घटते गुणनखंड लें।

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Question 6/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(7) अलग-अलग गीतों में से (3) गीतों को पहले, दूसरे और तीसरे स्थान पर बजाने के तरीके कितने हैं?

In how many ways can (3) songs from (7) different songs be played in first, second and third positions?

Explanation opens after your attempt

Correct Answer

B. (210)

Step 1

Concept

Playing order matters, so \({}^{7}P_{3}=7\times6\times5=210\). Changing the order changes the list.

Step 2

Why this answer is correct

The correct answer is B. (210). Playing order matters, so \({}^{7}P_{3}=7\times6\times5=210\). Changing the order changes the list.

Step 3

Exam Tip

बजाने का क्रम महत्त्वपूर्ण है इसलिए \({}^{7}P_{3}=7\times6\times5=210\)। क्रम बदलने से सूची बदल जाती है।

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Question 7/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (PAPER) में (P) दो बार आता है। इसके अलग-अलग अक्षर-क्रम कितने होंगे?

The word (PAPER) has (P) twice. How many distinct letter arrangements are possible?

Explanation opens after your attempt

Correct Answer

C. (60)

Step 1

Concept

Since (P) occurs twice, arrangements are \(\frac{5!}{2!}=60\). Divide by the factorial of the repeated letter.

Step 2

Why this answer is correct

The correct answer is C. (60). Since (P) occurs twice, arrangements are \(\frac{5!}{2!}=60\). Divide by the factorial of the repeated letter.

Step 3

Exam Tip

(P) दो बार समान है इसलिए व्यवस्थाएँ \(\frac{5!}{2!}=60\) होंगी। समान अक्षर के factorial से भाग दें।

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Question 8/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(6) अलग-अलग मेजों पर (2) विद्यार्थियों को बैठाने के तरीके कितने हैं?

In how many ways can (2) students be seated at (6) different desks?

Explanation opens after your attempt

Correct Answer

D. (30)

Step 1

Concept

The desks are distinct, so \({}^{6}P_{2}=6\times5=30\). Order is counted when seating at distinct places.

Step 2

Why this answer is correct

The correct answer is D. (30). The desks are distinct, so \({}^{6}P_{2}=6\times5=30\). Order is counted when seating at distinct places.

Step 3

Exam Tip

मेजें अलग हैं इसलिए \({}^{6}P_{2}=6\times5=30\)। अलग स्थानों पर बैठाने में क्रम गिना जाता है।

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Question 9/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{7}P_{7}\) का मान क्या होगा?

What will be the value of \({}^{7}P_{7}\)?

Explanation opens after your attempt

Correct Answer

A. (5040)

Step 1

Concept

When (r=n), \({}^{n}P_{n}=n!\), so \({}^{7}P_{7}=5040\). Use factorial when all are selected.

Step 2

Why this answer is correct

The correct answer is A. (5040). When (r=n), \({}^{n}P_{n}=n!\), so \({}^{7}P_{7}=5040\). Use factorial when all are selected.

Step 3

Exam Tip

जब (r=n) हो तो \({}^{n}P_{n}=n!\), इसलिए \({}^{7}P_{7}=5040\)। पूरा चयन होने पर factorial लें।

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Question 10/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (0,2,4,6,8) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बन सकती हैं?

How many (2)-digit even numbers can be formed from digits (0,2,4,6,8) without repetition?

Explanation opens after your attempt

Correct Answer

B. (16)

Step 1

Concept

If the unit digit is (0), there are (4) ways, and if it is a non-zero even digit, there are \(4\times3=12\) ways, total (16). Make cases for zero questions.

Step 2

Why this answer is correct

The correct answer is B. (16). If the unit digit is (0), there are (4) ways, and if it is a non-zero even digit, there are \(4\times3=12\) ways, total (16). Make cases for zero questions.

Step 3

Exam Tip

इकाई पर (0) हो तो (4) तरीके और इकाई पर गैर-शून्य सम अंक हो तो \(4\times3=12\) तरीके हैं, कुल (16)। शून्य वाले प्रश्न में cases बनाएं।

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Question 11/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(5) अलग-अलग पतंगों को (5) बच्चों में एक-एक देकर बाँटने के तरीके कितने हैं?

In how many ways can (5) different kites be distributed one each among (5) children?

Explanation opens after your attempt

Correct Answer

C. (120)

Step 1

Concept

(5) different kites can be distributed among (5) children in (5!=120) ways. Factorial appears in one-to-one distribution of distinct objects.

Step 2

Why this answer is correct

The correct answer is C. (120). (5) different kites can be distributed among (5) children in (5!=120) ways. Factorial appears in one-to-one distribution of distinct objects.

Step 3

Exam Tip

(5) अलग पतंग (5) बच्चों में (5!=120) तरीकों से बाँटी जा सकती हैं। अलग वस्तुओं के एक-एक वितरण में factorial आता है।

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Question 12/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(9) सवालों में से (3) सवालों को पहले, दूसरे और तीसरे क्रम में हल करने के तरीके कितने हैं?

In how many ways can (3) questions from (9) questions be solved in first, second and third order?

Explanation opens after your attempt

Correct Answer

D. (504)

Step 1

Concept

The solving order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Use permutation in ordered selection.

Step 2

Why this answer is correct

The correct answer is D. (504). The solving order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Use permutation in ordered selection.

Step 3

Exam Tip

हल करने का क्रम महत्त्वपूर्ण है इसलिए \({}^{9}P_{3}=9\times8\times7=504\)। ordered selection में permutation लगाएं।

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Question 13/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (GATE) के अक्षरों से ऐसे कितने क्रम बनेंगे जिनमें (G) पहला अक्षर हो?

How many arrangements of the letters of (GATE) are possible if (G) is the first letter?

Explanation opens after your attempt

Correct Answer

A. (6)

Step 1

Concept

The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.

Step 2

Why this answer is correct

The correct answer is A. (6). The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.

Step 3

Exam Tip

पहला स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed condition को पहले लागू करें।

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Question 14/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(13) अभ्यर्थियों में से अध्यक्ष और उपाध्यक्ष चुनने के तरीके कितने हैं?

In how many ways can a president and vice-president be chosen from (13) candidates?

Explanation opens after your attempt

Correct Answer

B. (156)

Step 1

Concept

The two posts are distinct, so \({}^{13}P_{2}=13\times12=156\). Do not use combination when posts are different.

Step 2

Why this answer is correct

The correct answer is B. (156). The two posts are distinct, so \({}^{13}P_{2}=13\times12=156\). Do not use combination when posts are different.

Step 3

Exam Tip

दो पद अलग हैं इसलिए \({}^{13}P_{2}=13\times12=156\)। पद अलग हों तो combination नहीं लगाएं।

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Question 15/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (1,5,9) से पुनरावृत्ति की अनुमति होने पर कितनी (3)-अंकीय संख्याएँ बनेंगी?

How many (3)-digit numbers can be formed from digits (1,5,9) if repetition is allowed?

Explanation opens after your attempt

Correct Answer

C. (27)

Step 1

Concept

Each place has (3) choices, so \(3^3=27\). When repetition is allowed, choices do not decrease.

Step 2

Why this answer is correct

The correct answer is C. (27). Each place has (3) choices, so \(3^3=27\). When repetition is allowed, choices do not decrease.

Step 3

Exam Tip

हर स्थान पर (3) विकल्प हैं इसलिए \(3^3=27\)। repetition allowed हो तो विकल्प कम नहीं होते।

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Question 16/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(6) अलग-अलग फोल्डरों को एक कतार में रखना है, पर एक विशेष फोल्डर हमेशा अंतिम स्थान पर रहेगा। कितने तरीके होंगे?

(6) different folders are to be arranged in a row, but one special folder always remains in the last position. How many ways are possible?

Explanation opens after your attempt

Correct Answer

D. (120)

Step 1

Concept

The last position is fixed and the remaining (5) folders can be arranged in (5!=120) ways. Arrange the remaining items after fixing the item.

Step 2

Why this answer is correct

The correct answer is D. (120). The last position is fixed and the remaining (5) folders can be arranged in (5!=120) ways. Arrange the remaining items after fixing the item.

Step 3

Exam Tip

अंतिम स्थान निश्चित है और शेष (5) फोल्डर (5!=120) तरीकों से लगेंगे। fixed item के बाद बचे items व्यवस्थित करें।

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Question 17/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{14}P_{1}\) का मान क्या है?

What is the value of \({}^{14}P_{1}\)?

Explanation opens after your attempt

Correct Answer

A. (14)

Step 1

Concept

Since \({}^{n}P_{1}=n\), \({}^{14}P_{1}=14\). For one position, count available objects directly.

Step 2

Why this answer is correct

The correct answer is A. (14). Since \({}^{n}P_{1}=n\), \({}^{14}P_{1}=14\). For one position, count available objects directly.

Step 3

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{14}P_{1}=14\)। एक स्थान के लिए सीधे उपलब्ध वस्तुएँ गिनें।

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Question 18/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (SMART) के सभी अक्षरों से कितने अलग क्रम बनेंगे?

How many distinct arrangements can be formed using all letters of (SMART)?

Explanation opens after your attempt

Correct Answer

B. (120)

Step 1

Concept

The word (SMART) has (5) distinct letters, so (5!=120). The full arrangement of distinct letters is factorial.

Step 2

Why this answer is correct

The correct answer is B. (120). The word (SMART) has (5) distinct letters, so (5!=120). The full arrangement of distinct letters is factorial.

Step 3

Exam Tip

(SMART) में (5) अलग अक्षर हैं इसलिए (5!=120)। distinct letters की पूरी arrangement factorial होती है।

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Question 19/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(10) अलग-अलग बैजों में से (4) बैजों को क्रम में लगाने के तरीके कितने हैं?

How many ways are there to arrange (4) badges from (10) different badges in order?

Explanation opens after your attempt

Correct Answer

C. (5040)

Step 1

Concept

\({}^{10}P_{4}=10\times9\times8\times7=5040\). For (4) places, take (4) decreasing factors.

Step 2

Why this answer is correct

The correct answer is C. (5040). \({}^{10}P_{4}=10\times9\times8\times7=5040\). For (4) places, take (4) decreasing factors.

Step 3

Exam Tip

\({}^{10}P_{4}=10\times9\times8\times7=5040\) है। (4) स्थानों के लिए (4) घटते गुणनखंड लें।

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Question 20/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अक्षरों (A,E,I,O,U) में से (2) अक्षरों का क्रम बनाना हो तो कितने तरीके होंगे?

If an order of (2) letters is to be made from (A,E,I,O,U), how many ways are possible?

Explanation opens after your attempt

Correct Answer

D. (20)

Step 1

Concept

This is \({}^{5}P_{2}=5\times4=20\). In ordered letters, (AE) and (EA) are considered different.

Step 2

Why this answer is correct

The correct answer is D. (20). This is \({}^{5}P_{2}=5\times4=20\). In ordered letters, (AE) and (EA) are considered different.

Step 3

Exam Tip

यह \({}^{5}P_{2}=5\times4=20\) है। ordered letters में (AE) और (EA) अलग माने जाते हैं।

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Question 21/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (DAD) के अलग-अलग arrangements कितने हैं?

How many distinct arrangements are there for the word (DAD)?

Explanation opens after your attempt

Correct Answer

A. (3)

Step 1

Concept

(D) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Do not count repeated letters as distinct.

Step 2

Why this answer is correct

The correct answer is A. (3). (D) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Do not count repeated letters as distinct.

Step 3

Exam Tip

(D) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements हैं। repeated letters को अलग-अलग न गिनें।

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Question 22/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(8) बच्चों में से (5) बच्चों को एक बेंच पर क्रम से बैठाने के तरीके कितने हैं?

In how many ways can (5) children from (8) children be seated on a bench in order?

Explanation opens after your attempt

Correct Answer

B. (6720)

Step 1

Concept

The answer is \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\). Left-right order on a bench matters.

Step 2

Why this answer is correct

The correct answer is B. (6720). The answer is \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\). Left-right order on a bench matters.

Step 3

Exam Tip

उत्तर \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\) है। बेंच पर बाएँ-दाएँ क्रम महत्त्वपूर्ण होता है।

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Question 23/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (2,4,6,8,9) से बिना पुनरावृत्ति कितनी (3)-अंकीय विषम संख्याएँ बनेंगी?

How many (3)-digit odd numbers can be formed from digits (2,4,6,8,9) without repetition?

Explanation opens after your attempt

Correct Answer

C. (12)

Step 1

Concept

Only (9) can be in the units place, and the other two places can be filled in \(4\times3=12\) ways. For odd numbers, check the units place first.

Step 2

Why this answer is correct

The correct answer is C. (12). Only (9) can be in the units place, and the other two places can be filled in \(4\times3=12\) ways. For odd numbers, check the units place first.

Step 3

Exam Tip

इकाई स्थान पर केवल (9) आएगा और बाकी दो स्थान \(4\times3=12\) तरीकों से भरेंगे। odd number में इकाई स्थान पहले देखें।

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Question 24/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{15}P_{0}\) का मान क्या है?

What is the value of \({}^{15}P_{0}\)?

Explanation opens after your attempt

Correct Answer

D. (1)

Step 1

Concept

\({}^{n}P_{0}=1\), so \({}^{15}P_{0}=1\). Choosing zero objects is counted in one way.

Step 2

Why this answer is correct

The correct answer is D. (1). \({}^{n}P_{0}=1\), so \({}^{15}P_{0}=1\). Choosing zero objects is counted in one way.

Step 3

Exam Tip

\({}^{n}P_{0}=1\) इसलिए \({}^{15}P_{0}=1\)। शून्य वस्तु चुनने का एक तरीका माना जाता है।

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Question 25/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(5) अलग-अलग लिफाफों पर (5) अलग-अलग टिकट चिपकाने के तरीके कितने हैं?

In how many ways can (5) different stamps be pasted on (5) different envelopes?

Explanation opens after your attempt

Correct Answer

A. (120)

Step 1

Concept

(5) different stamps can be pasted on (5) different envelopes in (5!=120) ways. Factorial is useful in one-to-one matching.

Step 2

Why this answer is correct

The correct answer is A. (120). (5) different stamps can be pasted on (5) different envelopes in (5!=120) ways. Factorial is useful in one-to-one matching.

Step 3

Exam Tip

(5) अलग टिकट (5) अलग लिफाफों पर (5!=120) तरीकों से लगेंगे। one-to-one matching में factorial उपयोगी है।

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Question 26/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (BANANA) के अलग-अलग अक्षर-क्रम कितने होंगे?

How many distinct letter arrangements are possible for the word (BANANA)?

Explanation opens after your attempt

Correct Answer

B. (60)

Step 1

Concept

(A) occurs three times and (N) twice, so \(\frac{6!}{3!2!}=60\). Divide by factorials of repeated groups.

Step 2

Why this answer is correct

The correct answer is B. (60). (A) occurs three times and (N) twice, so \(\frac{6!}{3!2!}=60\). Divide by factorials of repeated groups.

Step 3

Exam Tip

(A) तीन बार और (N) दो बार है इसलिए \(\frac{6!}{3!2!}=60\)। repeated groups के factorial से भाग दें।

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Question 27/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(6) अलग-अलग स्टेशनों में से शुरू और समाप्ति स्टेशन चुनने के तरीके कितने हैं?

In how many ways can a starting station and ending station be chosen from (6) different stations?

Explanation opens after your attempt

Correct Answer

C. (30)

Step 1

Concept

Starting and ending are ordered, so \({}^{6}P_{2}=30\). In a route, changing start and end changes the case.

Step 2

Why this answer is correct

The correct answer is C. (30). Starting and ending are ordered, so \({}^{6}P_{2}=30\). In a route, changing start and end changes the case.

Step 3

Exam Tip

शुरू और समाप्ति क्रम वाले हैं इसलिए \({}^{6}P_{2}=30\)। route में starting और ending बदलने से मामला बदलता है।

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Question 28/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (0,1,5,7) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बन सकती हैं?

How many (3)-digit numbers can be formed from digits (0,1,5,7) without repetition?

Explanation opens after your attempt

Correct Answer

D. (18)

Step 1

Concept

There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Zero is not allowed in the first place.

Step 2

Why this answer is correct

The correct answer is D. (18). There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Zero is not allowed in the first place.

Step 3

Exam Tip

सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प मिलते हैं, कुल (18)। पहले स्थान पर (0) की अनुमति नहीं है।

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Question 29/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(4) अलग-अलग दिशासूचक कार्डों को ऊपर से नीचे रखने के तरीके कितने हैं?

In how many ways can (4) different direction cards be placed from top to bottom?

Explanation opens after your attempt

Correct Answer

A. (24)

Step 1

Concept

The top-to-bottom arrangement of (4) distinct cards is (4!=24). Use factorial in linear order.

Step 2

Why this answer is correct

The correct answer is A. (24). The top-to-bottom arrangement of (4) distinct cards is (4!=24). Use factorial in linear order.

Step 3

Exam Tip

ऊपर से नीचे (4) अलग कार्डों की व्यवस्था (4!=24) है। linear order में factorial प्रयोग करें।

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Question 30/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(14) प्रतिभागियों में से प्रथम और द्वितीय वक्ता चुनने के तरीके कितने हैं?

In how many ways can first and second speakers be chosen from (14) participants?

Explanation opens after your attempt

Correct Answer

B. (182)

Step 1

Concept

First and second speakers are ordered positions, so \({}^{14}P_{2}=182\). Speaking order uses permutation.

Step 2

Why this answer is correct

The correct answer is B. (182). First and second speakers are ordered positions, so \({}^{14}P_{2}=182\). Speaking order uses permutation.

Step 3

Exam Tip

पहला और दूसरा वक्ता अलग क्रम वाले स्थान हैं इसलिए \({}^{14}P_{2}=182\)। speaking order में permutation लगता है।

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Question 31/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (CODE) के अक्षरों से ऐसे कितने क्रम बनेंगे जिनमें अंतिम अक्षर (E) हो?

How many arrangements of the letters of (CODE) are possible if the last letter is (E)?

Explanation opens after your attempt

Correct Answer

C. (6)

Step 1

Concept

The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate the fixed position first.

Step 2

Why this answer is correct

The correct answer is C. (6). The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate the fixed position first.

Step 3

Exam Tip

अंतिम स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed position को पहले अलग करें।

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Question 32/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{6}P_{5}\) का मान क्या है?

What is the value of \({}^{6}P_{5}\)?

Explanation opens after your attempt

Correct Answer

D. (720)

Step 1

Concept

\({}^{6}P_{5}=6\times5\times4\times3\times2=720\). For five places, take five decreasing factors.

Step 2

Why this answer is correct

The correct answer is D. (720). \({}^{6}P_{5}=6\times5\times4\times3\times2=720\). For five places, take five decreasing factors.

Step 3

Exam Tip

\({}^{6}P_{5}=6\times5\times4\times3\times2=720\) है। पाँच स्थान हों तो पाँच घटते गुणनखंड लें।

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Question 33/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(8) अलग-अलग प्लेटों में से (2) प्लेटों को ऊपरी और निचले रैक में रखने के तरीके कितने हैं?

How many ways are there to place (2) plates from (8) different plates on the upper and lower racks?

Explanation opens after your attempt

Correct Answer

A. (56)

Step 1

Concept

The upper and lower racks are distinct, so \({}^{8}P_{2}=56\). Order is counted for distinct positions.

Step 2

Why this answer is correct

The correct answer is A. (56). The upper and lower racks are distinct, so \({}^{8}P_{2}=56\). Order is counted for distinct positions.

Step 3

Exam Tip

ऊपरी और निचला रैक अलग हैं इसलिए \({}^{8}P_{2}=56\)। अलग स्थानों में order गिना जाता है।

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Question 34/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अक्षरों (P,Q,R,S,T,U) से बिना पुनरावृत्ति कितने (2)-अक्षरी कोड बनेंगे?

How many (2)-letter codes can be formed from (P,Q,R,S,T,U) without repetition?

Explanation opens after your attempt

Correct Answer

B. (30)

Step 1

Concept

\({}^{6}P_{2}=6\times5=30\) codes are possible. In a code, changing the position of letters changes the code.

Step 2

Why this answer is correct

The correct answer is B. (30). \({}^{6}P_{2}=6\times5=30\) codes are possible. In a code, changing the position of letters changes the code.

Step 3

Exam Tip

\({}^{6}P_{2}=6\times5=30\) कोड बनेंगे। कोड में अक्षर का स्थान बदलने से कोड बदलता है।

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Question 35/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (COCO) के अलग-अलग arrangements कितने होंगे?

How many distinct arrangements are possible for the word (COCO)?

Explanation opens after your attempt

Correct Answer

C. (6)

Step 1

Concept

(C) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated letters.

Step 2

Why this answer is correct

The correct answer is C. (6). (C) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated letters.

Step 3

Exam Tip

(C) दो बार और (O) दो बार हैं इसलिए \(\frac{4!}{2!2!}=6\)। दो repeated letters हों तो दोनों से भाग दें।

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Question 36/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(15) विद्यार्थियों में से प्रथम, द्वितीय और तृतीय rank देने के तरीके कितने हैं?

In how many ways can first, second and third ranks be given among (15) students?

Explanation opens after your attempt

Correct Answer

D. (2730)

Step 1

Concept

The three ranks are ordered, so \({}^{15}P_{3}=15\times14\times13=2730\). Rank questions use permutation.

Step 2

Why this answer is correct

The correct answer is D. (2730). The three ranks are ordered, so \({}^{15}P_{3}=15\times14\times13=2730\). Rank questions use permutation.

Step 3

Exam Tip

तीन rank क्रम वाले हैं इसलिए \({}^{15}P_{3}=15\times14\times13=2730\)। rank वाले प्रश्न permutation के होते हैं।

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Question 37/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (3,6,9) से पुनरावृत्ति की अनुमति होने पर कितनी (2)-अंकीय संख्याएँ बनेंगी?

How many (2)-digit numbers can be formed from digits (3,6,9) if repetition is allowed?

Explanation opens after your attempt

Correct Answer

A. (9)

Step 1

Concept

There are (3) choices at both places, so \(3^2=9\). With repetition allowed, choices remain the same.

Step 2

Why this answer is correct

The correct answer is A. (9). There are (3) choices at both places, so \(3^2=9\). With repetition allowed, choices remain the same.

Step 3

Exam Tip

दोनों स्थानों पर (3) विकल्प हैं इसलिए \(3^2=9\)। repetition allowed हो तो विकल्प वही रहते हैं।

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Question 38/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(6) अलग-अलग मोतियों को सीधी कढ़ाई में लगाने के तरीके कितने हैं?

How many ways are there to arrange (6) different beads in a straight embroidery line?

Explanation opens after your attempt

Correct Answer

B. (720)

Step 1

Concept

In a straight line, the arrangement of (6) distinct beads is (6!=720). Do not treat it as a circular arrangement.

Step 2

Why this answer is correct

The correct answer is B. (720). In a straight line, the arrangement of (6) distinct beads is (6!=720). Do not treat it as a circular arrangement.

Step 3

Exam Tip

सीधी रेखा में (6) अलग मोतियों की व्यवस्था (6!=720) है। इसे circular arrangement न मानें।

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Question 39/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{5}P_{0}+{}^{5}P_{2}\) का मान क्या है?

What is the value of \({}^{5}P_{0}+{}^{5}P_{2}\)?

Explanation opens after your attempt

Correct Answer

C. (21)

Step 1

Concept

\({}^{5}P_{0}=1\) and \({}^{5}P_{2}=20\), so the sum is (21). Find the values separately first.

Step 2

Why this answer is correct

The correct answer is C. (21). \({}^{5}P_{0}=1\) and \({}^{5}P_{2}=20\), so the sum is (21). Find the values separately first.

Step 3

Exam Tip

\({}^{5}P_{0}=1\) और \({}^{5}P_{2}=20\), इसलिए योग (21) है। पहले अलग-अलग values निकालें।

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Question 40/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(9) अलग-अलग पेनड्राइव में से (2) पेनड्राइव को पहले और दूसरे स्लॉट में लगाने के तरीके कितने हैं?

How many ways are there to place (2) pen drives from (9) different pen drives in the first and second slots?

Explanation opens after your attempt

Correct Answer

D. (72)

Step 1

Concept

There are (9) choices for the first slot and (8) for the second, so (72). Count permutation when slots are distinct.

Step 2

Why this answer is correct

The correct answer is D. (72). There are (9) choices for the first slot and (8) for the second, so (72). Count permutation when slots are distinct.

Step 3

Exam Tip

पहले स्लॉट के लिए (9) और दूसरे के लिए (8) विकल्प हैं इसलिए (72)। स्लॉट अलग हों तो permutation गिनें।

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Question 41/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (MOUSE) के सभी अक्षरों से कितने अलग शब्द बन सकते हैं?

How many distinct words can be formed using all letters of (MOUSE)?

Explanation opens after your attempt

Correct Answer

A. (120)

Step 1

Concept

The word (MOUSE) has (5) distinct letters, so (5!=120). For distinct letters, factorial gives total arrangements.

Step 2

Why this answer is correct

The correct answer is A. (120). The word (MOUSE) has (5) distinct letters, so (5!=120). For distinct letters, factorial gives total arrangements.

Step 3

Exam Tip

(MOUSE) में (5) अलग अक्षर हैं इसलिए (5!=120)। अलग अक्षरों में factorial ही कुल व्यवस्था देता है।

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Question 42/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(5) अलग-अलग घंटियों को (5) निर्धारित स्थानों पर लगाने के तरीके कितने हैं?

In how many ways can (5) different bells be placed at (5) fixed positions?

Explanation opens after your attempt

Correct Answer

B. (120)

Step 1

Concept

(5) different bells can be placed at (5) positions in (5!=120) ways. Arrangements at fixed positions use factorial.

Step 2

Why this answer is correct

The correct answer is B. (120). (5) different bells can be placed at (5) positions in (5!=120) ways. Arrangements at fixed positions use factorial.

Step 3

Exam Tip

(5) अलग घंटियाँ (5) स्थानों पर (5!=120) तरीकों से लगेंगी। निर्धारित स्थानों की arrangement factorial से होती है।

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Question 43/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (4,5,6,7,8,9) से बिना पुनरावृत्ति कितनी (4)-अंकीय संख्याएँ बनेंगी?

How many (4)-digit numbers can be formed from digits (4,5,6,7,8,9) without repetition?

Explanation opens after your attempt

Correct Answer

C. (360)

Step 1

Concept

For four places, there are \(6\times5\times4\times3=360\) ways. Without repetition, choices decrease.

Step 2

Why this answer is correct

The correct answer is C. (360). For four places, there are \(6\times5\times4\times3=360\) ways. Without repetition, choices decrease.

Step 3

Exam Tip

चार स्थानों के लिए \(6\times5\times4\times3=360\) तरीके हैं। बिना पुनरावृत्ति में विकल्प घटते हैं।

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Question 44/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(9) गायकों में से प्रथम, द्वितीय और तृतीय प्रदर्शन क्रम चुनने के तरीके कितने हैं?

In how many ways can first, second and third performance order be chosen from (9) singers?

Explanation opens after your attempt

Correct Answer

D. (504)

Step 1

Concept

Performance order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Changing order changes the programme.

Step 2

Why this answer is correct

The correct answer is D. (504). Performance order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Changing order changes the programme.

Step 3

Exam Tip

प्रदर्शन क्रम महत्त्वपूर्ण है इसलिए \({}^{9}P_{3}=9\times8\times7=504\)। order बदलने से कार्यक्रम बदल जाता है।

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Question 45/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (REFER) में (R) दो बार और (E) दो बार आता है। अलग व्यवस्थाएँ कितनी होंगी?

In the word (REFER), (R) occurs twice and (E) occurs twice. How many distinct arrangements are possible?

Explanation opens after your attempt

Correct Answer

A. (30)

Step 1

Concept

The arrangements are \(\frac{5!}{2!2!}=30\). Do not count identical letters as different.

Step 2

Why this answer is correct

The correct answer is A. (30). The arrangements are \(\frac{5!}{2!2!}=30\). Do not count identical letters as different.

Step 3

Exam Tip

व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। समान अक्षरों को अलग-अलग न गिनें।

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Question 46/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

\({}^{12}P_{0}+{}^{12}P_{1}\) का मान क्या होगा?

What will be the value of \({}^{12}P_{0}+{}^{12}P_{1}\)?

Explanation opens after your attempt

Correct Answer

B. (13)

Step 1

Concept

\({}^{12}P_{0}=1\) and \({}^{12}P_{1}=12\), so the sum is (13). Remember standard values.

Step 2

Why this answer is correct

The correct answer is B. (13). \({}^{12}P_{0}=1\) and \({}^{12}P_{1}=12\), so the sum is (13). Remember standard values.

Step 3

Exam Tip

\({}^{12}P_{0}=1\) और \({}^{12}P_{1}=12\), इसलिए योग (13) है। standard values याद रखें।

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Question 47/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(7) अलग-अलग पोस्टकार्डों को बाएँ से दाएँ सजाने के तरीके कितने हैं?

In how many ways can (7) different postcards be arranged from left to right?

Explanation opens after your attempt

Correct Answer

C. (5040)

Step 1

Concept

The left-to-right arrangement of (7) distinct postcards is (7!=5040). Use factorial in linear arrangement.

Step 2

Why this answer is correct

The correct answer is C. (5040). The left-to-right arrangement of (7) distinct postcards is (7!=5040). Use factorial in linear arrangement.

Step 3

Exam Tip

बाएँ से दाएँ (7) अलग पोस्टकार्डों की व्यवस्था (7!=5040) है। linear arrangement में factorial लें।

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Question 48/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

अंकों (1,2,3,4,6) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बनेंगी?

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

Explanation opens after your attempt

Correct Answer

D. (12)

Step 1

Concept

The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.

Step 2

Why this answer is correct

The correct answer is D. (12). The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.

Step 3

Exam Tip

इकाई स्थान पर (2,4,6) में से (3) विकल्प और दहाई पर (4) विकल्प हैं, कुल (12)। even number में इकाई स्थान पहले देखें।

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Question 49/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

(11) अलग-अलग चार्टों में से (3) चार्टों को क्रम से लगाने के तरीके कितने हैं?

How many ways are there to arrange (3) charts from (11) different charts in order?

Explanation opens after your attempt

Correct Answer

A. (990)

Step 1

Concept

\({}^{11}P_{3}=11\times10\times9=990\). Use \({}^{n}P_{r}\) in ordered selection.

Step 2

Why this answer is correct

The correct answer is A. (990). \({}^{11}P_{3}=11\times10\times9=990\). Use \({}^{n}P_{r}\) in ordered selection.

Step 3

Exam Tip

\({}^{11}P_{3}=11\times10\times9=990\) है। क्रम वाले चयन में \({}^{n}P_{r}\) लगाएं।

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Question 50/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (CIVIC) में (C) दो बार और (I) दो बार आता है। अलग arrangements कितने होंगे?

In the word (CIVIC), (C) occurs twice and (I) occurs twice. How many distinct arrangements are possible?

Explanation opens after your attempt

Correct Answer

B. (30)

Step 1

Concept

The arrangements are \(\frac{5!}{2!2!}=30\). It is necessary to divide by factorials of repeated letters.

Step 2

Why this answer is correct

The correct answer is B. (30). The arrangements are \(\frac{5!}{2!2!}=30\). It is necessary to divide by factorials of repeated letters.

Step 3

Exam Tip

व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। repeated letters के factorial से भाग देना जरूरी है।

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Class 11 Mathematics - Permutations And Combinations - Combinations Easy Quiz

(8) अलग-अलग ट्रॉफियों को एक शोकेस में सीधी पंक्ति में कितने तरीकों से सजाया जा सकता है?

Concept

Why this answer is correct

Exam Tip

(12) विद्यार्थियों में से कक्षा-प्रतिनिधि और सह-प्रतिनिधि चुनने के तरीके कितने हैं?

Concept

Why this answer is correct

Exam Tip

शब्द (TRAIN) के सभी अक्षरों से कितने अलग शब्द बनाए जा सकते हैं?

Concept

Why this answer is correct

Exam Tip

अंकों (2,3,5,7) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बनेंगी?

Concept

Why this answer is correct

Exam Tip

\({}^{11}P_{2}\) का मान क्या है?

Concept

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(7) अलग-अलग गीतों में से (3) गीतों को पहले, दूसरे और तीसरे स्थान पर बजाने के तरीके कितने हैं?

Concept

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शब्द (PAPER) में (P) दो बार आता है। इसके अलग-अलग अक्षर-क्रम कितने होंगे?

Concept

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(6) अलग-अलग मेजों पर (2) विद्यार्थियों को बैठाने के तरीके कितने हैं?

Concept

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\({}^{7}P_{7}\) का मान क्या होगा?

Concept

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अंकों (0,2,4,6,8) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बन सकती हैं?

Concept

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(5) अलग-अलग पतंगों को (5) बच्चों में एक-एक देकर बाँटने के तरीके कितने हैं?

Concept

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(9) सवालों में से (3) सवालों को पहले, दूसरे और तीसरे क्रम में हल करने के तरीके कितने हैं?

Concept

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शब्द (GATE) के अक्षरों से ऐसे कितने क्रम बनेंगे जिनमें (G) पहला अक्षर हो?

Concept

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(13) अभ्यर्थियों में से अध्यक्ष और उपाध्यक्ष चुनने के तरीके कितने हैं?

Concept

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अंकों (1,5,9) से पुनरावृत्ति की अनुमति होने पर कितनी (3)-अंकीय संख्याएँ बनेंगी?

Concept

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(6) अलग-अलग फोल्डरों को एक कतार में रखना है, पर एक विशेष फोल्डर हमेशा अंतिम स्थान पर रहेगा। कितने तरीके होंगे?

Concept

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\({}^{14}P_{1}\) का मान क्या है?

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शब्द (SMART) के सभी अक्षरों से कितने अलग क्रम बनेंगे?

Concept

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(10) अलग-अलग बैजों में से (4) बैजों को क्रम में लगाने के तरीके कितने हैं?

Concept

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अक्षरों (A,E,I,O,U) में से (2) अक्षरों का क्रम बनाना हो तो कितने तरीके होंगे?

Concept

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