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class11 MCQ Questions for Class 11

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1581 questions tagged with class11.

Question 721/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (1,4,7,8,9) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बनेंगी?

How many (3)-digit numbers can be formed from digits (1,4,7,8,9) without repetition?

Explanation opens after your attempt
Correct Answer

D. (60)

Step 1

Concept

For three places, there are \(5\times4\times3=60\) ways. Without repetition, each next choice decreases.

Step 2

Why this answer is correct

The correct answer is D. (60). For three places, there are \(5\times4\times3=60\) ways. Without repetition, each next choice decreases.

Step 3

Exam Tip

तीन स्थानों के लिए \(5\times4\times3=60\) तरीके हैं। बिना पुनरावृत्ति में हर अगला विकल्प कम होता है।

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Question 722/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) बच्चों में से (4) बच्चों को आगे की पंक्ति में खड़ा करने के तरीके कितने हैं?

In how many ways can (4) children from (6) children be made to stand in the front row?

Explanation opens after your attempt
Correct Answer

C. (360)

Step 1

Concept

The answer is \({}^{6}P_{4}=6\times5\times4\times3=360\). Order matters in a row.

Step 2

Why this answer is correct

The correct answer is C. (360). The answer is \({}^{6}P_{4}=6\times5\times4\times3=360\). Order matters in a row.

Step 3

Exam Tip

उत्तर \({}^{6}P_{4}=6\times5\times4\times3=360\) है। पंक्ति में क्रम महत्त्वपूर्ण होता है।

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Question 723/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

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

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

Explanation opens after your attempt
Correct Answer

A. (3)

Step 1

Concept

(M) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Repeated letters reduce total arrangements.

Step 2

Why this answer is correct

The correct answer is A. (3). (M) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Repeated letters reduce total arrangements.

Step 3

Exam Tip

(M) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements हैं। समान अक्षर होने पर कुल arrangements घट जाती हैं।

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Question 724/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

अक्षरों (A,B,C,D,E) में से (3) अक्षरों का क्रम बनाना हो तो कुल तरीके कितने होंगे?

If an order of (3) letters is to be made from (A,B,C,D,E), how many total ways are possible?

Explanation opens after your attempt
Correct Answer

B. (60)

Step 1

Concept

There will be \({}^{5}P_{3}=5\times4\times3=60\) ways. Changing the order of letters changes the arrangement.

Step 2

Why this answer is correct

The correct answer is B. (60). There will be \({}^{5}P_{3}=5\times4\times3=60\) ways. Changing the order of letters changes the arrangement.

Step 3

Exam Tip

\({}^{5}P_{3}=5\times4\times3=60\) तरीके होंगे। अक्षरों का क्रम बदलने से arrangement बदलती है।

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Question 725/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(8) अलग-अलग नोटबुकों में से (4) नोटबुकों को मेज पर क्रम से रखने के तरीके कितने हैं?

How many ways are there to arrange (4) notebooks from (8) different notebooks on a table in order?

Explanation opens after your attempt
Correct Answer

A. (1680)

Step 1

Concept

This is \({}^{8}P_{4}=8\times7\times6\times5=1680\). Take (4) decreasing factors for (4) positions.

Step 2

Why this answer is correct

The correct answer is A. (1680). This is \({}^{8}P_{4}=8\times7\times6\times5=1680\). Take (4) decreasing factors for (4) positions.

Step 3

Exam Tip

यह \({}^{8}P_{4}=8\times7\times6\times5=1680\) है। (4) स्थानों के लिए (4) घटते गुणनखंड लें।

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Question 726/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

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

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

Explanation opens after your attempt
Correct Answer

B. (120)

Step 1

Concept

The word (DELHI) has (5) distinct letters, so (5!=120). The number of distinct letters is the base of the factorial.

Step 2

Why this answer is correct

The correct answer is B. (120). The word (DELHI) has (5) distinct letters, so (5!=120). The number of distinct letters is the base of the factorial.

Step 3

Exam Tip

(DELHI) में (5) अलग अक्षर हैं इसलिए (5!=120)। अलग अक्षरों की संख्या ही factorial का आधार होती है।

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Question 727/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

\({}^{9}P_{1}\) का मान क्या है?

What is the value of \({}^{9}P_{1}\)?

Explanation opens after your attempt
Correct Answer

A. (9)

Step 1

Concept

Since \({}^{n}P_{1}=n\), \({}^{9}P_{1}=9\). For one position, count the available objects directly.

Step 2

Why this answer is correct

The correct answer is A. (9). Since \({}^{n}P_{1}=n\), \({}^{9}P_{1}=9\). For one position, count the available objects directly.

Step 3

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{9}P_{1}=9\)। एक स्थान के लिए सीधे उपलब्ध वस्तुएँ गिनें।

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Question 728/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग बैगों को एक कतार में रखना है, पर एक विशेष बैग हमेशा पहले स्थान पर रहेगा। कुल तरीके कितने हैं?

(5) different bags are to be arranged in a row, but one special bag always remains in the first position. How many ways are possible?

Explanation opens after your attempt
Correct Answer

D. (24)

Step 1

Concept

The first position is fixed and the remaining (4) bags can be arranged in (4!=24) ways. Arrange the remaining items after fixing the item.

Step 2

Why this answer is correct

The correct answer is D. (24). The first position is fixed and the remaining (4) bags can be arranged in (4!=24) ways. Arrange the remaining items after fixing the item.

Step 3

Exam Tip

पहला स्थान निश्चित है और शेष (4) बैग (4!=24) तरीकों से लगेंगे। fixed item के बाद बचे items की व्यवस्था करें।

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Question 729/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (2,5,8) से पुनरावृत्ति की अनुमति होने पर कितनी (2)-अंकीय संख्याएँ बन सकती हैं?

How many (2)-digit numbers can be formed from digits (2,5,8) if repetition is allowed?

Explanation opens after your attempt
Correct Answer

C. (9)

Step 1

Concept

There are (3) choices for each place, so \(3\times3=9\). With repetition, choices do not decrease.

Step 2

Why this answer is correct

The correct answer is C. (9). There are (3) choices for each place, so \(3\times3=9\). With repetition, choices do not decrease.

Step 3

Exam Tip

दोनों स्थानों पर (3) विकल्प हैं इसलिए \(3\times3=9\)। पुनरावृत्ति हो तो विकल्प कम नहीं होते।

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Question 730/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(10) खिलाड़ियों में से कप्तान, उपकप्तान और विकेटकीपर चुनने के तरीके कितने हैं?

In how many ways can a captain, vice-captain and wicketkeeper be chosen from (10) players?

Explanation opens after your attempt
Correct Answer

B. (720)

Step 1

Concept

The three posts are distinct, so \({}^{10}P_{3}=10\times9\times8=720\). Order matters for different posts.

Step 2

Why this answer is correct

The correct answer is B. (720). The three posts are distinct, so \({}^{10}P_{3}=10\times9\times8=720\). Order matters for different posts.

Step 3

Exam Tip

तीन पद अलग हैं इसलिए \({}^{10}P_{3}=10\times9\times8=720\)। अलग पदों में order महत्त्वपूर्ण होता है।

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Question 731/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

शब्द (NOTE) के अक्षरों से ऐसे कितने क्रम बनेंगे जिनमें पहला अक्षर (N) हो?

How many arrangements of the letters of (NOTE) are possible if the first letter is (N)?

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. Separate fixed positions 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. Separate fixed positions first.

Step 3

Exam Tip

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

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Question 732/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) रंगों में से (3) रंगों को ऊपर, बीच और नीचे के स्थान पर लगाने के तरीके कितने हैं?

How many ways are there to place (3) colours from (7) colours in top, middle and bottom positions?

Explanation opens after your attempt
Correct Answer

D. (210)

Step 1

Concept

The three positions are distinct, so \({}^{7}P_{3}=7\times6\times5=210\). Use permutation when positions have different names.

Step 2

Why this answer is correct

The correct answer is D. (210). The three positions are distinct, so \({}^{7}P_{3}=7\times6\times5=210\). Use permutation when positions have different names.

Step 3

Exam Tip

तीनों स्थान अलग हैं इसलिए \({}^{7}P_{3}=7\times6\times5=210\)। स्थानों के नाम अलग हों तो क्रमचय लें।

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Question 733/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(4) अलग-अलग फूलों को (4) फूलदानों में एक-एक करके रखने के तरीके कितने हैं?

In how many ways can (4) different flowers be placed one each in (4) vases?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

(4) distinct flowers can be placed in (4) distinct positions in (4!=24) ways. Placing one each counts arrangements.

Step 2

Why this answer is correct

The correct answer is C. (24). (4) distinct flowers can be placed in (4) distinct positions in (4!=24) ways. Placing one each counts arrangements.

Step 3

Exam Tip

(4) अलग फूल (4) अलग स्थानों पर (4!=24) तरीकों से रखे जा सकते हैं। एक-एक वस्तु रखने में व्यवस्था गिनी जाती है।

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Question 734/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (0,3,6,9) से बिना पुनरावृत्ति कितनी (2)-अंकीय संख्याएँ बन सकती हैं?

How many (2)-digit numbers can be formed from digits (0,3,6,9) without repetition?

Explanation opens after your attempt
Correct Answer

B. (9)

Step 1

Concept

Zero cannot be in the tens place, so \(3\times3=9\) numbers are possible. Check the first-place condition in number formation.

Step 2

Why this answer is correct

The correct answer is B. (9). Zero cannot be in the tens place, so \(3\times3=9\) numbers are possible. Check the first-place condition in number formation.

Step 3

Exam Tip

दहाई स्थान पर (0) नहीं आ सकता, इसलिए \(3\times3=9\) संख्याएँ बनेंगी। संख्या बनाते समय पहले स्थान की शर्त देखें।

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Question 735/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

\({}^{6}P_{6}\) का मान क्या होगा?

What will be the value of \({}^{6}P_{6}\)?

Explanation opens after your attempt
Correct Answer

A. (720)

Step 1

Concept

When (r=n), \({}^{n}P_{n}=n!\), so \({}^{6}P_{6}=720\). Use factorial when all are selected.

Step 2

Why this answer is correct

The correct answer is A. (720). When (r=n), \({}^{n}P_{n}=n!\), so \({}^{6}P_{6}=720\). Use factorial when all are selected.

Step 3

Exam Tip

जब (r=n) हो तो \({}^{n}P_{n}=n!\), इसलिए \({}^{6}P_{6}=720\)। पूरा चयन होने पर factorial लें।

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Question 736/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(5) अलग-अलग कुर्सियों पर (3) लोगों को कितने तरीकों से बैठाया जा सकता है?

In how many ways can (3) people be seated on (5) different chairs?

Explanation opens after your attempt
Correct Answer

D. (60)

Step 1

Concept

The chairs are distinct, so \({}^{5}P_{3}=5\times4\times3=60\). Changing seats changes the arrangement.

Step 2

Why this answer is correct

The correct answer is D. (60). The chairs are distinct, so \({}^{5}P_{3}=5\times4\times3=60\). Changing seats changes the arrangement.

Step 3

Exam Tip

कुर्सियाँ अलग हैं इसलिए \({}^{5}P_{3}=5\times4\times3=60\)। सीटों का क्रम बदलने से व्यवस्था बदलती है।

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Question 737/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

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

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

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 repeated letters.

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 repeated letters.

Step 3

Exam Tip

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

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Question 738/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(6) अलग-अलग पुरस्कारों में से (2) पुरस्कारों को पहले और दूसरे स्थान पर रखने के तरीके कितने हैं?

How many ways are there to place (2) prizes from (6) different prizes in first and second positions?

Explanation opens after your attempt
Correct Answer

B. (30)

Step 1

Concept

There are (6) choices for the first position and (5) for the second, so (30). Order matters in distinct positions.

Step 2

Why this answer is correct

The correct answer is B. (30). There are (6) choices for the first position and (5) for the second, so (30). Order matters in distinct positions.

Step 3

Exam Tip

पहले स्थान के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए (30)। अलग स्थानों में क्रम महत्त्वपूर्ण होता है।

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Question 739/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

\({}^{8}P_{2}\) का मान क्या है?

What is the value of \({}^{8}P_{2}\)?

Explanation opens after your attempt
Correct Answer

A. (56)

Step 1

Concept

\({}^{8}P_{2}=8\times7=56\). When (r=2), take two decreasing factors.

Step 2

Why this answer is correct

The correct answer is A. (56). \({}^{8}P_{2}=8\times7=56\). When (r=2), take two decreasing factors.

Step 3

Exam Tip

\({}^{8}P_{2}=8\times7=56\) होता है। (r=2) हो तो दो घटते गुणनखंड लें।

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Question 740/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

अंकों (1,2,4,5) से बिना पुनरावृत्ति कितनी (2)-अंकीय संख्याएँ बनेंगी?

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

Explanation opens after your attempt
Correct Answer

D. (12)

Step 1

Concept

There are (4) choices for tens and (3) for units, so \(4\times3=12\). Changing places changes the number.

Step 2

Why this answer is correct

The correct answer is D. (12). There are (4) choices for tens and (3) for units, so \(4\times3=12\). Changing places changes the number.

Step 3

Exam Tip

दहाई के लिए (4) और इकाई के लिए (3) विकल्प हैं इसलिए \(4\times3=12\)। स्थान बदलने से संख्या बदलती है।

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Question 741/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

शब्द (LAMP) के सभी अक्षरों से कितनी अलग व्यवस्थाएँ बनेंगी?

How many distinct arrangements can be formed using all letters of the word (LAMP)?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

The word (LAMP) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.

Step 2

Why this answer is correct

The correct answer is C. (24). The word (LAMP) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.

Step 3

Exam Tip

(LAMP) में (4) अलग अक्षर हैं इसलिए (4!=24)। अलग अक्षरों की पूरी व्यवस्था factorial से मिलती है।

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Question 742/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(9) छात्रों में से (2) छात्रों को प्रथम और द्वितीय स्थान देने के तरीके कितने हैं?

In how many ways can first and second positions be given to (2) students from (9) students?

Explanation opens after your attempt
Correct Answer

B. (72)

Step 1

Concept

The positions are ordered, so \({}^{9}P_{2}=9\times8=72\). Use permutation in ranking questions.

Step 2

Why this answer is correct

The correct answer is B. (72). The positions are ordered, so \({}^{9}P_{2}=9\times8=72\). Use permutation in ranking questions.

Step 3

Exam Tip

स्थान क्रम वाले हैं इसलिए \({}^{9}P_{2}=9\times8=72\)। रैंकिंग में क्रमचय का प्रयोग करें।

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Question 743/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

(7) अलग-अलग प्रमाणपत्रों को एक पंक्ति में सजाने के कुल तरीके कितने हैं?

How many total ways are there to arrange (7) different certificates in a row?

Explanation opens after your attempt
Correct Answer

A. (5040)

Step 1

Concept

The arrangement of (7) distinct objects is (7!=5040). Use factorial when all objects are distinct.

Step 2

Why this answer is correct

The correct answer is A. (5040). The arrangement of (7) distinct objects is (7!=5040). Use factorial when all objects are distinct.

Step 3

Exam Tip

(7) अलग वस्तुओं की व्यवस्था (7!=5040) होती है। सभी वस्तुएँ अलग हों तो factorial लगाएं।

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Question 744/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

अक्षरों (X,Y,Z,W) में से (2) अक्षर लेकर पासवर्ड के पहले दो स्थान कितने तरीकों से भरे जा सकते हैं?

In how many ways can the first two places of a password be filled using (2) letters from (X,Y,Z,W)?

Explanation opens after your attempt
Correct Answer

B. (12)

Step 1

Concept

There are (4) choices for the first place and (3) for the second, so \(4\times3=12\). In passwords, changing positions changes the arrangement.

Step 2

Why this answer is correct

The correct answer is B. (12). There are (4) choices for the first place and (3) for the second, so \(4\times3=12\). In passwords, changing positions changes the arrangement.

Step 3

Exam Tip

पहले स्थान के लिए (4) और दूसरे के लिए (3) विकल्प हैं, इसलिए \(4\times3=12\)। पासवर्ड में स्थान बदलने से arrangement बदलती है।

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Question 745/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(6) अलग-अलग मोबाइल कवर में से (3) कवर को दुकान की शेल्फ पर क्रम से रखने के तरीके कितने हैं?

How many ways are there to arrange (3) mobile covers from (6) different mobile covers on a shop shelf in order?

Explanation opens after your attempt
Correct Answer

A. (120)

Step 1

Concept

\({}^{6}P_{3}=6\times5\times4=120\). Use permutation for ordered arrangements.

Step 2

Why this answer is correct

The correct answer is A. (120). \({}^{6}P_{3}=6\times5\times4=120\). Use permutation for ordered arrangements.

Step 3

Exam Tip

\({}^{6}P_{3}=6\times5\times4=120\) होता है। क्रम वाली व्यवस्था में permutation लगाएं।

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Question 746/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(8) अलग-अलग पुस्तकों में से (3) पुस्तकों को मेज पर क्रम से रखने के तरीके कितने हैं?

How many ways are there to place (3) books from (8) different books on a table in order?

Explanation opens after your attempt
Correct Answer

B. (336)

Step 1

Concept

The answer is \({}^{8}P_{3}=8\times7\times6=336\). Changing the order of selected books changes the way.

Step 2

Why this answer is correct

The correct answer is B. (336). The answer is \({}^{8}P_{3}=8\times7\times6=336\). Changing the order of selected books changes the way.

Step 3

Exam Tip

उत्तर \({}^{8}P_{3}=8\times7\times6=336\) है। चुनी गई पुस्तकों का क्रम बदलने से तरीका बदलता है।

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Question 747/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{6}P_{2}\) का मान क्या है?

What is the value of \({}^{6}P_{2}\)?

Explanation opens after your attempt
Correct Answer

A. (30)

Step 1

Concept

\({}^{6}P_{2}=6\times5=30\). For two places, take (n) and then (n-1).

Step 2

Why this answer is correct

The correct answer is A. (30). \({}^{6}P_{2}=6\times5=30\). For two places, take (n) and then (n-1).

Step 3

Exam Tip

\({}^{6}P_{2}=6\times5=30\) है। दो स्थान हों तो पहले (n), फिर (n-1) लें।

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Question 748/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (MOON) में (O) दो बार आता है। इसके अलग-अलग arrangements कितने होंगे?

The word (MOON) has (O) twice. How many distinct arrangements are possible?

Explanation opens after your attempt
Correct Answer

D. (12)

Step 1

Concept

There are (4) letters and two identical (O)'s, so \(\frac{4!}{2!}=12\). Use a denominator for repeated letters.

Step 2

Why this answer is correct

The correct answer is D. (12). There are (4) letters and two identical (O)'s, so \(\frac{4!}{2!}=12\). Use a denominator for repeated letters.

Step 3

Exam Tip

कुल अक्षर (4) हैं और (O) दो समान हैं इसलिए \(\frac{4!}{2!}=12\)। repeated letters में denominator लगाएं।

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Question 749/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

अक्षरों (P,Q,R,S,T) में से (2) अक्षर लेकर कितने ordered pairs बनाए जा सकते हैं?

How many ordered pairs can be formed by taking (2) letters from (P,Q,R,S,T)?

Explanation opens after your attempt
Correct Answer

C. (20)

Step 1

Concept

For (2) ordered positions from (5) letters, \({}^{5}P_{2}=20\). In ordered pairs, ((P,Q)) and ((Q,P)) are different.

Step 2

Why this answer is correct

The correct answer is C. (20). For (2) ordered positions from (5) letters, \({}^{5}P_{2}=20\). In ordered pairs, ((P,Q)) and ((Q,P)) are different.

Step 3

Exam Tip

(5) अक्षरों से (2) ordered positions के लिए \({}^{5}P_{2}=20\)। ordered pairs में ((P,Q)) और ((Q,P)) अलग होते हैं।

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Question 750/1581 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(4) अलग-अलग चित्रों को एक दीवार पर बाएँ से दाएँ कितने तरीकों से लगाया जा सकता है?

In how many ways can (4) different pictures be placed on a wall from left to right?

Explanation opens after your attempt
Correct Answer

B. (24)

Step 1

Concept

Left-to-right order matters, so (4!=24). Changing the order changes the arrangement.

Step 2

Why this answer is correct

The correct answer is B. (24). Left-to-right order matters, so (4!=24). Changing the order changes the arrangement.

Step 3

Exam Tip

बाएँ से दाएँ क्रम महत्त्वपूर्ण है इसलिए (4!=24)। क्रम बदलते ही arrangement बदल जाती है।

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