Class 11 Mathematics Easy Quiz

Mathematics Easy Class 11 Level 59

Level 59 • 50/50 questions • 40 seconds per question.

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Question 1/50 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 2/50 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 3/50 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 4/50 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 5/50 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 6/50 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 7/50 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 8/50 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 9/50 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 10/50 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 11/50 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 12/50 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 13/50 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 14/50 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 15/50 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 16/50 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 17/50 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 18/50 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 19/50 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 20/50 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 21/50 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 22/50 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 23/50 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 24/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 59

\({}^{7}P_{0}\) का मान क्या है?

What is the value of \({}^{7}P_{0}\)?

Explanation opens after your attempt
Correct Answer

A. (1)

Step 1

Concept

\({}^{n}P_{0}=1\), so \({}^{7}P_{0}=1\). Choosing zero objects is counted in one way.

Step 2

Why this answer is correct

The correct answer is A. (1). \({}^{n}P_{0}=1\), so \({}^{7}P_{0}=1\). Choosing zero objects is counted in one way.

Step 3

Exam Tip

\({}^{n}P_{0}=1\) होता है इसलिए \({}^{7}P_{0}=1\)। शून्य वस्तु चुनने का एक ही तरीका माना जाता है।

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

(4) अलग-अलग डिब्बों पर (4) अलग-अलग लेबल चिपकाने के तरीके कितने हैं?

In how many ways can (4) different labels be pasted on (4) different boxes?

Explanation opens after your attempt
Correct Answer

B. (24)

Step 1

Concept

(4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.

Step 2

Why this answer is correct

The correct answer is B. (24). (4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.

Step 3

Exam Tip

(4) अलग लेबल (4) अलग डिब्बों पर (4!=24) तरीकों से लगेंगे। matching arrangements में factorial काम आता है।

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

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

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

Explanation opens after your attempt
Correct Answer

C. (360)

Step 1

Concept

There are (6) letters and two identical (O)'s, so \(\frac{6!}{2!}=360\). Use a denominator for a repeated letter.

Step 2

Why this answer is correct

The correct answer is C. (360). There are (6) letters and two identical (O)'s, so \(\frac{6!}{2!}=360\). Use a denominator for a repeated letter.

Step 3

Exam Tip

कुल (6) अक्षर हैं और (O) दो समान हैं इसलिए \(\frac{6!}{2!}=360\)। repeated letter के लिए denominator लगाएं।

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

(5) अलग-अलग शहरों में से (2) शहरों की यात्रा का क्रम चुनने के तरीके कितने हैं?

How many ways are there to choose an ordered trip of (2) cities from (5) different cities?

Explanation opens after your attempt
Correct Answer

D. (20)

Step 1

Concept

There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.

Step 2

Why this answer is correct

The correct answer is D. (20). There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.

Step 3

Exam Tip

पहले शहर के लिए (5) और दूसरे के लिए (4) विकल्प हैं इसलिए (20)। यात्रा क्रम बदलने से योजना बदलती है।

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

अंकों (0,1,2,3) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बन सकती हैं?

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

Explanation opens after your attempt
Correct Answer

A. (18)

Step 1

Concept

There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Do not put (0) in the first place.

Step 2

Why this answer is correct

The correct answer is A. (18). There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Do not put (0) in the first place.

Step 3

Exam Tip

सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प हैं, कुल (18)। पहले स्थान पर (0) न रखें।

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

(3) लाल, नीले और हरे अलग झंडों को एक डंडे पर ऊपर से नीचे लगाने के तरीके कितने हैं?

In how many ways can (3) distinct red, blue and green flags be placed on a pole from top to bottom?

Explanation opens after your attempt
Correct Answer

B. (6)

Step 1

Concept

The ordered arrangement of (3) distinct flags is (3!=6). Changing top-bottom order changes the arrangement.

Step 2

Why this answer is correct

The correct answer is B. (6). The ordered arrangement of (3) distinct flags is (3!=6). Changing top-bottom order changes the arrangement.

Step 3

Exam Tip

(3) अलग झंडों की क्रमबद्ध व्यवस्था (3!=6) है। ऊपर-नीचे का क्रम बदलने से arrangement बदलती है।

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

(11) अभ्यर्थियों में से अध्यक्ष और सचिव चुनने के तरीके कितने हैं?

In how many ways can a president and secretary be chosen from (11) candidates?

Explanation opens after your attempt
Correct Answer

C. (110)

Step 1

Concept

The two posts are distinct, so \({}^{11}P_{2}=11\times10=110\). Use permutation for distinct posts.

Step 2

Why this answer is correct

The correct answer is C. (110). The two posts are distinct, so \({}^{11}P_{2}=11\times10=110\). Use permutation for distinct posts.

Step 3

Exam Tip

दो पद अलग हैं इसलिए \({}^{11}P_{2}=11\times10=110\)। अलग पदों के लिए क्रमचय लगाएं।

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

शब्द (TEAM) के अक्षरों से कितने क्रम बनेंगे यदि अंतिम अक्षर (M) हो?

How many arrangements of the letters of (TEAM) are possible if the last letter is (M)?

Explanation opens after your attempt
Correct Answer

D. (6)

Step 1

Concept

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

Step 3

Exam Tip

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

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

\({}^{5}P_{4}\) का मान क्या है?

What is the value of \({}^{5}P_{4}\)?

Explanation opens after your attempt
Correct Answer

A. (120)

Step 1

Concept

\({}^{5}P_{4}=5\times4\times3\times2=120\). When (r=4), take (4) decreasing factors.

Step 2

Why this answer is correct

The correct answer is A. (120). \({}^{5}P_{4}=5\times4\times3\times2=120\). When (r=4), take (4) decreasing factors.

Step 3

Exam Tip

\({}^{5}P_{4}=5\times4\times3\times2=120\) है। (r=4) हो तो (4) घटते गुणनखंड लें।

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

(6) अलग-अलग किताबों में से (2) किताबों को दाएँ और बाएँ स्थान पर रखने के तरीके कितने हैं?

How many ways are there to place (2) books from (6) different books in right and left positions?

Explanation opens after your attempt
Correct Answer

B. (30)

Step 1

Concept

Right and left positions are distinct, so \({}^{6}P_{2}=30\). When positions are identified, order is counted.

Step 2

Why this answer is correct

The correct answer is B. (30). Right and left positions are distinct, so \({}^{6}P_{2}=30\). When positions are identified, order is counted.

Step 3

Exam Tip

दायाँ और बायाँ स्थान अलग हैं इसलिए \({}^{6}P_{2}=30\)। स्थानों की पहचान अलग हो तो क्रम गिना जाता है।

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

अक्षरों (K,L,M,N) से बिना पुनरावृत्ति कितने (3)-अक्षरी कोड बनेंगे?

How many (3)-letter codes can be formed from letters (K,L,M,N) without repetition?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

\({}^{4}P_{3}=4\times3\times2=24\) codes are possible. In a code, changing positions changes the code.

Step 2

Why this answer is correct

The correct answer is C. (24). \({}^{4}P_{3}=4\times3\times2=24\) codes are possible. In a code, changing positions changes the code.

Step 3

Exam Tip

\({}^{4}P_{3}=4\times3\times2=24\) कोड बनेंगे। कोड में अक्षरों का स्थान बदलने से कोड बदल जाता है।

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

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

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

Explanation opens after your attempt
Correct Answer

D. (3)

Step 1

Concept

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

Step 2

Why this answer is correct

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

Step 3

Exam Tip

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

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

(12) व्यक्तियों में से प्रथम वक्ता और द्वितीय वक्ता चुनने के तरीके कितने हैं?

How many ways are there to choose the first speaker and second speaker from (12) persons?

Explanation opens after your attempt
Correct Answer

A. (132)

Step 1

Concept

First and second speakers are ordered positions, so \({}^{12}P_{2}=132\). Speaking order uses permutation.

Step 2

Why this answer is correct

The correct answer is A. (132). First and second speakers are ordered positions, so \({}^{12}P_{2}=132\). Speaking order uses permutation.

Step 3

Exam Tip

पहला और दूसरा वक्ता क्रम वाले पद हैं इसलिए \({}^{12}P_{2}=132\)। बोलने के क्रम में permutation लगता है।

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

अंकों (4,5,6) से पुनरावृत्ति की अनुमति होने पर कितनी (3)-अंकीय संख्याएँ बनेंगी?

How many (3)-digit numbers can be formed from digits (4,5,6) if repetition is allowed?

Explanation opens after your attempt
Correct Answer

B. (27)

Step 1

Concept

There are (3) choices for each of the three places, so \(3^3=27\). With repetition allowed, choices remain the same at every place.

Step 2

Why this answer is correct

The correct answer is B. (27). There are (3) choices for each of the three places, so \(3^3=27\). With repetition allowed, choices remain the same at every place.

Step 3

Exam Tip

तीनों स्थानों पर (3) विकल्प हैं इसलिए \(3^3=27\)। repetition allowed हो तो हर स्थान पर समान विकल्प रहते हैं।

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

(5) अलग-अलग मोतियों को एक सीधी डोरी पर लगाने के तरीके कितने हैं?

How many ways are there to arrange (5) different beads on a straight string?

Explanation opens after your attempt
Correct Answer

C. (120)

Step 1

Concept

On a straight string, the arrangement of (5) distinct beads is (5!=120). Do not treat it as a circular arrangement.

Step 2

Why this answer is correct

The correct answer is C. (120). On a straight string, the arrangement of (5) distinct beads is (5!=120). Do not treat it as a circular arrangement.

Step 3

Exam Tip

सीधी डोरी पर (5) अलग मोतियों की व्यवस्था (5!=120) है। इसे circular arrangement न मानें।

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

\({}^{4}P_{0}+{}^{4}P_{1}\) का मान क्या है?

What is the value of \({}^{4}P_{0}+{}^{4}P_{1}\)?

Explanation opens after your attempt
Correct Answer

D. (5)

Step 1

Concept

\({}^{4}P_{0}=1\) and \({}^{4}P_{1}=4\), so the sum is (5). Remember standard values.

Step 2

Why this answer is correct

The correct answer is D. (5). \({}^{4}P_{0}=1\) and \({}^{4}P_{1}=4\), so the sum is (5). Remember standard values.

Step 3

Exam Tip

\({}^{4}P_{0}=1\) और \({}^{4}P_{1}=4\), इसलिए योग (5) है। standard values याद रखें।

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

(7) अलग-अलग पेंसिलों में से (2) पेंसिलों को पहले और दूसरे डिब्बे में रखने के तरीके कितने हैं?

How many ways are there to place (2) pencils from (7) different pencils in the first and second boxes?

Explanation opens after your attempt
Correct Answer

A. (42)

Step 1

Concept

There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.

Step 2

Why this answer is correct

The correct answer is A. (42). There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.

Step 3

Exam Tip

पहले डिब्बे के लिए (7) और दूसरे के लिए (6) विकल्प हैं इसलिए (42)। अलग डिब्बों में क्रमचय गिना जाता है।

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

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

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

Explanation opens after your attempt
Correct Answer

B. (120)

Step 1

Concept

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

Step 3

Exam Tip

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

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

(4) अलग-अलग विषयों को सोमवार के (4) पीरियड में लगाने के तरीके कितने हैं?

In how many ways can (4) different subjects be arranged in (4) periods on Monday?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

(4) different subjects can be arranged in (4) different periods in (4!=24) ways. Order matters in a timetable.

Step 2

Why this answer is correct

The correct answer is C. (24). (4) different subjects can be arranged in (4) different periods in (4!=24) ways. Order matters in a timetable.

Step 3

Exam Tip

(4) अलग विषय (4) अलग पीरियड में (4!=24) तरीकों से लगते हैं। timetable में order महत्त्वपूर्ण होता है।

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

अंकों (1,2,3,4,5) से बिना पुनरावृत्ति कितनी (4)-अंकीय संख्याएँ बनेंगी?

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

Explanation opens after your attempt
Correct Answer

D. (120)

Step 1

Concept

For four places, there are \(5\times4\times3\times2=120\) ways. Without repetition, choices keep decreasing.

Step 2

Why this answer is correct

The correct answer is D. (120). For four places, there are \(5\times4\times3\times2=120\) ways. Without repetition, choices keep decreasing.

Step 3

Exam Tip

चार स्थानों के लिए \(5\times4\times3\times2=120\) तरीके हैं। बिना पुनरावृत्ति में विकल्प घटते जाते हैं।

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

(8) चित्रकारों में से (3) को प्रथम, द्वितीय और तृतीय पुरस्कार देने के तरीके कितने हैं?

In how many ways can first, second and third prizes be given to (3) painters from (8) painters?

Explanation opens after your attempt
Correct Answer

A. (336)

Step 1

Concept

The prizes are distinct, so \({}^{8}P_{3}=8\times7\times6=336\). Order is counted for distinct prizes.

Step 2

Why this answer is correct

The correct answer is A. (336). The prizes are distinct, so \({}^{8}P_{3}=8\times7\times6=336\). Order is counted for distinct prizes.

Step 3

Exam Tip

पुरस्कार अलग हैं इसलिए \({}^{8}P_{3}=8\times7\times6=336\)। अलग पुरस्कारों में order गिना जाता है।

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

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

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

Explanation opens after your attempt
Correct Answer

B. (6)

Step 1

Concept

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

Step 2

Why this answer is correct

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

Step 3

Exam Tip

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

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

\({}^{10}P_{0}+{}^{10}P_{1}\) का मान क्या होगा?

What will be the value of \({}^{10}P_{0}+{}^{10}P_{1}\)?

Explanation opens after your attempt
Correct Answer

C. (11)

Step 1

Concept

\({}^{10}P_{0}=1\) and \({}^{10}P_{1}=10\), so the sum is (11). In simple sums, find the values separately first.

Step 2

Why this answer is correct

The correct answer is C. (11). \({}^{10}P_{0}=1\) and \({}^{10}P_{1}=10\), so the sum is (11). In simple sums, find the values separately first.

Step 3

Exam Tip

\({}^{10}P_{0}=1\) और \({}^{10}P_{1}=10\), इसलिए योग (11) है। आसान योग में पहले values अलग निकालें।

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

(6) अलग-अलग पोस्टरों को दीवार पर बाएँ से दाएँ लगाने के तरीके कितने हैं?

In how many ways can (6) different posters be arranged on a wall from left to right?

Explanation opens after your attempt
Correct Answer

D. (720)

Step 1

Concept

The left-to-right arrangement of (6) distinct posters is (6!=720). Use factorial in linear arrangement.

Step 2

Why this answer is correct

The correct answer is D. (720). The left-to-right arrangement of (6) distinct posters is (6!=720). Use factorial in linear arrangement.

Step 3

Exam Tip

बाएँ से दाएँ (6) अलग पोस्टरों की व्यवस्था (6!=720) है। linear arrangement में factorial लें।

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

अंकों (2,3,4,5,6) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बनेंगी?

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

Explanation opens after your attempt
Correct Answer

A. (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 A. (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 हैं और दहाई पर (4) विकल्प, 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 59

(9) अलग-अलग फाइलों में से (4) फाइलों को क्रम से रखने के तरीके कितने हैं?

How many ways are there to arrange (4) files from (9) different files in order?

Explanation opens after your attempt
Correct Answer

B. (3024)

Step 1

Concept

\({}^{9}P_{4}=9\times8\times7\times6=3024\). Use \({}^{n}P_{r}\) for ordered selection.

Step 2

Why this answer is correct

The correct answer is B. (3024). \({}^{9}P_{4}=9\times8\times7\times6=3024\). Use \({}^{n}P_{r}\) for ordered selection.

Step 3

Exam Tip

\({}^{9}P_{4}=9\times8\times7\times6=3024\) 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 59

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

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

Explanation opens after your attempt
Correct Answer

C. (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 C. (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 मानकर गिनती न करें।

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