Class 11 Mathematics Easy Quiz

Mathematics Easy Class 11 Level 58

Level 58 • 50/50 questions • 40 seconds per question.

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Question 1/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

यदि (5) अलग-अलग किताबों को एक पंक्ति में सजाना हो तो कितने तरीके होंगे?

If (5) different books are arranged in a row, how many ways are possible?

Explanation opens after your attempt
Correct Answer

A. (120)

Step 1

Concept

The arrangement of (5) distinct objects is (5!). In exams, use factorial for distinct objects.

Step 2

Why this answer is correct

The correct answer is A. (120). The arrangement of (5) distinct objects is (5!). In exams, use factorial for distinct objects.

Step 3

Exam Tip

(5) अलग वस्तुओं की व्यवस्था (5!) होती है। परीक्षा में अलग-अलग वस्तुओं के लिए factorial लगाएं।

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Question 2/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(6) विद्यार्थियों में से (3) विद्यार्थियों को अध्यक्ष, सचिव और कोषाध्यक्ष के पदों पर कितने तरीकों से चुना जा सकता है?

In how many ways can (3) students be chosen from (6) students for the posts of president, secretary and treasurer?

Explanation opens after your attempt
Correct Answer

B. (120)

Step 1

Concept

Here the posts are different, so order matters and the answer is \({}^{6}P_{3}=120\). In exams, use permutation when posts are distinct.

Step 2

Why this answer is correct

The correct answer is B. (120). Here the posts are different, so order matters and the answer is \({}^{6}P_{3}=120\). In exams, use permutation when posts are distinct.

Step 3

Exam Tip

यहाँ पद अलग हैं इसलिए क्रम महत्त्वपूर्ण है और उत्तर \({}^{6}P_{3}=120\) है। परीक्षा में पद अलग हों तो permutation लें।

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Question 3/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

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

How many different words can be formed using all letters of the word (CAT)?

Explanation opens after your attempt
Correct Answer

C. (6)

Step 1

Concept

The word (CAT) has (3) distinct letters, so (3!=6) words are possible. For small words, arrange all letters using factorial.

Step 2

Why this answer is correct

The correct answer is C. (6). The word (CAT) has (3) distinct letters, so (3!=6) words are possible. For small words, arrange all letters using factorial.

Step 3

Exam Tip

(CAT) में (3) अलग अक्षर हैं इसलिए (3!=6) शब्द बनेंगे। छोटे शब्दों में सभी अक्षरों की व्यवस्था सीधे factorial से करें।

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Question 4/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(4) अलग-अलग रंगों के झंडों को एक सीधी रेखा में कितने तरीकों से लगाया जा सकता है?

In how many ways can (4) different coloured flags be placed in a straight line?

Explanation opens after your attempt
Correct Answer

D. (24)

Step 1

Concept

The arrangement of (4) distinct flags is (4!=24). Placing in a straight line is a linear permutation.

Step 2

Why this answer is correct

The correct answer is D. (24). The arrangement of (4) distinct flags is (4!=24). Placing in a straight line is a linear permutation.

Step 3

Exam Tip

(4) अलग झंडों की व्यवस्था (4!=24) होती है। सीधी रेखा में लगाना linear permutation है।

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Question 5/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{7}P_{2}\) का मान क्या है?

What is the value of \({}^{7}P_{2}\)?

Explanation opens after your attempt
Correct Answer

A. (42)

Step 1

Concept

\({}^{7}P_{2}=7\times6=42\). For small (r), multiply (r) decreasing factors.

Step 2

Why this answer is correct

The correct answer is A. (42). \({}^{7}P_{2}=7\times6=42\). For small (r), multiply (r) decreasing factors.

Step 3

Exam Tip

\({}^{7}P_{2}=7\times6=42\) होता है। छोटे (r) के लिए घटते हुए (r) गुणनखंड लें।

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Question 6/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(8) खिलाड़ियों में से कप्तान और उपकप्तान कितने तरीकों से चुने जा सकते हैं?

In how many ways can a captain and vice-captain be chosen from (8) players?

Explanation opens after your attempt
Correct Answer

B. (56)

Step 1

Concept

Captain and vice-captain are distinct posts, so \({}^{8}P_{2}=56\). When roles differ, order matters.

Step 2

Why this answer is correct

The correct answer is B. (56). Captain and vice-captain are distinct posts, so \({}^{8}P_{2}=56\). When roles differ, order matters.

Step 3

Exam Tip

कप्तान और उपकप्तान अलग पद हैं इसलिए \({}^{8}P_{2}=56\) होगा। जब जिम्मेदारी अलग हो तो क्रम महत्त्वपूर्ण होता है।

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Question 7/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

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

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

Explanation opens after your attempt
Correct Answer

C. (6)

Step 1

Concept

The arrangement of three distinct digits is (3!=6). Without repetition, choices decrease place by place.

Step 2

Why this answer is correct

The correct answer is C. (6). The arrangement of three distinct digits is (3!=6). Without repetition, choices decrease place by place.

Step 3

Exam Tip

तीनों अलग अंकों की व्यवस्था (3!=6) है। बिना पुनरावृत्ति में हर स्थान के विकल्प घटते हैं।

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Question 8/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

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

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

Explanation opens after your attempt
Correct Answer

D. (120)

Step 1

Concept

The number of ways to seat (5) people in (5) places is (5!=120). Remember factorial for seating in a row.

Step 2

Why this answer is correct

The correct answer is D. (120). The number of ways to seat (5) people in (5) places is (5!=120). Remember factorial for seating in a row.

Step 3

Exam Tip

(5) लोगों को (5) स्थानों पर बैठाने के तरीके (5!=120) हैं। seating in a row में factorial याद रखें।

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Question 9/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{5}P_{1}\) का मान क्या होगा?

What will be the value of \({}^{5}P_{1}\)?

Explanation opens after your attempt
Correct Answer

A. (5)

Step 1

Concept

Since \({}^{n}P_{1}=n\), \({}^{5}P_{1}=5\). Choosing and arranging one object gives (n) ways.

Step 2

Why this answer is correct

The correct answer is A. (5). Since \({}^{n}P_{1}=n\), \({}^{5}P_{1}=5\). Choosing and arranging one object gives (n) ways.

Step 3

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{5}P_{1}=5\)। एक वस्तु चुनकर व्यवस्थित करने में (n) तरीके होते हैं।

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Question 10/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (DOG) के अक्षरों को कितने तरीकों से व्यवस्थित किया जा सकता है?

In how many ways can the letters of the word (DOG) be arranged?

Explanation opens after your attempt
Correct Answer

B. (6)

Step 1

Concept

The word (DOG) has (3) distinct letters, so there are (3!=6) arrangements. Use factorial when letters are distinct.

Step 2

Why this answer is correct

The correct answer is B. (6). The word (DOG) has (3) distinct letters, so there are (3!=6) arrangements. Use factorial when letters are distinct.

Step 3

Exam Tip

(DOG) में (3) अलग अक्षर हैं इसलिए (3!=6) व्यवस्थाएँ हैं। अक्षर अलग हों तो सीधे factorial प्रयोग करें।

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Question 11/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(9) वस्तुओं में से (2) वस्तुओं को क्रम में रखने के तरीके कितने हैं?

How many ways are there to arrange (2) objects selected from (9) objects in order?

Explanation opens after your attempt
Correct Answer

C. (72)

Step 1

Concept

The answer is \({}^{9}P_{2}=9\times8=72\). After the first place, the second choice decreases by one.

Step 2

Why this answer is correct

The correct answer is C. (72). The answer is \({}^{9}P_{2}=9\times8=72\). After the first place, the second choice decreases by one.

Step 3

Exam Tip

उत्तर \({}^{9}P_{2}=9\times8=72\) है। पहले स्थान के बाद दूसरा विकल्प एक कम हो जाता है।

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Question 12/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

किसी दौड़ में (7) धावकों में से प्रथम और द्वितीय स्थान कितने तरीकों से आ सकते हैं?

In a race with (7) runners, in how many ways can first and second places occur?

Explanation opens after your attempt
Correct Answer

D. (42)

Step 1

Concept

First and second places are ordered, so \({}^{7}P_{2}=42\). Ranking questions use permutation.

Step 2

Why this answer is correct

The correct answer is D. (42). First and second places are ordered, so \({}^{7}P_{2}=42\). Ranking questions use permutation.

Step 3

Exam Tip

प्रथम और द्वितीय स्थान क्रम वाले हैं इसलिए \({}^{7}P_{2}=42\) होगा। रैंकिंग में permutation लगता है।

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Question 13/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{6}P_{3}\) का मान क्या है?

What is the value of \({}^{6}P_{3}\)?

Explanation opens after your attempt
Correct Answer

B. (120)

Step 1

Concept

\({}^{6}P_{3}=6\times5\times4=120\). When (r=3), take three decreasing factors.

Step 2

Why this answer is correct

The correct answer is B. (120). \({}^{6}P_{3}=6\times5\times4=120\). When (r=3), take three decreasing factors.

Step 3

Exam Tip

\({}^{6}P_{3}=6\times5\times4=120\) है। (r=3) हो तो तीन घटते गुणनखंड लें।

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Question 14/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(3) अलग-अलग पुरस्कार (5) विद्यार्थियों में इस प्रकार बाँटने हैं कि एक विद्यार्थी को अधिकतम एक पुरस्कार मिले। कितने तरीके होंगे?

(3) different prizes are to be distributed among (5) students so that a student gets at most one prize. How many ways are possible?

Explanation opens after your attempt
Correct Answer

C. (60)

Step 1

Concept

The prizes are distinct and students cannot repeat, so \({}^{5}P_{3}=60\). Distribution of distinct prizes makes order important.

Step 2

Why this answer is correct

The correct answer is C. (60). The prizes are distinct and students cannot repeat, so \({}^{5}P_{3}=60\). Distribution of distinct prizes makes order important.

Step 3

Exam Tip

पुरस्कार अलग हैं और विद्यार्थी दोहराए नहीं जा सकते इसलिए \({}^{5}P_{3}=60\)। अलग पुरस्कारों के वितरण में क्रम महत्त्वपूर्ण होता है।

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Question 15/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (MATH) के सभी अक्षरों को कितने तरीकों से क्रमबद्ध किया जा सकता है?

In how many ways can all letters of the word (MATH) be ordered?

Explanation opens after your attempt
Correct Answer

D. (24)

Step 1

Concept

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

Step 3

Exam Tip

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

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Question 16/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

यदि \({}^{n}P_{0}\) पूछा जाए तो उसका मान क्या होता है?

If \({}^{n}P_{0}\) is asked, what is its value?

Explanation opens after your attempt
Correct Answer

A. (1)

Step 1

Concept

\({}^{n}P_{0}=1\) because there is one way to choose nothing. Remember this standard result.

Step 2

Why this answer is correct

The correct answer is A. (1). \({}^{n}P_{0}=1\) because there is one way to choose nothing. Remember this standard result.

Step 3

Exam Tip

\({}^{n}P_{0}=1\) होता है क्योंकि कोई वस्तु न चुनने का एक तरीका होता है। यह standard result याद रखें।

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Question 17/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(4) मित्रों को (2) विशेष सीटों पर कितने तरीकों से बैठाया जा सकता है?

In how many ways can (4) friends be seated on (2) special seats?

Explanation opens after your attempt
Correct Answer

B. (12)

Step 1

Concept

Order matters on two special seats, so \({}^{4}P_{2}=12\). If seats are distinct, use permutation.

Step 2

Why this answer is correct

The correct answer is B. (12). Order matters on two special seats, so \({}^{4}P_{2}=12\). If seats are distinct, use permutation.

Step 3

Exam Tip

दो विशेष सीटों पर क्रम महत्त्वपूर्ण है इसलिए \({}^{4}P_{2}=12\)। सीटें अलग हों तो permutation लें।

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Question 18/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

अक्षरों (A,B,C,D) में से (3) अक्षर लेकर कितने क्रम बनाए जा सकते हैं?

How many ordered arrangements can be made by taking (3) letters from (A,B,C,D)?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

The answer is \({}^{4}P_{3}=4\times3\times2=24\). Choosing and arranging in order is permutation.

Step 2

Why this answer is correct

The correct answer is C. (24). The answer is \({}^{4}P_{3}=4\times3\times2=24\). Choosing and arranging in order is permutation.

Step 3

Exam Tip

उत्तर \({}^{4}P_{3}=4\times3\times2=24\) है। चुनना और क्रम में रखना permutation है।

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Question 19/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

अंकों (1,3,5,7,9) से बिना पुनरावृत्ति कितनी (1)-अंकीय संख्याएँ बनेंगी?

How many (1)-digit numbers can be formed from digits (1,3,5,7,9) without repetition?

Explanation opens after your attempt
Correct Answer

D. (5)

Step 1

Concept

There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.

Step 2

Why this answer is correct

The correct answer is D. (5). There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.

Step 3

Exam Tip

एक स्थान के लिए (5) विकल्प हैं इसलिए उत्तर (5) है। (1)-स्थान वाली व्यवस्था में उपलब्ध विकल्प ही उत्तर होते हैं।

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Question 20/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(7) अलग-अलग पेन में से (4) पेन को क्रम से रखने के तरीके कितने हैं?

How many ways are there to arrange (4) pens selected from (7) different pens in order?

Explanation opens after your attempt
Correct Answer

A. (840)

Step 1

Concept

\({}^{7}P_{4}=7\times6\times5\times4=840\). For (r) places, multiply (r) decreasing factors.

Step 2

Why this answer is correct

The correct answer is A. (840). \({}^{7}P_{4}=7\times6\times5\times4=840\). For (r) places, multiply (r) decreasing factors.

Step 3

Exam Tip

\({}^{7}P_{4}=7\times6\times5\times4=840\)। (r) स्थानों के लिए (r) घटते गुणनखंड लें।

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Question 21/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (SUN) के अक्षरों से कितनी व्यवस्थाएँ बनती हैं?

How many arrangements are formed from the letters of the word (SUN)?

Explanation opens after your attempt
Correct Answer

B. (6)

Step 1

Concept

The word (SUN) has (3) distinct letters, so (3!=6). Count the letters and apply factorial for small words.

Step 2

Why this answer is correct

The correct answer is B. (6). The word (SUN) has (3) distinct letters, so (3!=6). Count the letters and apply factorial for small words.

Step 3

Exam Tip

(SUN) में (3) अलग अक्षर हैं इसलिए (3!=6)। छोटे शब्दों में अक्षर गिनकर factorial लगाएं।

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Question 22/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{10}P_{1}\) का मान क्या है?

What is the value of \({}^{10}P_{1}\)?

Explanation opens after your attempt
Correct Answer

C. (10)

Step 1

Concept

Since \({}^{n}P_{1}=n\), \({}^{10}P_{1}=10\). For one position, write (n) directly.

Step 2

Why this answer is correct

The correct answer is C. (10). Since \({}^{n}P_{1}=n\), \({}^{10}P_{1}=10\). For one position, write (n) directly.

Step 3

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{10}P_{1}=10\)। एक पद की व्यवस्था में सीधे (n) लिखें।

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Question 23/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(6) चित्रों में से (2) चित्रों को पहले और दूसरे स्थान पर लगाने के तरीके कितने हैं?

In how many ways can (2) pictures from (6) pictures be placed in first and second positions?

Explanation opens after your attempt
Correct Answer

D. (30)

Step 1

Concept

There are (6) choices for the first position and (5) for the second, so \(6\times5=30\). Count order when positions are distinct.

Step 2

Why this answer is correct

The correct answer is D. (30). There are (6) choices for the first position and (5) for the second, so \(6\times5=30\). Count order when positions are distinct.

Step 3

Exam Tip

पहले स्थान के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए \(6\times5=30\)। स्थान अलग-अलग हों तो क्रम गिनें।

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Question 24/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

यदि (n=5) और (r=2) हो तो \({}^{n}P_{r}\) का मान क्या है?

If (n=5) and (r=2), what is the value of \({}^{n}P_{r}\)?

Explanation opens after your attempt
Correct Answer

A. (20)

Step 1

Concept

\({}^{5}P_{2}=5\times4=20\). Start from (n) and take (r) decreasing factors.

Step 2

Why this answer is correct

The correct answer is A. (20). \({}^{5}P_{2}=5\times4=20\). Start from (n) and take (r) decreasing factors.

Step 3

Exam Tip

\({}^{5}P_{2}=5\times4=20\)। (n) से शुरू करके (r) तक घटते गुणनखंड लें।

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Question 25/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(5) अलग-अलग कार्डों में से (3) कार्डों को क्रम में लगाने के तरीके कितने हैं?

How many ways are there to arrange (3) cards selected from (5) different cards in order?

Explanation opens after your attempt
Correct Answer

B. (60)

Step 1

Concept

This is \({}^{5}P_{3}=5\times4\times3=60\). Changing the order of cards changes the arrangement.

Step 2

Why this answer is correct

The correct answer is B. (60). This is \({}^{5}P_{3}=5\times4\times3=60\). Changing the order of cards changes the arrangement.

Step 3

Exam Tip

यह \({}^{5}P_{3}=5\times4\times3=60\) है। कार्डों का क्रम बदलने से व्यवस्था बदलती है।

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Question 26/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (BOOK) में (4) अक्षर हैं और (O) दो बार आता है। इसके अलग-अलग arrangements कितने होंगे?

The word (BOOK) has (4) letters and (O) occurs twice. How many distinct arrangements are possible?

Explanation opens after your attempt
Correct Answer

C. (12)

Step 1

Concept

Because two (O)'s are identical, arrangements are \(\frac{4!}{2!}=12\). For identical objects, divide by their factorial.

Step 2

Why this answer is correct

The correct answer is C. (12). Because two (O)'s are identical, arrangements are \(\frac{4!}{2!}=12\). For identical objects, divide by their factorial.

Step 3

Exam Tip

दो समान (O) होने से व्यवस्थाएँ \(\frac{4!}{2!}=12\) होंगी। समान वस्तुओं के लिए factorial से भाग दें।

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Question 27/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(4) लोगों में से (3) लोगों को एक पंक्ति में कितने तरीकों से खड़ा किया जा सकता है?

In how many ways can (3) people be made to stand in a row from (4) people?

Explanation opens after your attempt
Correct Answer

D. (24)

Step 1

Concept

This is \({}^{4}P_{3}=4\times3\times2=24\). Standing in a row is an ordered arrangement.

Step 2

Why this answer is correct

The correct answer is D. (24). This is \({}^{4}P_{3}=4\times3\times2=24\). Standing in a row is an ordered arrangement.

Step 3

Exam Tip

यह \({}^{4}P_{3}=4\times3\times2=24\) है। पंक्ति में खड़ा करना ordered arrangement है।

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Question 28/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(6) अलग-अलग गेंदों को एक पंक्ति में रखने के तरीके कितने होंगे?

How many ways are possible to place (6) different balls in a row?

Explanation opens after your attempt
Correct Answer

A. (720)

Step 1

Concept

The arrangement of (6) distinct balls is (6!=720). If all objects are distinct, total arrangements are factorial.

Step 2

Why this answer is correct

The correct answer is A. (720). The arrangement of (6) distinct balls is (6!=720). If all objects are distinct, total arrangements are factorial.

Step 3

Exam Tip

(6) अलग गेंदों की व्यवस्था (6!=720) है। सभी वस्तुएँ अलग हों तो कुल व्यवस्था factorial होती है।

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Question 29/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

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

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

Explanation opens after your attempt
Correct Answer

B. (4)

Step 1

Concept

Zero cannot be in the tens place, so there are (2) choices for tens and (2) for units, total (4). In number formation, check zero in the first place.

Step 2

Why this answer is correct

The correct answer is B. (4). Zero cannot be in the tens place, so there are (2) choices for tens and (2) for units, total (4). In number formation, check zero in the first place.

Step 3

Exam Tip

दहाई स्थान पर (0) नहीं आ सकता इसलिए (2) विकल्प और इकाई पर (2) विकल्प हैं, कुल (4)। संख्या बनाते समय पहले स्थान पर (0) की जाँच करें।

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Question 30/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{8}P_{0}\) का मान क्या होगा?

What will be the value of \({}^{8}P_{0}\)?

Explanation opens after your attempt
Correct Answer

C. (1)

Step 1

Concept

\({}^{n}P_{0}=1\), so \({}^{8}P_{0}=1\). The arrangement of zero objects is counted as one.

Step 2

Why this answer is correct

The correct answer is C. (1). \({}^{n}P_{0}=1\), so \({}^{8}P_{0}=1\). The arrangement of zero objects is counted as one.

Step 3

Exam Tip

\({}^{n}P_{0}=1\) इसलिए \({}^{8}P_{0}=1\)। शून्य वस्तुओं की arrangement एक मानी जाती है।

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Question 31/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(3) अलग-अलग सिक्कों को एक क्रम में रखने के तरीके कितने हैं?

How many ways are there to place (3) different coins in an order?

Explanation opens after your attempt
Correct Answer

D. (6)

Step 1

Concept

The arrangement of (3) distinct coins is (3!=6). Use factorial for full arrangement of distinct objects.

Step 2

Why this answer is correct

The correct answer is D. (6). The arrangement of (3) distinct coins is (3!=6). Use factorial for full arrangement of distinct objects.

Step 3

Exam Tip

(3) अलग सिक्कों की व्यवस्था (3!=6) है। अलग objects की पूरी arrangement factorial से करें।

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Question 32/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(10) विद्यार्थियों में से मॉनिटर और सह-मॉनिटर चुनने के तरीके कितने हैं?

How many ways are there to choose a monitor and assistant monitor from (10) students?

Explanation opens after your attempt
Correct Answer

A. (90)

Step 1

Concept

There are two distinct posts, so \({}^{10}P_{2}=10\times9=90\). Order matters for different designations.

Step 2

Why this answer is correct

The correct answer is A. (90). There are two distinct posts, so \({}^{10}P_{2}=10\times9=90\). Order matters for different designations.

Step 3

Exam Tip

दो अलग पद हैं इसलिए \({}^{10}P_{2}=10\times9=90\)। अलग designation में order महत्त्वपूर्ण होता है।

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Question 33/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (RAVI) के अक्षरों की कुल व्यवस्थाएँ कितनी हैं?

What is the total number of arrangements of the letters of the word (RAVI)?

Explanation opens after your attempt
Correct Answer

B. (24)

Step 1

Concept

The word (RAVI) has (4) distinct letters, so (4!=24). Apply factorial when all letters are distinct.

Step 2

Why this answer is correct

The correct answer is B. (24). The word (RAVI) has (4) distinct letters, so (4!=24). Apply factorial when all letters are distinct.

Step 3

Exam Tip

(RAVI) में (4) अलग अक्षर हैं इसलिए (4!=24)। सभी distinct letters हों तो factorial लगाएं।

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Question 34/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(6) विषयों में से (2) विषयों को पहले और दूसरे पीरियड में लगाने के तरीके कितने हैं?

In how many ways can (2) subjects from (6) subjects be placed in the first and second periods?

Explanation opens after your attempt
Correct Answer

C. (30)

Step 1

Concept

There are (6) choices for the first period and (5) for the second, so (30). Timetable order uses permutation.

Step 2

Why this answer is correct

The correct answer is C. (30). There are (6) choices for the first period and (5) for the second, so (30). Timetable order uses permutation.

Step 3

Exam Tip

पहले पीरियड के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए (30)। timetable order में permutation लगता है।

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Question 35/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{4}P_{4}\) का मान क्या है?

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

Explanation opens after your attempt
Correct Answer

D. (24)

Step 1

Concept

\({}^{4}P_{4}=4!=24\). When (r=n), \({}^{n}P_{n}=n!\).

Step 2

Why this answer is correct

The correct answer is D. (24). \({}^{4}P_{4}=4!=24\). When (r=n), \({}^{n}P_{n}=n!\).

Step 3

Exam Tip

\({}^{4}P_{4}=4!=24\) होता है। जब (r=n) हो तो \({}^{n}P_{n}=n!\)।

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Question 36/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

अंकों (3,4,5,6) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बन सकती हैं?

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

Explanation opens after your attempt
Correct Answer

A. (24)

Step 1

Concept

For three places, there are \(4\times3\times2=24\) ways. Without repetition, choices decrease at each next place.

Step 2

Why this answer is correct

The correct answer is A. (24). For three places, there are \(4\times3\times2=24\) ways. Without repetition, choices decrease at each next place.

Step 3

Exam Tip

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

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Question 37/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(7) अलग-अलग फाइलों में से (3) फाइलों को शेल्फ पर क्रम में रखने के तरीके कितने हैं?

How many ways are there to place (3) files from (7) different files on a shelf in order?

Explanation opens after your attempt
Correct Answer

B. (210)

Step 1

Concept

This is \({}^{7}P_{3}=7\times6\times5=210\). Changing order on a shelf changes the arrangement.

Step 2

Why this answer is correct

The correct answer is B. (210). This is \({}^{7}P_{3}=7\times6\times5=210\). Changing order on a shelf changes the arrangement.

Step 3

Exam Tip

यह \({}^{7}P_{3}=7\times6\times5=210\) है। शेल्फ पर क्रम बदलने से व्यवस्था बदलती है।

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Question 38/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (LEVEL) में (5) अक्षर हैं, (L) दो बार और (E) दो बार आता है। अलग व्यवस्थाएँ कितनी होंगी?

The word (LEVEL) has (5) letters, with (L) twice and (E) 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 forget to divide by factorials of identical letters.

Step 2

Why this answer is correct

The correct answer is C. (30). The arrangements are \(\frac{5!}{2!2!}=30\). Do not forget to divide by factorials of identical letters.

Step 3

Exam Tip

व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। समान अक्षरों के factorial से भाग देना न भूलें।

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Question 39/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

किसी परीक्षा में (5) विद्यार्थियों में से प्रथम, द्वितीय और तृतीय स्थान कितने तरीकों से मिल सकते हैं?

In an exam, in how many ways can first, second and third ranks be obtained among (5) students?

Explanation opens after your attempt
Correct Answer

D. (60)

Step 1

Concept

The three ranks are ordered, so \({}^{5}P_{3}=60\). In rank questions, use permutation, not combination.

Step 2

Why this answer is correct

The correct answer is D. (60). The three ranks are ordered, so \({}^{5}P_{3}=60\). In rank questions, use permutation, not combination.

Step 3

Exam Tip

तीन rank क्रम वाले हैं इसलिए \({}^{5}P_{3}=60\)। rank वाले प्रश्नों में combination नहीं, permutation लें।

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Question 40/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

यदि (3) अलग-अलग खिलौनों को (3) बच्चों के सामने एक पंक्ति में रखा जाए तो व्यवस्थाएँ कितनी होंगी?

If (3) different toys are placed in a row before (3) children, how many arrangements are possible?

Explanation opens after your attempt
Correct Answer

A. (6)

Step 1

Concept

The arrangements of (3) distinct toys are (3!=6). If all objects are distinct, total arrangements are factorial.

Step 2

Why this answer is correct

The correct answer is A. (6). The arrangements of (3) distinct toys are (3!=6). If all objects are distinct, total arrangements are factorial.

Step 3

Exam Tip

(3) अलग खिलौनों की arrangements (3!=6) हैं। सभी objects अलग हों तो total arrangements factorial होती हैं।

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Question 41/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(9) बच्चों में से (3) बच्चों को मंच पर क्रम से खड़ा करने के तरीके कितने हैं?

How many ways are there to make (3) children from (9) children stand on a stage in order?

Explanation opens after your attempt
Correct Answer

B. (504)

Step 1

Concept

The answer is \({}^{9}P_{3}=9\times8\times7=504\). Left-right order on a stage is important.

Step 2

Why this answer is correct

The correct answer is B. (504). The answer is \({}^{9}P_{3}=9\times8\times7=504\). Left-right order on a stage is important.

Step 3

Exam Tip

उत्तर \({}^{9}P_{3}=9\times8\times7=504\) है। मंच पर बायाँ-दायाँ क्रम महत्त्वपूर्ण होता है।

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Question 42/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

\({}^{3}P_{2}\) का मान क्या है?

What is the value of \({}^{3}P_{2}\)?

Explanation opens after your attempt
Correct Answer

C. (6)

Step 1

Concept

\({}^{3}P_{2}=3\times2=6\). In small permutations, write decreasing factors directly.

Step 2

Why this answer is correct

The correct answer is C. (6). \({}^{3}P_{2}=3\times2=6\). In small permutations, write decreasing factors directly.

Step 3

Exam Tip

\({}^{3}P_{2}=3\times2=6\) है। छोटे permutation में सीधे घटते गुणनखंड लिखें।

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Question 43/50 Easy Mathematics Permutations and Combinations Permutations Class 11 Level 58

(5) अलग-अलग चाबियों को एक की-रिंग में सीधी सूची के रूप में लिखने पर कितने क्रम होंगे?

If (5) different keys are written as a straight list, how many orders are possible?

Explanation opens after your attempt
Correct Answer

D. (120)

Step 1

Concept

In a straight list, the arrangement of (5) distinct objects is (5!=120). Do not use circular counting here.

Step 2

Why this answer is correct

The correct answer is D. (120). In a straight list, the arrangement of (5) distinct objects is (5!=120). Do not use circular counting here.

Step 3

Exam Tip

सीधी सूची में (5) अलग वस्तुओं की व्यवस्था (5!=120) है। यहाँ circular counting नहीं करनी है।

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Question 44/50 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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Question 45/50 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 46/50 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 47/50 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 48/50 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 49/50 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 50/50 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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