Class 11 Chapter Practice

Mathematics Probability MCQ Questions for Class 11

Related questions grouped automatically for chapter-wise practice. Topics include Permutations.

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Class 11 Mathematics Probability Practice

Related questions grouped automatically for chapter-wise practice.

Probability - Topics Covered

Mathematics Probability ke topic-wise MCQs yahan grouped context me milenge. jo aap ko Exam ki preparation me madad milegi. Ye questions exam-oriented hai and students ko concept clarity, quick revision aur board exam preparation kaafi madad karenge. Sabhi se jude MCQs important topics ke anusar arranged hai, taaki aap Probability ko easy tarike se practice aur revise kar sake.

  1. Permutations
    5 MCQs

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Mathematics Probability MCQ Questions

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Question 1/5 Hard Mathematics Permutations and Combinations Permutations Class 11 Level 58

शब्द (PROBABILITY) के अक्षरों की अलग-अलग व्यवस्थाएं कितनी होंगी?

How many distinct arrangements are possible using the letters of (PROBABILITY)?

Explanation opens after your attempt
Correct Answer

A. (9979200)

Step 1

Concept

There are (11) letters with (B) twice and (I) twice. Hence \(\frac{11!}{2!2!}=9979200\) arrangements are possible.

Step 2

Why this answer is correct

The correct answer is A. (9979200). There are (11) letters with (B) twice and (I) twice. Hence \(\frac{11!}{2!2!}=9979200\) arrangements are possible.

Step 3

Exam Tip

(11) अक्षरों में (B) दो और (I) दो बार हैं। इसलिए \(\frac{11!}{2!2!}=9979200\) व्यवस्थाएं होंगी।

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Question 2/5 Hard Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (PROBABILITY) के अक्षरों की कुल भिन्न व्यवस्थाएं कितनी हैं?

What is the total number of distinct arrangements of the letters of (PROBABILITY)?

Explanation opens after your attempt
Correct Answer

A. (9979200)

Step 1

Concept

Here (B) and (I) are repeated twice, so the count is (11!/(2!2!)). Identifying repeated letters is the first step.

Step 2

Why this answer is correct

The correct answer is A. (9979200). Here (B) and (I) are repeated twice, so the count is (11!/(2!2!)). Identifying repeated letters is the first step.

Step 3

Exam Tip

इस शब्द में (B) और (I) दो-दो बार हैं, इसलिए संख्या (11!/(2!2!)) है। दोहराए अक्षर पहचानना पहला कदम है।

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Question 3/5 Hard Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (PROBABILITY) की व्यवस्थाओं में सभी स्वर साथ हों, ऐसी कितनी भिन्न व्यवस्थाएं हैं?

In arrangements of (PROBABILITY), how many distinct arrangements have all vowels together?

Explanation opens after your attempt
Correct Answer

C. (241920)

Step 1

Concept

Inside the vowel block there are (4!/2!) arrangements, and outside there are (8!/2!) arrangements. Divide for repeated consonants and vowels.

Step 2

Why this answer is correct

The correct answer is C. (241920). Inside the vowel block there are (4!/2!) arrangements, and outside there are (8!/2!) arrangements. Divide for repeated consonants and vowels.

Step 3

Exam Tip

स्वर-खंड के भीतर (4!/2!) और बाहर (8!/2!) व्यवस्थाएं हैं। दोहराए व्यंजन और स्वर दोनों से भाग दें।

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Question 4/5 Expert Mathematics Permutations and Combinations Permutations Class 11 Level 59

शब्द (PROBABILITY) के अक्षरों से अलग-अलग व्यवस्थाओं की संख्या क्या है?

What is the number of distinct arrangements of the letters of (PROBABILITY)?

Explanation opens after your attempt
Correct Answer

A. (1814400)

Step 1

Concept

There are (11) letters and (B,I) each repeat (2) times, so the count is \(\frac{11!}{2!2!}\). Count letter frequencies carefully in long words.

Step 2

Why this answer is correct

The correct answer is A. (1814400). There are (11) letters and (B,I) each repeat (2) times, so the count is \(\frac{11!}{2!2!}\). Count letter frequencies carefully in long words.

Step 3

Exam Tip

कुल (11) अक्षर हैं और (B,I) प्रत्येक (2) बार आते हैं, इसलिए \(\frac{11!}{2!2!}\) है। बड़े शब्दों में अक्षर आवृत्ति सावधानी से गिनें।

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Question 5/5 Expert Mathematics Permutations and Combinations Permutations Class 11 Level 60

शब्द (PROBABILITY) के अक्षरों की कुल अलग-अलग व्यवस्थाओं की संख्या कितनी है?

How many distinct arrangements are possible using all letters of the word (PROBABILITY)?

Explanation opens after your attempt
Correct Answer

A. (19958400)

Step 1

Concept

There are (11) letters, with (B) and (I) repeated twice, so the count is \(\frac{11!}{2!,2!}\). Exam tip: divide by the factorial of each repeated letter.

Step 2

Why this answer is correct

The correct answer is A. (19958400). There are (11) letters, with (B) and (I) repeated twice, so the count is \(\frac{11!}{2!,2!}\). Exam tip: divide by the factorial of each repeated letter.

Step 3

Exam Tip

कुल (11) अक्षर हैं और (B,I) दो-दो बार आते हैं, इसलिए संख्या \(\frac{11!}{2!,2!}\) है। परीक्षा में हर repeated letter के factorial से अलग-अलग भाग दें।

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Related questions grouped automatically for chapter-wise practice. Topics include Permutations.

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