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

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

Explanation opens after your attempt
Correct Answer

A. (9979200)

Step 1

Concept

There are (11) letters with (M,A,T) each repeated twice, so \(\frac{11!}{2!2!2!}=4989600\). In exams, divide by factorials of all repeated letters.

Step 2

Why this answer is correct

The correct answer is A. (9979200). There are (11) letters with (M,A,T) each repeated twice, so \(\frac{11!}{2!2!2!}=4989600\). In exams, divide by factorials of all repeated letters.

Step 3

Exam Tip

(11) अक्षरों में (M,A,T) दो-दो बार आते हैं, इसलिए \(\frac{11!}{2!2!2!}=4989600\)। परीक्षा में सभी repeated letters के factorial से भाग दें।

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शब्द (MATHEMATICS) के अक्षरों की अलग-अलग व्यवस्थाएं कितनी होंगी? / How many distinct arrangements are possible using the letters of (MATHEMATICS)?

Correct Answer: A. (9979200). Explanation: (11) अक्षरों में (M,A,T) दो-दो बार आते हैं, इसलिए \(\frac{11!}{2!2!2!}=4989600\)। परीक्षा में सभी repeated letters के factorial से भाग दें। / There are (11) letters with (M,A,T) each repeated twice, so \(\frac{11!}{2!2!2!}=4989600\). In exams, divide by factorials of all repeated letters.

Which concept should I revise for this Mathematics MCQ?

There are (11) letters with (M,A,T) each repeated twice, so \(\frac{11!}{2!2!2!}=4989600\). In exams, divide by factorials of all repeated letters.

What exam hint can help solve this Mathematics question?

(11) अक्षरों में (M,A,T) दो-दो बार आते हैं, इसलिए \(\frac{11!}{2!2!2!}=4989600\)। परीक्षा में सभी repeated letters के factorial से भाग दें।