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

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

Explanation opens after your attempt
Correct Answer

A. (45360)

Step 1

Concept

There are (9) letters with (M,T,E) each repeated twice, so \(\frac{9!}{2!2!2!}=45360\). In exams, put each repeated letter factorial in the denominator.

Step 2

Why this answer is correct

The correct answer is A. (45360). There are (9) letters with (M,T,E) each repeated twice, so \(\frac{9!}{2!2!2!}=45360\). In exams, put each repeated letter factorial in the denominator.

Step 3

Exam Tip

(9) अक्षरों में (M,T,E) दो-दो बार हैं, इसलिए \(\frac{9!}{2!2!2!}=45360\)। परीक्षा में हर repeated letter का factorial हर में रखें।

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

Correct Answer: A. (45360). Explanation: (9) अक्षरों में (M,T,E) दो-दो बार हैं, इसलिए \(\frac{9!}{2!2!2!}=45360\)। परीक्षा में हर repeated letter का factorial हर में रखें। / There are (9) letters with (M,T,E) each repeated twice, so \(\frac{9!}{2!2!2!}=45360\). In exams, put each repeated letter factorial in the denominator.

Which concept should I revise for this Mathematics MCQ?

There are (9) letters with (M,T,E) each repeated twice, so \(\frac{9!}{2!2!2!}=45360\). In exams, put each repeated letter factorial in the denominator.

What exam hint can help solve this Mathematics question?

(9) अक्षरों में (M,T,E) दो-दो बार हैं, इसलिए \(\frac{9!}{2!2!2!}=45360\)। परीक्षा में हर repeated letter का factorial हर में रखें।