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

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

Explanation opens after your attempt
Correct Answer

A. (19958400)

Step 1

Concept

There are (11) letters with (T) repeated twice and all others distinct. Hence \(\frac{11!}{2!}=19958400\).

Step 2

Why this answer is correct

The correct answer is A. (19958400). There are (11) letters with (T) repeated twice and all others distinct. Hence \(\frac{11!}{2!}=19958400\).

Step 3

Exam Tip

(11) अक्षरों में (T) दो बार है और बाकी अलग हैं। इसलिए \(\frac{11!}{2!}=19958400\) है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. (19958400). Explanation: (11) अक्षरों में (T) दो बार है और बाकी अलग हैं। इसलिए \(\frac{11!}{2!}=19958400\) है। / There are (11) letters with (T) repeated twice and all others distinct. Hence \(\frac{11!}{2!}=19958400\).

Which concept should I revise for this Mathematics MCQ?

There are (11) letters with (T) repeated twice and all others distinct. Hence \(\frac{11!}{2!}=19958400\).

What exam hint can help solve this Mathematics question?

(11) अक्षरों में (T) दो बार है और बाकी अलग हैं। इसलिए \(\frac{11!}{2!}=19958400\) है।