( \frac{\frac{(n+2)!}{(n-2)!}}{\frac{(n+1)!}{(n-3)!}} ) का सरल रूप क्या है?

What is the simplified form of ( \frac{\frac{(n+2)!}{(n-2)!}}{\frac{(n+1)!}{(n-3)!}} )?

Explanation opens after your attempt
Correct Answer

A. \(\frac{n+2}{n-2}\)

Step 1

Concept

The common factors ((n+1)n(n-1)) cancel out. Therefore \( \frac{n+2}{n-2} \) remains.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{n+2}{n-2}\). The common factors ((n+1)n(n-1)) cancel out. Therefore \( \frac{n+2}{n-2} \) remains.

Step 3

Exam Tip

समान गुणक ((n+1)n(n-1)) कट जाते हैं। इसलिए \( \frac{n+2}{n-2} \) बचता 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

( \frac{\frac{(n+2)!}{(n-2)!}}{\frac{(n+1)!}{(n-3)!}} ) का सरल रूप क्या है? / What is the simplified form of ( \frac{\frac{(n+2)!}{(n-2)!}}{\frac{(n+1)!}{(n-3)!}} )?

Correct Answer: A. \(\frac{n+2}{n-2}\). Explanation: समान गुणक ((n+1)n(n-1)) कट जाते हैं। इसलिए \( \frac{n+2}{n-2} \) बचता है। / The common factors ((n+1)n(n-1)) cancel out. Therefore \( \frac{n+2}{n-2} \) remains.

Which concept should I revise for this Mathematics MCQ?

The common factors ((n+1)n(n-1)) cancel out. Therefore \( \frac{n+2}{n-2} \) remains.

What exam hint can help solve this Mathematics question?

समान गुणक ((n+1)n(n-1)) कट जाते हैं। इसलिए \( \frac{n+2}{n-2} \) बचता है।