( \frac{\frac{(n+4)!}{n!}}{\frac{(n+3)!}{(n-1)!}} ) का सरल रूप क्या है?

What is the simplified form of ( \frac{\frac{(n+4)!}{n!}}{\frac{(n+3)!}{(n-1)!}} )?

Explanation opens after your attempt
Correct Answer

D. \( \frac{n+4}{n} \)

Step 1

Concept

After canceling the common factors in numerator and denominator, \( \frac{n+4}{n} \) remains. First convert large ratios into products.

Step 2

Why this answer is correct

The correct answer is D. \( \frac{n+4}{n} \). After canceling the common factors in numerator and denominator, \( \frac{n+4}{n} \) remains. First convert large ratios into products.

Step 3

Exam Tip

ऊपर और नीचे के समान गुणक कटने पर \( \frac{n+4}{n} \) बचता है। बड़े अनुपात को पहले गुणनफल में बदलें।

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Mathematics Answer, Explanation and Revision Hints

( \frac{\frac{(n+4)!}{n!}}{\frac{(n+3)!}{(n-1)!}} ) का सरल रूप क्या है? / What is the simplified form of ( \frac{\frac{(n+4)!}{n!}}{\frac{(n+3)!}{(n-1)!}} )?

Correct Answer: D. \( \frac{n+4}{n} \). Explanation: ऊपर और नीचे के समान गुणक कटने पर \( \frac{n+4}{n} \) बचता है। बड़े अनुपात को पहले गुणनफल में बदलें। / After canceling the common factors in numerator and denominator, \( \frac{n+4}{n} \) remains. First convert large ratios into products.

Which concept should I revise for this Mathematics MCQ?

After canceling the common factors in numerator and denominator, \( \frac{n+4}{n} \) remains. First convert large ratios into products.

What exam hint can help solve this Mathematics question?

ऊपर और नीचे के समान गुणक कटने पर \( \frac{n+4}{n} \) बचता है। बड़े अनुपात को पहले गुणनफल में बदलें।