यदि ( \frac{(n+3)!}{(n-1)!(n+2)(n+1)}=108 ), तो (n) का मान क्या है?

If ( \frac{(n+3)!}{(n-1)!(n+2)(n+1)}=108 ), what is the value of (n)?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

C. (9)

Explanation

Simple Explanation

सरल रूप (n(n+3)) है। \(9\cdot12=108\), इसलिए (n=9)। / The simplified form is (n(n+3)). Since \(9\cdot12=108\), (n=9).

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि ( \frac{(n+3)!}{(n-1)!(n+2)(n+1)}=108 ), तो (n) का मान क्या है? / If ( \frac{(n+3)!}{(n-1)!(n+2)(n+1)}=108 ), what is the value of (n)?

Correct Answer: C. (9). Explanation: सरल रूप (n(n+3)) है। \(9\cdot12=108\), इसलिए (n=9)। / The simplified form is (n(n+3)). Since \(9\cdot12=108\), (n=9).