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)?
Explanation opens after your attempt
C. (9)
Concept
The simplified form is (n(n+3)). Since \(9\cdot12=108\), (n=9).
Why this answer is correct
The correct answer is C. (9). The simplified form is (n(n+3)). Since \(9\cdot12=108\), (n=9).
Exam Tip
सरल रूप (n(n+3)) है। \(9\cdot12=108\), इसलिए (n=9)।
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