यदि ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), तो (n) का मान क्या है?
If ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), what is the value of (n)?
Explanation opens after your attempt
B. (6)
Concept
Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.
Why this answer is correct
The correct answer is B. (6). Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.
Exam Tip
सरल रूप में (n=6) रखने पर \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \) मिलता है। विकल्प जांचते समय कटौती करें।
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