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If ( \frac{(n!)^2}{(n-2)!(n+2)!}=\frac{10}{21} ), what is the value of (n)?यदि ( \frac{(n!)^2}{(n-2)!(n+2)!}=\frac{10}{21} ), तो (n) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (5)(5)
The simplified form is ( \frac{n(n-1)}{(n+1)(n+2)} ). Putting (n=5) gives ( \frac{20}{42}=\frac{10}{21} ).सरल रूप ( \frac{n(n-1)}{(n+1)(n+2)} ) है। (n=5) रखने पर ( \frac{20}{42}=\frac{10}{21} ) मिलता है।
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