If frac{n!}{2!(n-2)!} and frac{n!}{4!(n-4)!} | vedclass

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(C) Given the ratio $\frac{\frac{n!}{2!(n-2)!}}{\frac{n!}{4!(n-4)!}} = \frac{2}{1}$.This simplifies to $\frac{n!}{2!(n-2)!} 	imes \frac{4!(n-4)!}{n!}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the ratio $\frac{\frac{n!}{2!(n-2)!}}{\frac{n!}{4!(n-4)!}} = \frac{2}{1}$.This simplifies to $\frac{n!}{2!(n-2)!} 	imes \frac{4!(n-4)!}{n!} = 2$.Canceling $n!$ and expanding the factorials,we get $\frac{4 	imes 3 	imes 2!}{2! 	imes (n-2)(n-3)(n-4)!} 	imes (n-4)! = 2$.$\frac{12}{(n-2)(n-3)} = 2$.$(n-2)(n-3) = 6$.$n^2 - 5n + 6 = 6$.$n^2 - 5n = 0$.$n(n-5) = 0$.Since $n \ge 4$ for the term $\frac{n!}{4!(n-4)!}$ to be defined,we have $n = 5$.

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$,then $n=$

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