Question

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$, find the value of $n$.

• A. $1$
• B. $5$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$5$

We have,
$\frac{n!}{2!(n-2)!}:\frac{n!}{4!(n-2)!}=2:1$
$\Rightarrow \frac{n!}{2!(n-2)!}\times \frac{4!(n-4)!}{n!}=\frac{2}{1}$
$\Rightarrow \frac{4!(n-4)!}{2!(n-2)\times (n-3)\times (n-4)!}=\frac{2}{1}$
$\Rightarrow \frac{4\times 3\times 2!}{2!(n-2)(n-3)}=\frac{2}{1}$
$\Rightarrow (n-2)(n-3)=6$
$\Rightarrow n^2-5n=0$
$\Rightarrow n(n-5)=0$ $\Rightarrow n=5$
But, for $n=0,(n-2)!$ and $(n-4)!$ are not meaningful. Therefore, $n=5.$