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. 2
• C. 3
• D. None of these

Answer 1

$\frac{n!}{2!(n-2)!}:\frac{n!}{4!(n-4)!}=2:1\Rightarrow \frac{n!}{2!(n-2)!}\times \frac{n!}{4!(n-4)!}=\frac{2}{1}\Rightarrow \frac{4!(n-4)!}{2!(n-2)!}=2\Rightarrow \frac{4\times 3\times 2}{2!}\times \frac{(n-4)!}{(n-2)(n-3)(n-4)!}=2\Rightarrow 12=2(n-2)(n-3)\Rightarrow 6=n^2-5n+6\Rightarrow n^2-5n=0\Rightarrow n(n-5)=0\Rightarrow n=0or5.$