Question

If ${}^n{P}_5=20$. ${}^n{P}_3$, then n=

• A. 4
• B. 8
• C. 6
• D. 7

Answer 1

The correct option is B 8

$\frac{n!}{(n-5)!}\times \frac{(n-3)!}{n!}=20$
$\Rightarrow (n-3)(n-4)=20\Rightarrow n=-1,8$
But -1 is not exceptable.