Question

If ${}^n{P}_4=20{\times }^n{P}_2$ then find the value of n.

Answer 1

${}^n{P}_4=20{\times }^n{P}_2$

$\Rightarrow \frac{n!}{(n-4)!}=20\frac{n!}{(n-2)!}$

$\Rightarrow (n-2)(n-3)=20$

$\Rightarrow n^2-5n-114=0$

$\Rightarrow (n-7)(n+2)=0$

$n=7,-2$ As $n>0$

$\Rightarrow n=7$