Question

Find the value of $n$ such that
${}^n{P}_5=42{}^n{P}_3$, $n>4$

Answer 1

It is given that,

${}^n{P}_5={42.}^n{P}_3$

$⟹\frac{n!}{(n-5)!}=42.\frac{n!}{(n-3)!}$

$⟹\frac{(n-3)!}{(n-5)!}=42$

$⟹\frac{(n-3)(n-4)(n-5)!}{(n-5)!}=42$

$⟹(n-3)(n-4)=42$

$⟹n=-3,10$

As, $n>4$

Hence, $n=10$