Question

If ${}^n{P}_2=90$, find '$n$'.

Answer 1

Given that:

${}^n{P}_2=90$

$⟹\frac{n!}{(n-2)!}=90$ .... $[\because {}^n{P}_r=\frac{n!}{(n-r)!}]$
$⟹n(n-1)=90$

$⟹n^2-n-90=0$

$⟹(n+9)(n-10)=0$

$⟹n=10$ or $n=-9$

But $n$ can't be negative
$\therefore n=10$