Question

If ${}^n{P}_2=90$ then find the value of $n$.

Answer 1

Solution:

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

or, $\frac{n!}{(n-2)!}=90$

or, $n(n-1)=90$

or, $n^2-n-90=0$

or, $n^2-10n+9n-90=0$

or, $n(n-10)+9(n-10)=0$

or, $(n-10)(n+9)=0$

or, $n=10$ or $=n=-9$

$n$ can't be negative so, $n=10.$