Question

Find $n$ if ${}^{n-1}{P}_3:{}^n{P}_4=1:9.$

Answer 1

Solving ${}^{n-1}{P}_3\text{and}{}^n{P}_4$ separately.

${}^{n-1}{P}_3=\frac{(n-1)!}{(n-1-3)!}=\frac{(n-1)!}{(n-4)!}$
$=\frac{(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}$
${}^{n-1}{P}_3=(n-1)(n-2)(n-3)\cdots \left(i\right)$

And ${}^n{P}_4=\frac{n!}{(n-4)!}$
$=\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}$
${}^n{P}_4=n(n-1)(n-2)(n-3)\cdots \left(ii\right)$

${}^{n-1}{P}_3:{}^n{P}_4=1:9$
$9\left({}^{n-1}{P}_3\right)={}^n{P}_4$
Putting values from $\left(i\right)$ and $\left(ii\right)$
$9(n-1)(n-2)(n-3)=n(n-1)(n-2)(n-3)$
$\Rightarrow n=9$.