Question

Find $n$, if ${}^n{P}_5$ $=42$ ${}^n{P}_3$, $n\ge 5.$

Answer 1

${}^n{P}_5=\frac{n!}{(n-5)!}$

${}^n{P}_3=\frac{n!}{(n-3)!}$

Given:${}^n{P}_5={42}^n{P}_3$

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

$\frac{1}{(n-5)!}=42\times \frac{1}{(n-3)!}$

$\frac{1}{(n-5)!}=42\times \frac{1}{(n-3)\times (n-4)\times (n-5)!}$

$(n-3)(n-4)=42$

$n^2-7n+12=42$

$n^2-7n-30=0$

$n=\frac{7\pm \sqrt{7^2-4\left(1\right)(-30)}}{2}$

$n=\frac{7\pm \sqrt{169}}{2}$

$n=10$ or $n=-3$

n cannot be negative.

So $n=10$