Question

Solve for n if $\frac{\left(15-n\right)!}{\left(12-n\right)!}=5!$

Answer 1

$\frac{(15-n)!}{(12-n)!}=5!$

on expansion we get

$\frac{(15-n)(14-n)(13-n)(12-n)!}{(12-n)!}=120$

$(210-15n-14n+n^2)(13-n)=120$

$n^3-42n^2+587n-2610=0$

$(n-9)(n^2-33n+290)=0$

$(n=9)$ $n^2-33n+290=0$

$n=\frac{33\pm \sqrt{1089-1160}}{2}=\frac{33\pm \sqrt{-71}}{2}$

Hence other roots are imaginary.

$\therefore$ value of $[n=9]$