Question

$\frac{A_{n}^{n-5}*(5!)^2}{P_{n}}$

Answer 1

$A^{n-5}_{n}=\frac{n!}{(n-(n-5))!}=\frac{n!}{5!} \\\\ P_{n}=n!\\\\\frac{A^{n-5}_{n}\cdot (5!)^2}{P_{n}}=\frac{n!\cdot (5!)^2}{5!\cdot n!}=5!=1\cdot 2\cdot 3\cdot 4\cdot 5=120$