Question

(n+2)!(n^2-9) / (n+4)!

Answer 1

$\frac{(n+2)!(n^2-9) }{(n+4)! } = \frac{(n+2)!(n^2-9) }{(n+2)!(n+3)(n+4) } = \frac{((n-3)(n+3)) }{(n+3)(n+4) } = \frac{n-3 }{n+4 }$