Given $ (n+2)! = 60 (n-1)!$

$\Rightarrow (n+2)! = (n+2) (n+1) n!$

Hint: Now, $n! = n (n-1)!$

$\Rightarrow (n+2)! = (n+2) (n+1) n (n-1)!$

The equation now resolves to $(n+2) (n+1) n (n-1)! = 60 (n-1)!$

Canceling out $(n-1)$ on both sides, $(n+2) (n+1) n = 60$

$\Rightarrow (n^2 + 2n + n + 2)n = 60$

$\Rightarrow n^3 + 3n^2 + 2n = 60$

$\Rightarrow n^3 + 3n^2 + 2n - 60 = 0$

Factoring the left hand side, this equation resolves to $(n-3) (n^2+6n+20) = 0$

Since it can be shown that $(n^2 + 6n + 20) = 0$ has no real solution, $n = 3$

**Solution: $n = 3$**