Simplify the expression (combination and factorial) Simplify the following expression:

$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$

My attempt:

$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$

and this is where I get stuck... How to continue?

When I put $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{n}{6}$

When I put $\frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{1}{6} * (n^{3} - n +1)$
Well, for starters you can cancel the two $(n+1)!$s from the top and the bottom of the fraction.

Also note that $(n-1)! = (n-1)((n-2)!)$ and then you can cancel an $(n-2)!$ from the top and the bottom.
$$n/6$$
as ${n+1\choose 3}$ can be reduced to $(n+1)(n)(n-1)/3!$, now $(n+1)(n)(n-1)$ is canceled from the denominator and the $(n-2)!$ is canceled too so we are left with $(n-1) + 1$ numerator and $3!$ denominator. So the answer is $n/6 $.