<s>[INST] 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: $\frac{1}{6} * (n^{3} - n +1)$
 [/INST] You must have made a typo with your second entrance into WA. To confirm your first answer:
\begin{align}\require{cancel}
\binom{n+1}{3}\cdot\frac{(n-1)!+(n-2)!}{(n+1)!}&= \frac{\cancel{(n+1)!}}{3!(n-2)!}\cdot\frac{(n-1)!+(n-2)!}{\cancel{(n+1)!}}\\[1em]
&= \frac{(n-1)!+(n-2)!}{3!(n-2)!}\\[1em]
&= \frac{\cancel{(n-2)!}\cdot[(n-1)+1]}{3!\cancel{(n-2)!}}\\[1em]
&= \frac{n}{3!}\\[1em]
&= \frac{n}{6}
\end{align}
This verifies your first simplification by WA. There must have been a typo in the second one. 
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