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: $\frac{1}{6} * (n^{3} - n +1)$
 A: 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.
A: Oh, lol!
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n-1)! + (n-2)!}{3!(n-2)!} = \frac{(n-2)!((n-1) + 1)}{3!(n-2)!}=\frac{n}{3!} = \frac{n}{6}$
Correct? :)
How come Wolfram Alpha gives two different results?
A: $$n/6$$
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 $.
A: 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. 
