How to solve $\sum\limits_{k=0}^{n-a-b}\binom{n-a-b}{k}(a+k-1)!(n-a-k)!$ I'm solving a probability problem, and I've ended up with this sum:
$$\sum\limits_{k=0}^{n-a-b}\binom{n-a-b}{k}(a+k-1)!(n-a-k)!$$
WolframAlpha says I should get the answer $\frac{n!}{a\binom{a+b}{a}}$, but I don't see how to get there. I tried to get to something containing $\binom{a+k-1}{k}$ so that I could use the Hockey Stick Theorem, but I wasn't successful.
So any hints would be very welcome, thanks for any help

We seek to find a closed form of
$$(n-1)! \sum_{k=0}^{n-a-b} 
{n-a-b\choose k} {n-1\choose n-a-k}^{-1}$$
where $n\gt a+b$ and $a,b\ge 1.$
Recall from MSE
4316307  the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We get for our sum
$$(n-1)! \sum_{k=0}^{n-a-b} {n-a-b\choose k}
(n-a-k) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{a-1+k}
\\ = (n-1)! \sum_{k=0}^{n-a-b} {n-a-b\choose k}
(k+b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b-k}.$$
We get two pieces, the first is
$$b [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\sum_{k=0}^{n-a-b} {n-a-b\choose k} (z-1)^{-k}
\\ = b [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\left[1+\frac{1}{z-1}\right]^{n-a-b}
\\ = b [z^{a+b-1}] \log\frac{1}{1-z} (z-1)^{a-1}
\\ = {a+b-1\choose b}^{-1}.$$
The second is
$$ (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\sum_{k=1}^{n-a-b} {n-a-b-1\choose k-1} (z-1)^{-k}
\\ =  (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-2-b}
\sum_{k=0}^{n-a-b-1} {n-a-b-1\choose k} (z-1)^{-k}
\\ =  (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-2-b}
\left[1+\frac{1}{z-1}\right]^{n-a-b-1}
\\ = (n-a-b) [z^{a+b}] \log\frac{1}{1-z} (z-1)^{a-1}
\\ = \frac{n-a-b}{b+1} {a+b\choose b+1}^{-1}.$$
Collecting everything we find
$$(n-1)! {a+b\choose b}^{-1}
\left[ \frac{a+b}{a} + \frac{n-a-b}{b+1} \frac{b+1}{a} \right]
\\ = \frac{n!}{a} {a+b\choose b}^{-1}.$$
This is the claim.
