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
 A: 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.
A: It seems to be the result of taking $x := a$, $y := b+1$, $z := 1$ and "$n$" $:= n - a - b$ in the Rothe-Hagen identity (thank you for indirectly making me discover this complicated yet fascinating identity today).
This gives you, using $\displaystyle \frac{c}{d} \binom{d}{d - c} = \frac{c \cdot d!}{d \cdot c!(d-c)!} = \binom{d-1}{d - c}$ to simplify the formula:
$$\sum_{k = 0}^{n - a - b} \binom{a + k - 1}{k}\binom{n - a - k}{n - a - b - k} = \binom{n}{n - a - b}$$
Let's reorder all the factorials hidden in the binomials:
$$\begin{split}1 =& \frac{1}{\binom{n}{n - a - b}} \sum_{k = 0}^{n - a - b} \binom{a + k - 1}{k}\binom{n - a - k}{n - a - b - k}\\& = \frac{(a+b)!(n - a - b)!}{n!}\sum_{k = 0}^{n - a - b} \frac{(a + k - 1)!}{k!(a-1)!}\frac{(n - a - k)!}{(n - a - b - k)!b!}\\
& = \frac{1}{n!}\frac{(a+b)!}{(a-1)!b!} \sum_{k = 0}^{n - a - b} \frac{(n - a - b)!}{(n - a - b - k)!k!}(a + k - 1)!(n - a - k)!\\
& = \frac{a\binom{a+b}{a}}{n!} \sum_{k = 0}^{n - a - b} \binom{n - a - b}{k}(a + k - 1)!(n - a - k)!\end{split}$$
This finally grants:
$$\sum_{k = 0}^{n - a - b} \binom{n - a - b}{k}(a + k - 1)!(n - a - k)! = \frac{n!}{a\binom{a+b}{a}}$$
A: 
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{n-a-b}}&\color{blue}{\binom{n-a-b}{k}(a+k-1)!(n-a-k)!}\\
&=(n-1)!\sum_{k=0}^{n-a-b}\binom{n-a-b}{k}\binom{n-1}{n-a-k}^{-1}\tag{1}\\
&=n!\sum_{k=0}^{n-a-b}\binom{n-a-b}{k}\int_{0}^1z^{n-a-k}(1-z)^{a+k-1}\,dz\tag{2}\\
&=n!\int_{0}^1z^{n-a}(1-z)^{a-1}\sum_{k=0}^{n-a-b}\binom{n-a-b}{k}\left(\frac{1-z}{z}\right)^k\,dz\\
&=n!\int_{0}^1z^{n-a}(1-z)^{a-1}\left(1+\frac{1-z}{z}\right)^{n-a-b}\,dz\\
&=n!\int_{0}^1z^b(1-z)^{a-1}\,dz\\
&=\frac{n!}{a+b}\binom{a+b-1}{b}^{-1}\tag{2}\\
&\,\,\color{blue}{=\frac{n!}{a}\binom{a+b}{a}^{-1}}\tag{3}
\end{align*}
in accordance with the claim and other answers.

Comment:

*

*In (1) we use a representation using binomial coefficients.


*In (2) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}


*In (3) we use the binomial identity
\begin{align*}
(a+b)\binom{a+b-1}{b}=(a+b)\binom{a+b-1}{a-1}=a\binom{a+b}{a}
\end{align*}
