Closed form of $\sum\limits_{p = k}^{n - k + 1} C^{k - 1}_{p - 1} C^{k - 1}_{n - p}$ Does anyone has an idea how to prove that $$\sum\limits_{p = k}^{n - k + 1} \binom{p - 1}{k - 1} \binom{n - p}{k - 1} = \frac{n!}{(2 k - 1)! (n - 2 k + 1)!}?$$
 A: In the notation that I prefer, this is
$$\sum_{p=k}^{n-k+1}\binom{p-1}{k-1}\binom{n-p}{k-1}=\binom{n}{2k-1}\;.$$
We can allow $p$ to run over all the integers if we wish, since $\binom{p-1}{k-1}=0$ if $p<k$, and $\binom{n-p}{k-1}=0$ if $p>n-k+1$.
The righthand side is clearly the number of subsets of $[n]=\{1,\ldots,n\}$ of cardinality $2k-1$. Each $(2k-1)$-subset $A$ of $[n]$ has a middle element, i.e., an element $p$ such that $k-1$ elements of $A$ are less than $p$, and $k-1$ are greater than $p$. $\binom{p-1}{k-1}\binom{n-p}{k-1}$ is the number of ways to choose $k-1$ elements of $[n]$ that are less than $p$ and $k-1$ that are greater than $p$, so it’s the number of $(2k-1$-subsets of $[n]$ whose middle element is $p$. The identity is now clear.
A: Suppose we seek to evaluate
$$\sum_{p=k}^{n-k+1} {p-1\choose k-1} {n-p\choose k-1}.$$
Introduce
$${n-p\choose k-1} = {n-p\choose n-k+1-p} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2-p}} (1+z)^{n-p} \; dz.$$
Observe that this  controls the range being zero  when $p\gt n-k+1$ so
we may extend $p$ to infinity to obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2}} (1+z)^{n}
\sum_{p\ge k} {p-1\choose k-1} \frac{z^p}{(1+z)^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2}} (1+z)^{n} \frac{z^k}{(1+z)^k}
\sum_{p\ge 0} {p+k-1\choose k-1} \frac{z^p}{(1+z)^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+2}} (1+z)^{n} \frac{1}{(1+z)^k}
\frac{1}{(1-z/(1+z))^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+2}} (1+z)^{n}
\frac{1}{(1+z-z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+2}} (1+z)^{n} \; dz
\\ = {n\choose n-2k+1} = {n\choose 2k-1}.$$
This completes the proof.
