Identity on the sum $\sum_{i=1}^{n-2}{i-1 \choose k}{n-2-i \choose k-1}={n-2 \choose 2k}$ I need to evaluate the first summation for $k\geq1$ and $n\geq2$.
Some computer calculations allow the identification with the expression ${n-2 \choose 2k}$
above but, how can it be derived? Are there different methods or any
combinatorial interpretation?
 A: Hint: From the set $\{1, 2, \dots, n-2\}$, choose $2k$ elements and let $i$ be the $(k+1)$-th element chosen, using the standard ordering. The first $k$ elements are chosen from $\{1, 2, \dots, i-1\}$ and the last $k - 1$ elements are chosen from $\{i+1, i+2, \dots, n-2\}$.
A: An Extension of Vandermonde's Identity
$$
\begin{align}
\sum_{\substack{k\\n-k\ge a\\m+k\ge b}}\binom{n-k}{a}\binom{m+k}{b}
&=\sum_{k}\binom{n-k}{n-k-a}\binom{m+k}{m+k-b}\tag1\\
&=\sum_{k}(-1)^{n-k-a}\binom{-a-1}{n-k-a}(-1)^{m+k-b}\binom{-b-1}{m+k-b}\tag2\\[9pt]
&=(-1)^{n+m-a-b}\binom{-a-b-2}{n+m-a-b}\tag3\\[9pt]
&=\binom{n+m+1}{n+m-a-b}\tag4\\[9pt]
&=\binom{n+m+1}{a+b+1}\tag5
\end{align}
$$
Explanation:
$(1)$: symmetry of Pascal's Triangle
$(2)$: negative binomial coefficients
$(3)$: Vandermonde's Identity
$(4)$: negative binomial coefficients
$(5)$: symmetry of Pascal's Triangle
Applying $(5)$ gives
$$
\sum_{i=1}^{n-2}\binom{i-1}{k}\binom{n-2-i}{k-1}=\binom{n-2}{2k}\tag6
$$
A: Trying to verify that
$$\sum_{q=1}^n {q-1\choose k} {n-q\choose k-1} =
{n\choose 2k}$$
using that $k\ge 1$ and $n\ge 1$ we get
$q-1\ge k$ and $q\le n-k+1$ and may re-write as
$$\sum_{q=k+1}^{n-k+1} {q-1\choose k} {n-q\choose k-1}
= \sum_{q=0}^{n-2k} {q+k\choose k} {n-(k+1)-q\choose k-1}
\\ = \sum_{q=0}^{n-2k} {q+k\choose k} {n-k-1-q\choose n-2k-q}
= [z^{n-2k}] (1+z)^{n-k-1}
\sum_{q=0}^{n-2k} {q+k\choose k} \frac{z^q}{(1+z)^q}.$$
Now the coefficient extractor enforces the upper limit of the sum
and we may extend $q$ to infinity, getting
$$[z^{n-2k}] (1+z)^{n-k-1}
\frac{1}{(1-z/(1+z))^{k+1}}
= [z^{n-2k}] (1+z)^{n-k-1} (1+z)^{k+1}
\\ = [z^{n-2k}] (1+z)^n = {n\choose n-2k} = {n\choose 2k}.$$
This is the claim.
