Closed form for $\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}$ How to get a closed form for $$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}\;?$$
I tried to write binominal in term of gamma function but I got no result. What is your suggestion to solve the problem?
 A: For this kind of problems it is best to have the summation variable running in opposite directions; also note that you can remove the distracting upper bound on $k$ since terms for $k>n-p$ will be zero anyway due to the second binomial coefficient. So apply symmetry either in the first or the second binomial coefficient, giving respectively
$$
  \sum_{k\geq0}\binom{n}{n-k}\binom{n}{p+k}
\qquad\text{or}\qquad
  \sum_{k\geq0}\binom{n}{k}\binom{n}{n-p-k}.
$$
The first summation can be interpreted as the counting way to choose $n+p$ elements out of a set of$~2n$, with $n-k$ coming from the first half, and the remaining $p+k$ from the second half. The second summation can be interpreted as the counting way to choose $n-p$ elements out of a set of$~2n$, with $k$ coming from the first half, and the remaining $n-p-k$ from the second half. The results are the same, since
$$
  \binom{2n}{n+p} = \binom{2n}{n-p}
$$
by symmetry. This summation is a specialisation of the Vandermonde identity.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{k = 0}^{n - p}{n \choose k}{n \choose p + k}:\ {\large ?}}$.

$$
\mbox{Note that}\quad{n \choose p + k}_{k\ >\ n - p} = 0\quad\mbox{such that}\quad
\sum_{k = 0}^{n - p}{n \choose k}{n \choose p + k}
=\sum_{k = 0}^{\color{#c00000}{\LARGE n}}{n \choose k}{n \choose p + k}
$$

$$
\mbox{Hereafter, we'll use the identity}\quad
{m \choose s} =\oint_{\verts{z}\ =\ a\ >\ 0}{\pars{1 + z}^{m} \over z^{s + 1}}
\,{\dd z \over 2\pi\ic}
$$

\begin{align}&\color{#66f}{\large%
\sum_{k = 0}^{n - p}{n \choose k}{n \choose p + k}}
=\sum_{k = 0}^{n}{n \choose k}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{p + k + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{p + 1}}
\sum_{k = 0}^{n}{n \choose k}\pars{1 \over z}^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{p + 1}}
\,\pars{1 + {1 \over z}}^{n}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{p + 1 + n}}
\,{\dd z \over 2\pi\ic}
=\color{#66f}{\large{2n \choose n + p}}
\end{align}

A: $$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}=\sum_{k=0}^{n-p}\binom{n}{n-k}\binom{n}{p+k}$$
Using Vandermonde's convolution follows:
$$=\binom{2n}{n+p}$$
