Closed form for $\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}$

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{k = 0}^{n - p}{n \choose k}{n \choose p + k}:\ {\large ?}}$.

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

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

\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}

Answer 3

$$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}=\sum_{k=0}^{n-p}\binom{n}{n-k}\binom{n}{p+k}$$

Using Vandermonde's convolution follows: $$=\binom{2n}{n+p}$$