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Prove that $\sum\limits_{k = 0}^n k{m \choose k}{n \choose k}= n{m+n-1 \choose n}$

We can write ${m \choose k} = m!/(m-k)!(k)!$ similarly ${n \choose k}$ and ${m+n-1 \choose n}$ can also be written but I am confused how to proceed further.

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    if you can please share the link, that would be very helful. – 2017-02-24

3 Answers 3

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{n}k{m \choose k}{n \choose k} & = \sum_{k = 0}^{n}k{n \choose k}{m \choose m - k} = \sum_{k = 0}^{n}k{n \choose k}\bracks{z^{m - k}}\pars{1 + z}^{m} \\[5mm] & = \sum_{k = 0}^{n}k{n \choose k}\braces{\vphantom{\Large A}\bracks{z^{m}}z^{k}\pars{1 + z}^{m}} = \bracks{z^{m}}\braces{\vphantom{\Large A}\pars{1 + z}^{m}\overbrace{\sum_{k = 0}^{n}{n \choose k}kz^{k}} ^{\ds{nz\,\pars{1 + z}^{n - 1}}}} \\[5mm] & = n\bracks{z^{m - 1}}\pars{1 + z}^{m + n - 1} = \bbx{\ds{n\,{m + n - 1 \choose m - 1}}} \end{align}

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    thanks a lot for the solution – 2017-02-24
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    @LalitPathak You're welcome. I'm glad it was useful. Thanks. – 2017-02-24
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For a combinatorial proof:

Let there be $m$ labeled red balls and $n$ labeled blue balls in an urn. We wish to count how many ways we can pull $n$ balls out of potentially mixed color and then afterwards place one of the blue balls from those we had left behind into our hand.

On the left, break into cases based on how many red balls were selected, $k$. Pick which $k$ red balls were taken, pick which $k$ blue balls were left behind (thus picking which $n-k$ blue balls were taken). From there, pick which specific blue ball from those left behind was placed into hand. This yields the amount $\sum\limits_{k=0}^nk\binom{m}{k}\binom{n}{k}$

On the right, first pick the special blue ball. Then pick the $n$ balls from those remaining. This gives $n\binom{m+n-1}{n}$

As the two methods correctly count the same scenario, they must be equal.

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Look at the coefficient of $x^n$ in the product $((1+x)^m)'(1+x)^n$, and try to compute it otherwise.