Please help me with this combination problem 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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\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}
