Prove that for m > 0, the following identity holds: $\sum_{k=0}^m (-1)^{m-k}{n+k-1\choose k}{n\choose m-k} = 0$ How shall I tackle this proof?  I would brute force it because my knowledge is minimal.  Can this one be done combinatorially?  How to proceed with the alternating sign has stumped me as well. 
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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)}
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\begin{align}
&\bbox[#ffd,10px]{\ds{%
\sum_{k = 0}^{m}\pars{-1}^{m - k}{n + k - 1 \choose k}{n \choose m - k}}}
=
\sum_{k = 0}^{\infty}\pars{-1}^{m - k}\
\overbrace{\bracks{{-n \choose k}\pars{-1}^{k}}}
^{\begin{array}{l}
\mbox{Negating}
\\
\mbox{the Binomial}
\end{array}}\
{n \choose m - k}
\\[5mm] = &\
\pars{-1}^{m}\sum_{k = 0}^{\infty}
{-n \choose k}\bracks{z^{m - k}}\pars{1 + z}^{n} =
\pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{n}\sum_{k = 0}^{\infty}
{-n \choose k}z^{k}
\\[5mm] = &\
\pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{n}\pars{1 + z}^{-n} =
\pars{-1}^{m}\bracks{z^{m}}z^{0} = \bbx{\delta_{m0}}
\end{align}
A: Start by manipulating the summand
\begin{eqnarray*}
\binom{n+k-1}{k} \binom{n}{m-k}= \frac{n}{m} \binom{m}{k} \binom{n-k-1}{m-1}.
\end{eqnarray*}
Now express the second binomial coefficient as the coefficient of the a function
\begin{eqnarray*}
  \binom{n-k-1}{m-1} = [x^{m-1}]: (1+x)^{n-k-1}
\end{eqnarray*}
So
\begin{eqnarray*}
\sum_{k=0}^{m} (-1)^{m-k} \binom{m}{k} \binom{n-k-1}{m-1} &=& \frac{(-1)^{m}n}{m} \sum_{k=0}^{m} (-1)^{k}\binom{m}{k} \binom{n-k-1}{m-1} \\
&=& \frac{(-1)^{m}n}{m} [x^{m-1}]: \sum_{k=0}^{m} (-1)^{k}\binom{m}{k} (1+x)^{n-k-1} \\
&=& \frac{(-1)^{m}n}{m} [x^{m-1}]: (1+x)^{n-1} \left( 1- \frac{1}{1+x} \right)^m \\
&=& \frac{(-1)^{m}n}{m} [x^{m-1}]: (1+x)^{n-m-1} x^m =\color{red}{0}. \\
\end{eqnarray*}
