Alternating sum of a part of a row of Pascal's triangle $\displaystyle \sum_{r=0}^m(-1)^r{n \choose r}=(-1)^m{n-1 \choose m}$ if $m$ is less than $n$.
This question actually consists of two part that is when $m$ is less than $n$ and when $m$ is equal to $n$. I can solve the second part but not the first part.
 A: Hint: Use induction on $m$. This boils down to proving
$$ (-1)^m \binom{n-1}{m} + (-1)^{m+1} \binom{n}{m+1} = (-1)^{m+1} \binom{n-1}{m+1}, $$
which follows from Pascal's identity.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{r\ =\ 0}^{m}\pars{-1}^{r}{n \choose r}
     =\pars{-1}^{m}{n - 1 \choose m}:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\sum_{r\ =\ 0}^{m}\pars{-1}^{r}{n \choose r}}
=\sum_{r\ =\ 0}^{m}\pars{-1}^{r}\ \overbrace{\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n} \over z^{r + 1}}\,{\dd z \over 2\pi\ic}}^{\ds{=\ {n \choose r}}}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n} \over z}\sum_{r\ =\ 0}^{m}\pars{-\,{1 \over z}}^{r}
\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n} \over z}{\pars{-1/z}^{m + 1} - 1 \over -1/z - 1}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n} \over z}\,{z^{m + 1} + \pars{-1}^{m} \over z^{m}\pars{1 + z}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\ \underbrace{\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n - 1}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ 0}}\ +\
\pars{-1}^{m}\
\underbrace{\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 1} \over z^{m + 1}}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ {n - 1 \choose m}}}
\end{align}

$$\color{#66f}{\large\sum_{r\ =\ 0}^{m}\pars{-1}^{r}{n \choose r}
=\pars{-1}^{m}{n - 1 \choose m}}
$$
