Let $m$ and $n$ be two positive integers satisfying $m\le n$. Find the value of $$S_{m,n} = \displaystyle\sum_{k=0}^{m}(-1)^k\binom{m}{n}$$
When $m=n$, the result will be $0$ by the conclusion of # 3158.
When $m < n$, applying the Pascal identity $$\binom{n}{k}=\displaystyle\binom{n-1}{k-1} + \binom{n-1}{k}$$
gives $$\begin{align*} S_{m,n} =\ &\binom{n}{0} + \displaystyle\sum_{k=1}^{m}(-1)^k\binom{n}{k} \\ \\=\ & 1 + \sum_{k=1}^{m}(-1)^k\left(\binom{n-1}{k-1} + \binom{n-1}{k}\right) \\ \\=\ & 1 - \binom{n-1}{0} + (-1)^m\binom{n-1}{m} \\ \\=\ & (-1)^m\displaystyle\binom{n-1}{m} \end{align*}$$
Therefore, in summary, the answer is $$ \displaystyle\sum_{k=0}^{m}(-1)^k\binom{m}{n}\left\{ \begin{array}{ll} 0 &, m=n \\ (-1)^m\displaystyle\binom{n-1}{m} &, m < n \end{array} \right.$$