Problem - 4302
Let $m$ and $n$ be two positive integers satisfying $m < n$. Show that $$S_{m,n}=\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0$$
The conclusion of # 2682 states $$\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}$$
Therefore, $$S_{m,n}=\binom{n}{m}\sum_{k=m}^{n}(-1)^k\binom{n-m}{k-m}$$
Let $l=k-m$, the above expression is equivalent to $$\binom{n}{m}\sum_{l=0}^{n-m}(-1)^{m+l}\binom{n-m}{l} = (-1)^{m}\binom{n}{m}\sum_{l=0}^{n-m}(-1)^l\binom{n-m}{l}=0$$
The last step utilized the conclusion of # 3158.