Problem - 3289
Let positive integers $m$ and $n$ satisfy $m\le n$. Prove $$\sum_{k=m}^n\binom{n}{k}\binom{k}{m}=2^{n-m}\binom{n}{m}$$
Applying the conclusion of # 2682 leads to $$\sum_{k=m}^n\binom{n}{k}\binom{k}{m}=\binom{n}{m}\sum_{k=m}^n\binom{n-m}{k-m}=\binom{n}{m}\sum_{i=0}^{n-m}\binom{n-m}{i} =2^{n-m}\binom{n}{m}$$