# Combinatorial Argument for identity [duplicate]

Prove $$\sum_{k=0}^n\binom nk\binom{m+k}n=\sum_{k=0}^n\binom nk\binom{m}k2^k$$

I know LHS is the number of ways to divide group of $m+k$ people to three groups of $k,n-k,m+k-n$ people each but since the interative is also involved it doesnt feel right.

Algebraic solutions are also welcome