provethat $\sum_{k}\binom{m}{k}\binom{n+k}{m} = \sum_{j}\binom{n}{j}\binom{m}{j}2^{j}$ Prove that $$\sum_{k=0}^m\binom{m}{k}\binom{n+k}{m} = \sum_{j=0}^{\min(m,n)}\binom{n}{j}\binom{m}{j}2^{j}.$$ So, I know the left side of the identity counts the Delannoy paths and the right side counts the lattice ball. But I want to get a direct combinatorial proof of the identity by showing that both sides count the following set $S$. Let $M$ and $N$ be sets with size $m$ and $n$, Let $S$ be the family of the ordered pairs $(A,B)$ such that $A$ is a subset of $M$ and $B$ is an $m$-subset of $N \cup A$.
So, I know how to get the left side of the identity by counting the set $S$, but I am stuck at figuring out the right side of the identity, I am not sure what index $j$ in the right side identity means combinatorially in the set $S$.

Left Hand Side
We have $m$ junior students and $n$ senior students. Some junior students are member of a football team while all senior students are in the team. From the team, $m$ students are starter. Say the number of junior students in the team is $k$. The number of total possibilities is given below:
$$
\sum_{k}{\binom{m}{k}\binom{n+k}{m}}
$$
Right Hand Side
Still the same but with different method. This time we select some ($j$) senior students to be starters, then we select $m-j$ junior students to complete the starters. The remaining $j$ junior students may or may not be in the team. The number of total possibilities is given below:
$$
\sum_{j}{\binom{n}{j}\binom{m}{m-j}\cdot 2^{j}}
$$
Since we count the same objects, the two expressions must be equal
