Show that$\displaystyle \sum_{t \ge 0} \binom{n}{j+t} \binom{m}{k+t} = \binom {m+n}{n+k-j}$

Question

$$\sum_{t \ge 0} \binom{n}{j+t} \binom{m}{k+t} = \binom {m+n}{n+k-j}$$

Can anybody provide me an idea of a combinatorical proof of this identity. I can deduce it algebraically, but I could not find a suitable combinatorical proof. I want a practical scenario to prove this.

Answer 1

If we remove the restriction on $t$, the LHS becomes

$$\sum_{t} \binom{n}{j+t} \binom{m}{k+t} = \sum_{t} \binom{n}{n-j-t} \binom{m}{k+t}$$

Then the LHS counts the number of ways of selecting $(n-j-t) +(k+t) = n-j+k$ items from $n+m$ items.