$$\sum_{0 \le k \le a}{a \choose k}{b \choose k} = {a+b \choose a}$$
Is there any way to prove it directly?
Using that $\displaystyle{a \choose k}=\frac{a!}{k!(a-k)!}$?
$$\sum_{0 \le k \le a}{a \choose k}{b \choose k} = {a+b \choose a}$$
Is there any way to prove it directly?
Using that $\displaystyle{a \choose k}=\frac{a!}{k!(a-k)!}$?