TITLE: Proof of $\sum_{0 \le k \le a} {a \choose k} {b \choose k} = {a+b \choose a}$
QUESTION [8 upvotes]: $$\sum_{0 \le k \le a}{a \choose k}{b \choose k} = {a+b \choose a}$$
Is there any way to prove it directly?
Using that $\displaystyle{a \choose k}=\frac{a!}{k!(a-k)!}$?

REPLY [2 votes]: How about this proof? (Actually an extended version of your identity.)

http://en.wikipedia.org/wiki/Chu-Vandermonde_identity#Algebraic_proof

I don't think it is "direct" enough, though...