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

How about this proof? (Actually an extended version of your identity.)


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*http://en.wikipedia.org/wiki/Chu-Vandermonde_identity#Algebraic_proof
I don't think it is "direct" enough, though...
