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$\sum_{0 \le k \le a}{a \choose k}{b \choose k} = {a+b \choose a}$

Is there any way to prove it directly?

Using that $\displaystyle{a \choose k}=\frac{a!}{k!(a-k)!}$?

2 Answers 2

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There is a very nice combinatorical proof without any calculations.

Note that ${a \choose k} = {a \choose a-k}$. So your sum is $\sum_{0 \le k \le a}{a \choose a-k}{b \choose k}.$ What does it mean? (what combinatorical objects does it count?) You choose $a-k$ elements from a $a$-element set and $k$ elements from a $b$-element set (you can do it in ${a \choose a-k}{b \choose k}$ ways). You do it with all possible $k$ ($\sum{a \choose a-k}{b \choose k}$). It's just choosing $a$ elements from the sum of those sets which has $a+b$ elements (${a+b \choose a}$).

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

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

  • 0
    What I mean by direct is using (ab)=a!/(a-b)!b! I am wondering is there any way to prove it by going to $\sum_{0 \le k \le a} {a \choose k} {b \choose k} = \sum_{0 \le k \le a} \frac{a!}{k!(a-k)!} \frac{b!}{k!(b-k)!}$2012-02-27