# Evaluate the following sum using a combinatorial argument [duplicate]

Evaluate the following sum using a combinatorial argument:

$$\sum\limits_{k=0}^n {n \choose k} {m \choose k}$$

Can someone push me in the right direction with this? I thought for combinatorial proofs there has to be a left side and a right side where one side can be used to form a question? (if that makes sense? haha)

Is there a difference with combinatorial arguments? Any help would be greatly appreciated.

• You are supposed to find the right hand side yourself :) – darij grinberg Feb 24 '15 at 20:12
• Is $m$ a constant, or is it somehow related to $n$? – Théophile Feb 24 '15 at 20:13
• There is nothing saying that it is related to n in anyways, so I am assuming it is a constant. – MathyMatherson Feb 24 '15 at 20:16
• I think we can assume $m \ge n$, so that the problem makes sense. – DanielV Feb 24 '15 at 20:20
• Try a few examples: choose $n=3$, say, and try different values for $m$. Do you see the pattern? – Théophile Feb 24 '15 at 20:20

$$\sum_{k=0}^n\binom{n}k\binom{m}{m-k}\;.$$