# proof of a combinatorial identity

How to prove the following using inclusion exclusion

$$\sum _{k=m} ^{n} (-1)^{k-m} {n \choose k} = {n-1 \choose m-1}$$

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Hint: $\binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}$ –  Thomas Andrews Apr 9 '13 at 14:02
But, by the tag "inclusion-exclusion" perhaps you are looking for a counting proof? –  Thomas Andrews Apr 9 '13 at 14:05
yes, I want a counting proof –  RIchard Williams Apr 9 '13 at 16:05