Combinatorial proof to the identity $\sum_{k=0}^n\binom{x+k}{k}=\binom{x+n+1}{n}$? [duplicate]

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Combinatorial proof for two identities

The identity $$\sum_{k=0}^n\binom{x+k}{k}=\binom{x+n+1}{n}$$ can be verified by a straightforward induction, but is there some nice combinatorial argument which proves it? Here $x$ is a nonnegative integer, by the way.

I've always found combinatorial arguments make it easier to remember such identities and intuitively grasp them so I would appreciate seeing one. Thank you.

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Sorry, it was hard to tell out of the questions with similar titles if one was a duplicate, since the identities weren't in the title. I tried to include it in my title. – Threepio Jan 24 '12 at 21:35

marked as duplicate by Arturo Magidin, Byron Schmuland, Aryabhata, Fabian, Grumpy ParsnipJan 24 '12 at 22:02

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1 Answer

Rewrite it as $$\sum_{k=0}^n\binom{x+k}x=\binom{x+n+1}{x+1}\;.$$ The righthand side clearly counts the subsets of $[x+n+1]$ of size $x+1$. The lefthand side breaks that count down according to the largest element of the subset. That is, if the largest element is $x+k+1$, the other $x$ elements must be chosen from the set $[x+k]$, and this can be done in $\binom{x+k}k$ ways. (Here I use $[n]$ for $\{1,\dots,n\}$.)

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