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I'm trying to figure out a combinatorial proof for:

$$\displaystyle \sum_{k \mathop = 0}^n \binom {r+k} k = \binom {r + n + 1} n$$

I've tried the committee counting thing, but that didn't work.

Martin Sleziak
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OhDear
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  • See http://math.stackexchange.com/questions/357063/proving-that-sum-k-0n-mk-choosem-mn1-choose-m1, http://math.stackexchange.com/questions/833451/prove-sum-i-0n-binomik-1k-1-binomnkk and other posts linked there. – Martin Sleziak Jun 20 '15 at 09:17
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    This question asks specifically about combinatorial proofs: http://math.stackexchange.com/questions/1408642/combinatorial-proof-for-binomial-identity-sum-k-0n-binomkp-binom – Martin Sleziak Aug 29 '15 at 09:12

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A combinatorial proof is by counting bit strings of length $r+n+1$ with exactly $n$ ones. There are $\binom{r+n+1}{n}$ such strings.

Such strings have $r+1$ zeros. Now count all bit strings of length $r+n+1$ with exactly $r+1$ zeros in which the last $0$ is at $(r+k+1)$-st position. Here $k \in \{0,1,\ldots,n\}$. There are exactly $\binom{r+k}{r} = \binom{r+k}{k}$ such strings. Since all possible positions for the last zero are $r+1, \ldots, r+n+1$, and since the sets of bit strings with the last zero at different positions are disjoint, this proves the given identity.

wdacda
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