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In answering someone else's question (Permutation query involving 3 people with a limit of 45 as the sum.) I came across an interesting result that wasn't immediately obvious to me:

$$\binom{n}{k} = \sum_{m=k}^{n} \binom{m-1}{k-1}$$

It basically says that the number of positive integer $k$-tuples whose sum is less than or equal to $n$ is given as $\tbinom{n}{k}$. I have not actually proven that this relationship holds, but I threw an handful of different test values at it, and the equality held.

If this is true, is there any kind of intuitive explanation as to why? I imagine that it can be proven mathematically without great effort, but I am mostly interested in a "balls and urns" or "stars and bars" type explanation, if one exists.

Community
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nispio
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2 Answers2

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This is known as the hockey stock identity.

There is an immediate algebraic proof by looking at the terms in Pascal's triangle.

There is an immediate combinatorial proof by considering the number of subsets of n with size k, and double counting conditional on the smallest element.

Calvin Lin
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  • Thanks for pointing me to the hockey stick identity. There is something fun about independently discovering something that already exists. – nispio Nov 22 '13 at 04:56
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Since you mentioned stars-and-bars, note that $\sum_{m=k}^n\binom{m-1}{k-1}$ is the number of solutions in positive integers to

$$k\le x_1+x_2+\ldots+x_k\le n\;.\tag{1}$$

For any solution $\langle x_1,\ldots,x_k\rangle$ to $(1)$ let $x_{k+1}=n+1-\sum_{i=1}^kx_i\ge 1$; then

$$x_1+x_2+\ldots+x_k+x_{k+1}=n+1\tag{2}\;.$$

Conversely, if $\langle x_1,\ldots,x_{k+1}\rangle$ is a solution to $(2)$ in positive integers, then $\langle x_1,\ldots,x_k\rangle$ is a solution to $(1)$ in positive integers. Since $(2)$ has $\binom{n}k$ solutions in positive integers, we must have

$$\sum_{m=k}^n\binom{m-1}{k-1}=\binom{n}k\;.$$

Brian M. Scott
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  • Not exactly balls and urns, but a very nice proof. :) Thanks. – nispio Nov 22 '13 at 04:54
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    @nispio: Well, you can always translate it: the sum is the number of ways of putting between $k$ and $n$ identical balls into $k$ labelled urns, with at least one ball per urn, and the righthand side is the number of ways of putting $n+1$ identical balls into $k+1$ urns with at least one ball per urn, thereby putting between $k$ and $n$ into the first $k$ urns. :-) You’re welcome. – Brian M. Scott Nov 22 '13 at 15:13