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Prove that $$\sum_{k=0}^r {m \choose k} {n \choose r-k} = {m+n \choose r}.$$

This problem is in the chapter about discrete random variables, but I have no idea what to go about substituting.

I can't get it to be a featured formula without screwing stuff up, sorry about that.

Michael Hardy
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Nick
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4 Answers4

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Standard Binomial Coefficients

Use the Binomial Theorem $$ \begin{align} (1+x)^m(1+x)^n &=\sum_{k=0}^m\binom{m}{k}x^k\sum_{j=0}^n\binom{n}{j}x^j\\ &=\sum_{k=0}^m\sum_{j=0}^n\binom{m}{k}\binom{n}{j}x^{k+j}\\ &=\sum_{k=0}^m\sum_{r=k}^{k+n}\binom{m}{k}\binom{n}{r-k}x^r&&j=r-k\\ &=\sum_{r=0}^{m+n}\color{#C00000}{\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}}x^r\tag{1}\\ (1+x)^{m+n} &=\sum_{r=0}^{m+n}\color{#C00000}{\binom{m+n}{r}}x^r\tag{2} \end{align} $$ Compare coefficients of $x^r$ in $(1)$ and $(2)$.

Generalized Binomial Coefficients

As Michael Hardy mentions, the formula is true, even when $m$ and $n$ are not integers. The binomial coefficients can be generalized to any real number in the top argument: $$ \binom{x}{k}=\frac{x(x-1)(x-2)\dots(x-k+1)}{k!}\tag{3} $$ When $x$ is a negative integer, $(3)$ gives the formula for the negative binomial coefficients: $$ \begin{align} \binom{-n}{k} &=\frac{-n(-n-1)(-n-2)\dots(-n-k+1)}{k!}\\ &=(-1)^k\frac{(n+k-1)(n+k-2)(n+k-3)\dots n}{k!}\\ &=(-1)^k\binom{n+k-1}{k}\tag{4} \end{align} $$ The Generalized Binomial Theorem states that for any real $m$, $$ (1+x)^m=\sum_{k=0}^\infty\binom{m}{k}x^k\tag{5} $$ Note that when $m$ is a non-negative integer, $\binom{m}{k}=0$ for $k\gt m$ and so $(5)$ is a polynomial in that case. Using $(5)$, we can imitate the proof for the standard binomial coefficients: $$ \begin{align} (1+x)^m(1+x)^n &=\sum_{k=0}^\infty\binom{m}{k}x^k\sum_{j=0}^\infty\binom{n}{j}x^j\\ &=\sum_{k=0}^\infty\sum_{j=0}^\infty\binom{m}{k}\binom{n}{j}x^{k+j}\\ &=\sum_{k=0}^\infty\sum_{r=k}^\infty\binom{m}{k}\binom{n}{r-k}x^r&&j=r-k\\ &=\sum_{r=0}^\infty\color{#C00000}{\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}}x^r\tag{6}\\ (1+x)^{m+n} &=\sum_{r=0}^\infty\color{#C00000}{\binom{m+n}{r}}x^r\tag{7} \end{align} $$ Compare coefficients of $x^r$ in $(6)$ and $(7)$.

robjohn
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Notice that we set ${i \choose j} = 0$ whenever $0 \leq j \leq i$ is false.

\begin{align} \color{#ff0000}{\large\sum_{k = 0}^{\infty}{m \choose k}{n \choose r - k}} &= \sum_{k = 0}^{\infty}{m \choose k}\sum_{\ell = 0}^{n}{n \choose \ell} \delta_{\ell, r - k} = \sum_{k = 0}^{\infty}{m \choose k}\sum_{\ell = 0}^{n}{n \choose \ell} \oint_{\left\vert z\right\vert = 1}{{\rm d} z \over 2\pi{\rm i}}\, {1 \over z^{\ell - r + k + 1}} \\[3mm]&= \oint_{\left\vert z\right\vert = 1}{{\rm d} z \over 2\pi{\rm i}}\, {1 \over z^{1 - r}} \sum_{k = 0}^{m}{m \choose k}\left(1 \over z\right)^{k} \sum_{\ell = 0}^{n}{n \choose \ell}\left(1 \over z\right)^{\ell} \\[3mm]&= \oint_{\left\vert z\right\vert = 1}{{\rm d} z \over 2\pi{\rm i}}\, {1 \over z^{1 - r}}\,\left(1 + {1 \over z}\right)^{m} \left(1 + {1 \over z}\right)^{n} \\[3mm]&= \oint_{\left\vert z\right\vert = 1}{{\rm d} z \over 2\pi{\rm i}}\, {1 \over z^{m + n -r + 1}}\,\left(1 + z\right)^{m + n} = \oint_{\left\vert z\right\vert = 1}{{\rm d} z \over 2\pi{\rm i}}\, {1 \over z^{m + n -r + 1}}\sum_{k = 0}^{r + n}{r + n \choose k}z^{k} \\[3mm]&= \sum_{k = 0}^{m + n}{m + n \choose k} \oint_{\left\vert z\right\vert = 1}{{\rm d} z \over 2\pi{\rm i}}\, {1 \over z^{m + n - r + 1 - k}} = \sum_{k = 0}^{m + n}{m + n \choose k}\delta_{m + n - r,k} \\[3mm]&= {m + n \choose m + n - r} = \color{#ff0000}{\large{m + n \choose r}} \end{align}

Felix Marin
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    this seems unnecessarily complicated. is there really no easier way? not to sound ungrateful, there's just no way this is the kind of answer we're expected to crank out on homework. – Nick Sep 24 '13 at 01:23
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    There are many easier ways. This is one of the funnier ones I've seen, though! (This is basically the binomial-identity method 'disguised' through contour integrals) – Steven Stadnicki Sep 24 '13 at 15:37
  • +1 I didn't try to follow your proof but I like the symbols and the colours – miracle173 Sep 24 '13 at 16:51
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    Drat! I need to add more symbols to my answer... – robjohn Sep 24 '13 at 16:56
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Think about picking a subsets of size $r$ from a set of size $m+n$. You can simply choose it, and you can do it in

$${m+n} \choose r$$

ways, or you can first choose a subset of size $k$ of $m$ and then a subset of size $r-k$ of $n$. And you can do this for $k=0,1,\ldots,r$ in

$${m \choose k }{n \choose r-k}$$

ways.

Pedro
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  • +1 but I am no really satisifed with the formulation, e.g. a soubset of m, but here m is a number , not a set. And "you can do this for k=": you have to distinguish this different cases. basically we have two disjoint sets of size m and n and we pick r elements from the union or we pick from each set individually. – miracle173 Sep 24 '13 at 22:11
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    @miracle173 There is a natural way to think of $m$ as ${1,2,\ldots,m}$ or ${0,1,2,\ldots,m-1}$ if you're a set theorist. – Pedro Sep 24 '13 at 22:16
  • @miracle173 Indeed, when I write "you can do this for $k=0,1,2,\ldots$ I am precisely distinguishing the cases. You are to fix the numbers $m,n$, then move along. – Pedro Sep 24 '13 at 22:31
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Vandermonde's identity $$ \sum_{k=0}^r {m \choose k} {n \choose r-k} = {m+n \choose r} $$ is actually true even when $m$ and $n$ are not integers. But just suppose you have a committee in the Senate consisting of $m$ Democrats and $n$ Republicans. The number of ways to choose a subcommittee of $r$ senators is $$ \sum_{k=0}^r\Big(\text{the number of ways to choose $k$ Democrats AND $r-k$ Republicans}\Big). $$ You should associate the word "and" with multiplication.

As far as I know, one must resort to other methods when $m$ or $n$ is not an integer or is negative.

PS: When $m$ is not an integer or is negative then $\dbinom m k$ is defined as $$ \frac{m(m-1)(m-2)\cdots(m-k+1)}{k!}. $$ When $m$ is a nonnegative integer, then this is the same as $\dfrac{m!}{k!(m-k)!}$.

Michael Hardy
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