Vandermonde's identity

In combinatorial mathematics, Vandermonde's identity, named after Alexandre-Théophile Vandermonde, states that

${n+m \choose r}=\sum _{k=0}^{r}{n \choose k}{m \choose r-k}.$

Proof

This may be proved by simple algebra relying on the fact that

${n \choose k}={n! \over k!(n-k)!},$

(see factorial) but it also admits a more combinatorics-flavored bijective proof, as follows. Suppose a committee in the US Senate consists of n Democrats and m Republicans. In how many ways can a subcommittee of r members be formed? The answer is of course

${n+m \choose r}.$

But on the other hand, the answer is the sum over all possible values of k, of the number of subcommittees consisting of k Democrats and r − k Republicans.

The hypergeometric probability distribution

When both sides have been divided by the expression on the left, so that the sum is 1, then the terms of the sum may be interpreted as probabilities. The resulting probability distribution is the hypergeometric distribution. That is the probability distribution of the number of red marbles in r draws without replacement from an urn containing n red and m blue marbles. For example, suppose one chooses a subcommittee of r members randomly from a committee of n Democrats and m Republicans. What then is the probability that the number of Democrats on the subcommittee is k? The answer is the appropriate term in the sum.

Chu-Vandermonde identity

The identity generalizes to non-integer arguments. In this case, it is known as the Chu-Vandermonde identity and takes the form

${s+t \choose n}=\sum _{k=0}^{\infty }{s \choose k}{t \choose n-k}$

for general complex-valued s and t. This identity may be re-written in terms of the falling Pochhammer symbols as

$(s+t)_{n}=\sum _{k=0}^{\infty }{n \choose k}(s)_{k}(t)_{n-k}$

in which form it is clearly recognizable as an umbral variant of the binomial theorem. The Chu-Vandermonde identity can also be seen to be a special case of Gauss's hypergeometric theorem, which states that

$\;_{2}F_{1}(a,b;c;1)={\frac {\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}}$

where $\;_{2}F_{1}$ is the hypergeometric series and $\Gamma (n+1)=n!$ is the gamma function. One regains the Chu-Vandermonde identity by taking a = −n and applying the identity

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

liberally.