Series evaluation

Question

I'm wondering how following series from Jaynes' "Probability Theory: The Logic of Science" can be evaluated:

$$ \sum_{R=0}^{N}\binom{R}{r}\binom{N-R}{n-r}\binom{N}{R}\binom{N}{n}^{-1}g^R(1-g)^{N-R} = \binom{n}{r}g^r(1-g)^{n-r} $$

I just can't belive it's true. I would be grateful for clarification!

Answer 1

Let us start by tidying up part of the summand \begin{eqnarray*} &\binom{R}{r} \binom{N-R}{n-r} \binom{N}{R} \binom{N}{n}^{-1} \\ &=\frac{R!}{r!(R-r)!} \frac{(N-R)!}{(n-r)!(N-R-n-r)!} \frac{N!}{R!(n-R)!} \frac{n!(N-n)!}{N!} \\&= \binom{n}{r} \binom{N-n}{R-r}. \end{eqnarray*} Now the sum can be rewritten \begin{eqnarray*} & &\sum_{R=0}^{N} \binom{R}{r} \binom{N-R}{n-r} \binom{N}{R} \binom{N}{n}^{-1} g^R(1-g)^{N-R} \\&=& \binom{n}{r} g^r(1-g)^{n-r} \sum_{R=0}^{N} \binom{N-n}{R-r}g^{R-r}(1-g)^{N-R-n-r} \\ &=& \binom{n}{r} g^r(1-g)^{n-r} \underbrace{\sum_{R=r}^{N-n+r} \binom{N-n}{R-r}g^{R-r}(1-g)^{N-R-n-r}}_{[g+(1-g)]^{N-n}=1} \\ &=& \binom{n}{r} g^r(1-g)^{n-r}. \end{eqnarray*}