I encountered the following formula while working out a problem in a different context and could not figure out a way to simplify it. After spending a fair amount of time on it and not really making headway, I'm hoping someone here can help me out.

$$E(S) = \sum_{k=0}^{m} \dfrac{k}{2m-k} \dfrac{{m\choose k}\times{n-m\choose m-k}}{n\choose m}$$

If anyone needs to know where this formula came up from, here's a link http://infolab.stanford.edu/~ullman/mmds/ch3.pdf Ex 3.1.3


If we ignore the binomial $\binom{n}{m}$ in the denominator, because it is constant across the $k$-sum, conversion to $\Gamma$-functions yields $$\binom{n}{m}E(s)=$$ $$\frac{m}{2m-1}\binom{n-m}{m-1}{}_3F_2(1-2m,1-m,1-m;2-2m,2-2m+n;1)$$. This might have a simpler representation if the hypergeometric series can be reduced with one of the formulas in http://arxiv.org/abs/1105.3126 ,which I did not check.

  • $\begingroup$ Thanks but this solution seems to be way more complicated than I was expecting. I'll hold out for a while in the hope of a simpler solution. In the meantime, could you give me some pointers to hypergeometric series so that I can follow how you arrived at the answer. $\endgroup$ – krypto07 Feb 28 '14 at 19:13

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