# Closed form of binomial sum

I want to find closed form of the following expression. Will you kindly help me?

$\binom{e}{0}\binom{n-e}{i}+\binom{e}{2}\binom{n-e}{i-2}+\ldots+\binom{e}{i}\binom{n-e}{0}$ for an even positive integer $i(<n)$. Also $e<n$.

It is clear that if all terms like $\binom{e}{1}\binom{n-e}{i-1}$ are present in the above expression, one can find the closed form.

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Do you have a reason for thinking that there is a closed form? –  Gerry Myerson Sep 28 '12 at 6:07
Yes, for some other calculation I require this. –  user12290 Oct 20 '12 at 22:33
That's a reason for hoping there is a closed form, not a reason for thinking there is one. –  Gerry Myerson Oct 20 '12 at 22:55

This isn't exactly a closed form, but depending on what you need this for it might help.

Let's consider

$$\sum_{k=0}^i(-1)^k\binom ek\binom{n-e}{i-k}\;,$$

from which you can obtain your sum by taking the average with

$$\sum_{k=0}^i\binom ek\binom{n-e}{i-k}=\binom ni\;.$$

This is the excess of the number of selections of $i$ items from $n$ such that an even number of items are in some subset of size $e$ over the number of choices where an odd number are in the subset.

This is the coefficient of $q^i$ in $(1-q)^e(1+q)^{n-e}$. That's not particularly helpful as it stands, but if $e$ is near $n/2$ we can get a sum with fewer terms from it. Assume $e\le n/2$ (the case $e\gt n/2$ can be treated analogously). Then the coefficient of $q^i$ in $(1-q)^e(1+q)^{n-e}=(1-q^2)^e(1+q)^{n-2e}$ is

$$\sum_{j=0}^e(-1)^j\binom ej\binom{n-2e}{i-2j}\;,$$

where the coefficients with $i-2j\lt0$ or $i-2j\gt n-2e$ are $0$, so the sum contains few terms for small $i$ (like the original sum) and for small $n-2e$ (unlike the original sum).

In particular, for $n=2e$, we get $(-1)^{i/2}\binom e{i/2}$.

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