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.  
 A: 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 $(-1)^{i/2}\binom e{i/2}$.
