# Is there a simpler (preferably closed-form) way to compute this sum of binomial coefficients?

From my years browsing math.SE, I've come to learn there's always some useful--albeit obscure--identity for every conceivable sum of combinatoric functions.

Mine is as simple as they come: $$S(n,m)=\sum _{k=0} ^m {n \choose k}$$ for $m \leq n$.

That is, a partial sum of binomial coefficients. Practically: the number of ways one can choose a subset of $m$ or fewer objects out of a set of $n$ objects.

Apologies if this is already answered elsewhere, but a search for "sum of binomial coefficients" turns up hundreds of permutations (no pun intended) of the question, none of which seem to be relevant to this specific case.

• I doubt there are nice closed-form expressions for those, as they are essentially unnormalized tails of a binomial distribution. However, there are some very tight bounds: look up Chernoff bounds for unbiased coin flips. – amakelov Apr 7 '17 at 19:09
• @amakelov: will do; thanks. – COTO Apr 7 '17 at 19:12

$${n\choose n-m}{\mbox{_2F_1}(1,-m;\,1+n-m;\,-1)}$$
where $\mbox{$_2$F$_1$}$ is a hypergeometric function.