# Truncated alternating sum of binomial coefficients

Are there any nice counting arguments that identify the sum

$$S_{n,k} = {n \choose k} - {n \choose k-1} + {n \choose k-2} + \cdots + (-1)^k {n \choose 0}$$

with a simpler, more conceptually-appealing expression?

This expression comes up as the type of rank of a homology group of a somewhat complicated object and I'm hopeful I might be inspired by a counting argument.

• $S_{n,0}=1$ doesn't it? – JMoravitz Oct 10 '18 at 5:47
• Thanks. I did not think about those properties as carefully as I should have. I'll just edit them out. – Ryan Budney Oct 10 '18 at 19:46
• @RyanBudney Could you help me with this please? where we are using that $f,g:(X,A)\to (Y,B)$ are homotopic in math.stackexchange.com/questions/104795/…. – Nash Jul 3 at 20:16

$$S_{n,k}=\binom{n-1}k.$$ Look at the subsets of $$[n]=\{1,\ldots,n\}$$ of size $$\le k$$, and pair off $$A$$ with $$A\cup\{n\}$$ for $$A\subseteq[n-1]$$. The unpaired sets are the $$k$$-element subsets of $$[n-1]$$.
• It seems to be the binomial expansion of $(1-1)^n$. – sirous Oct 10 '18 at 7:52