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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.

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  • $\begingroup$ $S_{n,0}=1$ doesn't it? $\endgroup$ – JMoravitz Oct 10 '18 at 5:47
  • $\begingroup$ Thanks. I did not think about those properties as carefully as I should have. I'll just edit them out. $\endgroup$ – Ryan Budney Oct 10 '18 at 19:46
  • $\begingroup$ @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/…. $\endgroup$ – Nash Jul 3 at 20:16
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$$S_{n,k}=\binom{n-1}k.$$ 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 $[n-1]$.

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  • $\begingroup$ It seems to be the binomial expansion of $(1-1)^n$. $\endgroup$ – sirous Oct 10 '18 at 7:52
  • $\begingroup$ @sirous No, since it only goes part of the way. $\endgroup$ – Lord Shark the Unknown Oct 10 '18 at 17:25
  • $\begingroup$ Thanks, that makes sense. $\endgroup$ – Ryan Budney Oct 10 '18 at 20:05

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