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I am looking for a combinatorial interpretation to $f(n)=\sum\limits_{k=0}^{n}(-1)^{n-k}\binom{n}{k}g(k)$, where

1.) $g(n)=\binom{nr}{s}$

2.) $g(n)=\binom{\binom{n}{r}}{s}$

3.) $g(n)=2^{\binom{n}{2}}$

bronko
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    You phrase it as if you are giving someone an order --- that's not appreciated. Also, you give no indication that you have put the slightest effort into the problem, no indication of what you do or don't know about it. You don't make it easy to help you. – Gerry Myerson Nov 22 '12 at 12:04
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    I am sorry for that. I did not mean to be rude. Well, the $f(n)$ part looks to me like Inclusion-Exclusion principle. So I guess it is about counting the intersections of some kind of sets. Other than that I can not really think of anything. Thank you! – bronko Nov 22 '12 at 12:30
  • You might find the answers to this question helpful. – Mike Spivey Nov 23 '12 at 04:36

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Hints: $$f(n)=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}g(k)=\sum_{m=0}^{n}(-1)^{m}\binom{n}{m}g(n-m)$$ Try to use the Inclusion–exclusion principle: Find sets $A_1,...,A_n$ such that the intersection of any $m$ of them is of size $g(n-m)$. Give a combinatorial interpretation (as bad properties) to $A_1,...,A_k$, and then $f(n)$ is the amount of elements with no 'bad' properties.

For example:
In $(3)$ you can take $A_k$ to be the set of all labeled graphs on $n$ vertices in which the vertex $k$ is isolated (not connected to any other vertex). Then $f(n)$ will be the number of graphs on $n$ vertices without isolated vertices.

Dennis Gulko
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  • Thank you, @DennisGulko for your answer. I do not understand what you mean by recombination interpretation. Would you give me an example, please! – bronko Nov 22 '12 at 12:38
  • Stupid auto-correct! I will edit my answer now. – Dennis Gulko Nov 22 '12 at 15:05
  • So in the case of 3.) it would mean that $f(n)$ is the number of elements with no two 'bad properties', right? – bronko Nov 24 '12 at 07:26
  • No - still without any bad properties. For example, you can take $A_k$ to be the set of all labeled graphs on $n$ vertices in which the vertex $k$ is isolated (not connected to any other vertex). Then $f(n)$ will be the number of graphs on $n$ vertices without isolated vertices. – Dennis Gulko Nov 24 '12 at 09:12
  • So $f(n)$ is the number of graphs on n vertices without isolated vertices because $2^{\binom{k}{2}}$ is the number of elements in the intersection of all sets of labeled graphs on $n$ vertices with an isolated vertex?! – bronko Nov 25 '12 at 10:05
  • $2^{\binom{n-m}{2}}$ is the number of graphs on $n$ vertices in which the vertices $1,...,m$ are isolated. – Dennis Gulko Nov 25 '12 at 10:13
  • Thank you, Dennis. However, I do not have much experience with combinatorics so I am stuck with coming up with ideas for the other two. I know that 1.) it is choosing $s$ elements out of a $(n-m)r$-set and 2.) choosing $r$ elements from a $n-m$-set and then choosing $s$ elements from those. Still I dont see how this can be intersections of sets with bad properties. – bronko Nov 26 '12 at 06:43