The question is this:
In how many partitions of $\{1,2,\ldots,n\}$ into three disjoint sets, $A$, $B$, and $C$, are there no two consecutive numbers in the set $A$?
thanks!
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Do you consider these ordered partitions, so that, for instance, $A={1,2}$, $B={3}$, $C={4}$ does not qualify, but $A={3}$, $B={1,2}$, $C={4}$ is okay? – Arturo Magidin Feb 11 '11 at 21:29
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exactly, second exmpale is okay. – Feb 11 '11 at 21:32
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Number of ways of selecting $k$ numbers from $1,2,\dots,n$ such that no two are consecutive is
$\displaystyle \binom{n-k+1}{k}$, for instance see my answer here: Consecutive birthdays probability
Once you do that, count the number of partitions of the remaining into $B$ and $C$.
Which is given by $2S(n-k,2) + 2$ where $S(a,b)$ are the Stirling numbers of the second kind (for and explanation see the answer linked above).
Now I guess your answer is
$$\sum_{k=0}^{n} \binom{n-k+1}{k} (2S(n-k,2) + 2)$$
(Note: I haven't tried to confirm it, I leave that to you)
Hope that helps.
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Having pulled out $k$ items for $A$, aren't there $2^{(n-k)}$ subsets that you could choose for $B$ so you would have $$\sum_{k=0}^{n} \binom{n-k+1}{k}2^{(n-k)}$$? It looks like your answer doesn't allow $B$ or $C$ to be empty and considers swapping $B$ and $C$ to be the same configuration. – Ross Millikan Feb 11 '11 at 22:08
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@Ross: $S(n,2) = 2^{n-1} -1$. I mentioned Strirling numbers because it generalizes to more sets :-) See here: http://en.wikipedia.org/wiki/Stirling_number_of_the_second_kind#Simple_identities – Aryabhata Feb 11 '11 at 22:17
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@Moron: yes, I had seen that, which is how I figured out the differences between the approaches. It depends what you want to count. – Ross Millikan Feb 11 '11 at 23:18
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@Ross: Multiplying by 2 and adding 2 is supposed to take care of swapping B and C, and empty sets. The formula($2^{n-1}-1$) then gives what you wrote in your comment, so it does seem to match... Am I missing something, or do we agree now? – Aryabhata Feb 11 '11 at 23:21
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@Moron: sorry for that. I just read it differently, perhaps prompted by Arturo's question to OP. Depending upon what you want to count, I think we were both right, and I should have said that earlier. – Ross Millikan Feb 12 '11 at 03:05