# probability and permutation and combination

7 friends went to see a movie, at the time of interval they went away, in how many ways when they come back can seat that no 2 adjacent people will not seat together?

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Please state the question more clearly and also tell us what you have tried. –  Trevor Wilson Oct 10 '12 at 1:18
I take it the friends were seated one next to the other in a single row, and what's wanted is the number of ways to reseat them so no two who were adjacent before will be adjacent now. Is that right? –  Gerry Myerson Oct 10 '12 at 1:23
Why the urgency, by the way? –  Gerry Myerson Oct 10 '12 at 1:26
@GerryMyerson: It certainly appears to be asked on a cell phone for a multiple-choice exam. –  Noah Snyder Oct 10 '12 at 9:09
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EDIT: Looking at that page (and assuming that I have the correct interpretation of the question), it seems unlikely to me that this could be set as an exam question, and the answer doesn't seem to be any of the choices given by OP in the first comment. The recurrence is $$a(n) = (n+1)a(n-1)-(n-2)a(n-2)-(n-5)a(n-3)+(n-3)a(n-4)$$ with $a(0)=a(1)=1$, $a(2)=a(3)=0$, not the kind of thing I'd expect someone to find under test conditions. The closest thing to a closed-form formula is $$a(n)=n!+\sum_{k=1}^n(-1)^k\sum_{t=1}^k{k-1\choose t-1}{n-k\choose t}2^t(n-k)!$$ which again I wouldn't expect to see on an exam. There's an asymptotic expansion $${a(n)\over n!}\sim e^{-2}\bigl(1-2n^{-2}-(10/3)n^{-3}-6n^{-4}-(154/15)n^{-5}\bigr)$$
Of course, the question doesn't ask for a general formula or asymptotics, just for $a(7)$, but I don't see any easy way to do even that under test conditions.