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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The answer, and much more, is available here at the Online Encyclopedia of Integer Sequences. 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. MORE EDIT: This seems to me to be an interesting question. I don't see why it was closed. |
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