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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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closed as off-topic by Did, user91500, Grigory M, amWhy, Hakim May 18 at 13:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user91500, Grigory M, amWhy, Hakim
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please i need this question answer urgently,the answer are given 7!,3120,4080,0............please help me out –  gaurav hazra Oct 10 '12 at 0:37
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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
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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

1 Answer 1

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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