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How many ways to arrange $n\geq2$ people around a circular table, given two specific people who cannot stand next to each other?

I've observed that when $n=2$ and $n=3$ there exists no way to arrange them so that the two specific people aren't standing next to each other. I think for when $n=4$ there are $3!$ ways to arrange them, but I've had trouble coming up with a formula that can describe all values of $n$.

Perhaps I can attack the problem by first arranging the other people in the circle, and then placing either one of the two specific people in between the gaps?

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  • $\begingroup$ @saulspatz Thanks for the info. Updated the body with what I've come up with so far. $\endgroup$ – ivyleaf57 Feb 6 at 17:51
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    $\begingroup$ Just think of the number of ways you can seat the two next to each other for $N$ people. Subtract that from the total number of ways to arrange $N$ people. $\endgroup$ – Bridgeburners Feb 6 at 18:01
  • $\begingroup$ Does rotated things count? And reflected? $\endgroup$ – Math Lover Feb 6 at 18:15
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We can fix one of the 2 pearson. Then we have $n-3$ positions for the second pearson (since one is already ocupied and it can not be seated near to it), and then we can arrange the other on $(n-2)!$ ways, so the answer is $(n-3)\cdot (n-2)!$

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Well there are $n$ seats. Assuming the seats are not numbered, so the first mandatory person can be placed anywhere.

Then, place the second mandatory person, there are $n-3$ choices (why?)

Place the rest of $n-2$ people into $n-2$ seats, so $(n-2)!$ ways

Total: $(n-3)(n-2)!$ ways

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  • $\begingroup$ It's a circular table, so it doesn't matter where the first person is seated. $\endgroup$ – jvdhooft Feb 6 at 18:09
  • $\begingroup$ I see, guess I assumed the seats are numbered so person 1 has the best view of the TV $\endgroup$ – NazimJ Feb 6 at 18:37

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