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In how many ways can $n$ couples sit at a round table such that no couple is sitting opposite each other?

I know that if we were to only arrange $n$ couples around a table we get $(2n-1)!$. But I don't know how to do this question.

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  • $\begingroup$ So you want $2n$ people in a circle such that no two people from the same couple are sitting opposite each other? What exactly is the $(n-1)!$ counting? $\endgroup$ – TMM Apr 27 '13 at 14:43
  • $\begingroup$ (n-1)! is counting if we arrange the couples around the circular table but allowing couples to sit opposite to each other. $\endgroup$ – please delete me Apr 27 '13 at 15:37
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    $\begingroup$ How is that $(n - 1)!$? There are $2n$ people, so I suppose the answer would be closer to $2n!$. Or should couples sit on the same seat? $\endgroup$ – TMM Apr 27 '13 at 16:11
  • $\begingroup$ it would be $(2n-1)!$ because it's a circular table. $\endgroup$ – pad Apr 27 '13 at 16:20
  • $\begingroup$ Sorry, yes I meant $$(2n-1)!$$ But this isn't the solution to this question. How do I do this question? $\endgroup$ – please delete me Apr 27 '13 at 16:51
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If we have n couples, we have 2n people. And we can arrange 2n people around a circular table in $\frac{2n!}{2n}$ = $(2n-1)!$ For n-1 couples say we have $a_{n-1}$ arrangements. When we go to the n case we can put the first spouse in $2n-2 $ places and then we can put the second in $ 2n -2$ places also. so we get $a_n = a_{n-1}(2n-2)^2$ Rolling back we get $a_n = a_1(2n-2)^2(2n-4)^2...(2)^2 =4^{n-1}(n-1)^2(n-2)^2...(1)^2 =4^{n-1}(n-1!)^2$

...and sorry about the former mess up.

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  • $\begingroup$ Thank you so much! Probability is the hardest topic for me, I can do calculus, algebra and analysis (HS level) but probability, combinatorics and binomial probability can be very tricky, do you have any advice for me? $\endgroup$ – please delete me Apr 27 '13 at 17:41
  • $\begingroup$ My favourite 2 books on probability are both problem books(and come equipped with in depth solutions). These are Fifth Challenging Problems in probability by Frederick Mosteller and 102 Combinatorial Problems by Titu Andreescu. They are both rather witty. A more introductory book is understanding probability by henk tijms. the first half of this book gives intuition on probability and the second half is the theory behind the first half. it's a very "fun" book as well. $\endgroup$ – pad Apr 27 '13 at 18:00
  • $\begingroup$ Thank you, I will try to find them, if not I will buy them. $\endgroup$ – please delete me Apr 27 '13 at 18:05
  • $\begingroup$ @Alexander: This isn’t right, I’m afraid. Suppose that there are $3$ couples, $aA,bB$, and $cC$. If $a$ and $A$ sit opposite each other, there are $4!$ ways to seat the remaining two couples relative to $a$ and $A$. Similarly, there are $4!$ seatings with $b$ and $B$ opposite each other, and $4!$ with $c$ and $C$ opposite each other. If two couples are seated opposite each other, so is the third, and there are $2^3$ such seatings, each of which has been counted $3$ times, so the net number of unacceptable seatings is $3\cdot4!-2\cdot2^3=56$, not $4!$. $\endgroup$ – Brian M. Scott Apr 27 '13 at 18:38
  • $\begingroup$ sorry. I just realised and came back now that what I said assumes some kind of non distinct couples which is nonsense. i'll rewrite what I have at the moment. and if people aren't happy i'll just delete the answer. clearly I need to spend more time with those books... $\endgroup$ – pad Apr 27 '13 at 19:21

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