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I am trying to solve the following question -

A total of $2n$ people, consisting of $n$ married couples (a man and a woman), are randomly seated (all possible orderings being equally likely) at a round table such that men and women alternate. Let $C_i$ denote the event that the members of $i^{th}$ couple are seated next to each other.

I have to find the number of ways where couple $i$ sit together and couple $j$ sit together (i.e. for the event $C_i\cap C_j$). (This is an extension of classic 2n people around a table problem with the restriction that men and women must sit alternately.)

I started by taking any two couples $i$ and $j$ and seating them. Let the members of a couple $i$ be $M_i$ and $W_i$. First fix couple $i$'s seating and let the man sits on the right side of his partner (lets say this combination $M_iW_i$). So, there are $2n-2$ positions remaining, and if the man of couple $j$ sits on the right of his partner, the number of ways of seating couple $j$ are $n-1$ and if he sits on the left, the number of ways are $n-2$. The remaining $n-2$ couples can be seated in $(n-2)!$ ways. So number of ways for $M_iW_i = (n-1+n-2)*(n-2)! = (2n-3)*(n-2)!$.

Similarly number of ways for combination $W_iM_i = (2n-3)*(n-2)!$.

So total ways of seating couples $i$ and $j = 2(2n-3)*(n-2)!$

No I have to multiply this with the number of ways in which I can select couples $i$ and $j$ (which is $\binom {n}{2}$).

So, number of ways of $C_i\cap C_j$ = $\binom {n}{2}2(2n-3)*(n-2)!$

This is where I am stuck as I am not getting correct answer. According to the answer, it seems I am overcounting but I don't understand what.

Ankit Seth
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  • For example, if we have $M_iW_iM_j$ (extending your notation a little) then $W_jM_j$ is not possible. – user10354138 Sep 19 '21 at 13:32
  • I have already taken this into account. Thats why I calculated $M_iW_i$ first with $M_jW_j$ (which is $n-1$) and then with $W_jM_j$ (which is $n-2$) and then summed it up. – Ankit Seth Sep 19 '21 at 13:38
  • Is the question asking for two specific couples to seat together or any two couples? If you find ways for any two couples to seat together, won't you overcount when you multiply by ${n \choose 2}$ as you are counting arrangements where at least couple $i$ and $j$ seat together and then multiplying by ${n \choose 2}$? – Math Lover Sep 19 '21 at 13:50
  • @MathLover Any two couples who sit together. Sit together in the sense that the members of any specific couple selected (2 people) have to sit together. Its not asking about seating two different couples (4 people) together. – Ankit Seth Sep 19 '21 at 13:57
  • @AnkitSeth yes understood, it is not $4$ people seated together. If it is two specific couples $i, j$ then the answer should be $4 (n-2)! (n-1)!$. If any two couples then it is better to find ways in which no wife-husband sits together and in which exactly one husband-wife sits together. – Math Lover Sep 19 '21 at 14:00
  • @MathLover How would I overcount by multiplying by $\binom{n}{2}$? Approaching in a different way, I first select two couples in $\binom{n}{2}$ ways, then fix the seating of one couple (lets say i) and proceed as usual. I will get $2(2n-3)(n-2)!$ ways, so total number of ways should be same again $\binom{n}{2}2(2n-3)(n-2)!$ . – Ankit Seth Sep 19 '21 at 14:21
  • I think you should clarify the problem statement. If it is any two couples and not specific $i, j$ couples, the problem statement is to find number of ways that at least two couples sit together or exactly two couples sit together. By the way I do not think any of them is a closed form. One of the approaches is the application of principle of inclusion exclusion. – Math Lover Sep 19 '21 at 14:35
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    @AnkitSeth in your earlier comment, what is $2 (2n-3) (n-2)!$? Does it not count arrangements where at least two chosen couples are seated together? When you choose couple $i, j$ to seat together as part of ${n \chosoe 2}$, the arrangements count couples $i, j, k, l$ seated together. Now when you choose $k, l$ couples as part of ${n \choose 2}$, it will again count cases where $i, j, k, l$ couples are seated together. There is clearly overcounting without application of Principle of Inclusion Exclusion. – Math Lover Sep 19 '21 at 14:42

1 Answers1

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  • We shall count in "double seats" , so $n$ such seats

  • First assume a $MWMWM...$ configuration looking clockwise

  • Seat couple $i$ at the "head" of the table, ie as reference

  • Seat couple $j$ in any of the remaining $n-1$ seats

  • Seat the remaining men and women in $(n-2)!*(n-2)!$ ways in the slots for men and women respectively

  • Multiply by $2$ for the $WMW..$ configuration

Putting it all together, ans = $\boxed{2*(n-1)(n-2)!*(n-2)!}$

Illustration added for $10$ seats, ie $5$ "double seats"

$\boxed{2*4*3!*3!}$

Note

The wording isn't crystal clear. I have assumed that two particular couples are seated together, and that the others may or may not be together.

true blue anil
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  • You can have this configuration as well - $M_iW_iM_kW_jM_j....$ i.e., in $MW$ configuration, $j^{th}$ couple can sit as $WM$. Similarly for other case well. – Ankit Seth Sep 23 '21 at 02:16