Number of ways of seating 2 out of n couples (man and woman) around circular table

Question

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.

Answer 1

• 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.