I am not able to figure out where I am going wrong. The question is this -

A total of $2n$ people, consisting of $n$ married couples, are randomly seated (all possible orderings being equally likely) at a round table where men and women are to sit alternately. Let $C_i$ denote the event that the members of couple $i$ are seated next to each other, $i = 1, . . . , n$.

(a) Find $P(C_i)$.

(b) For $j \neq i$, find $P(C_j|C_i)$.

For part (a), my reasoning goes as follows - Pick a man and place him at a particular position. Then, his partner can be choosen in $1$ way and she can be seated in $2$ ways. Then remaining $n-1$ men and $n-1$ women can be seated in $(n-1)!(n-1)!$ ways. The total number of ways of seating $n$ men and $n$ women is $n!(n-1)!$. So, probability is $2/n$.

For part (b), I am starting in same way - Pick a man and place him at a particular position. Then, his partner can be choosen in $1$ way and she can be seated in $2$ ways. Now pick another man, which can be done in $n-1$ ways and his partner, and they can be seated in $n-1$ positions (as there are $n-1$ remaining positions for men and women to sit in alternate positions). Remaining $n-2$ men and $n-2$ women can be seated in $(n-2)!(n-2)!$ ways. So,

$P(C_i\cap C_j) = 2(n-1)^2(n-2)!(n-2)!/(n!(n-1)!) = 2/n$

and $P(C_i) = 2/n$ (from part (a) )

So, $P(C_j|C_i) = P(C_i\cap C_j)/P(C_i) = (2/n)/(2/n) = 1$

I don't understand here where I am going wrong.

  • $\begingroup$ you see arranging n-1 men and n-1 women have (n-2)!(n-1)!.But after that there are 2n-2 gaps for married couple in which they can be put in only one way(so that they sit alternatively).So the answer was right , but method was wrong! $\endgroup$ Feb 8, 2020 at 18:19
  • $\begingroup$ First of all your interpretation of question in 2nd part is wrong because we have fixed the 2 couples who need to together (j is a fixed number like i). Secondly if it was correct then the way you have counted will contain a lot of repeated cases. Let's say first you selected 1st couple , then selected 2nd couple and made them 2 sits right of first couple . Then arranging n-2 couples it is possible 3rd couple sits together and put them 4 seats left of first couple.This is also possible if you selected 3rd couple before 2nd and while arranging n-2 couples made 2nd couple sit together. $\endgroup$ Feb 8, 2020 at 18:31

1 Answer 1


(a) Fix one of the partners in the couple. The other one is equally likely to be in any of the remaining $2n-1$ places, of which $2$ are next to the first partner, so $P(C_i)=\frac2{2n-1}$.

(b) If couple $i$ are seated next to each other, that leaves $2n-2$ seats in a row. All selections of two of these seats for couple $j$ are equally likely. There are $\binom{2n-2}2$ such selections, and $2n-3$ of them have adjacent seats, so $P(C_j\mid C_i)=\frac{2n-3}{\binom{2n-2}2}=\frac{2(2n-3)}{(2n-2)(2n-3)}=\frac1{n-1}$.

  • $\begingroup$ Have you accounted for the fact stated in the problem that seating alternates men and women around the table? $\endgroup$
    – paw88789
    Feb 8, 2020 at 18:12
  • $\begingroup$ @paw88789: I don't see how that's relevant (unless we have some prior for the gender composition of couples, but none is specified). $\endgroup$
    – joriki
    Feb 8, 2020 at 18:14
  • $\begingroup$ @joriki women need to sit in alternate manner so total available seats are n for a woman.It's not possible that her partner sits seat away from him $\endgroup$ Feb 8, 2020 at 18:15
  • 1
    $\begingroup$ @paw88789: Well, I didn't feel like making old-fashioned (a.k.a. heteronormative) assumptions that the OP may or may not have forgotten to state – I answered the question exactly as posed. $\endgroup$
    – joriki
    Feb 8, 2020 at 18:18
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    $\begingroup$ @joriki I just want a solution that gives me the answer that matches in the book (of course assuming that it is correct). If you can get the answer using your assumptions, I will be very happy to accept the answer and also my fault for not considering those assumptions. $\endgroup$
    – Ankit Seth
    Feb 10, 2020 at 13:30

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