# Probability of 2 married couples sitting in a circle

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.

• 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! Feb 8, 2020 at 18:19
• 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. Feb 8, 2020 at 18:31

(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 $$P(C_j\mid C_i)=\frac{2n-3}{\binom{2n-2}2}=\frac{2(2n-3)}{(2n-2)(2n-3)}=\frac1{n-1}$$.