Question: A total of 2n people, consisting of n married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let C_i denote the event that couple i are seated next to each other, i = 1, 2, ... n.(adsbygoogle = window.adsbygoogle || []).push({});

(a) Find P(C_i).

There are two different ways to seat C_i together on a round table and the possible orderings of the rest of the couples is (2n - 2)!, hence we have:

P(C_i) = (2 * (2n - 2)!)/(2n - 1)!) = 2/(2n - 1)

The the (2n - 1)! being the total number of orderings (C_i included) on a round table. The book gives an answer of 2/(2n + 1).

What am I doing wrong?

(b) For j <> i, find P(C_j | C_i)

Instead of using the P(C_j | C_i) = P(C_j C_i) / P(C_i) formula, we may cut to the chase and say that since a couple has already been selected, we can view the round table as a straight line. So the number of ways to arrange C_j are:

i) 2 choices on which of the couple to sit

ii) Pick a place among (2n - 3) seats

iii) Place the other couple on the other side (1 choice)

2 (2n - 3)

So P(C_j | C_i) = ( 2 (2n - 3) (2n - 4)! )/ (2n - 2)! = 2 / (2n - 2)

Which is in the back of the book

(c) When n is large, approximate the probability that there are no married couples who are seated next to each other.

I imagine we're supposed to use the Poisson distribution?

**Physics Forums - The Fusion of Science and Community**

# Sitting n married couples on a round table

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Sitting n married couples on a round table

Loading...

**Physics Forums - The Fusion of Science and Community**