marriage probability problem

Question

$N$ marriged couples are lined up randomly.( so there'are 2n people).

What's the probability that A and B (who are a couple) sit next to each other?

What's the probability that A and B (who are a couple) sit next to each other, and C and D(who are a couple as well) sit next to each other?

Answer 1

You may note that this generalizes to $k$ couples as follows (the problem is only about $k=1$ and $k=2$): There are $(2n)!$ possible arrangements in total. To count the "successful" arrangements, send one member of each of the $k$ special couples outside and remove $k$ chairs. There are $(2n-k)!$ possible arrangements now. Ask the other $k$ people back in and let each select whether they want to sit to the left or to the right of their partner. Thus we arrive at $$p_k =\frac{2^k(2n-k)!}{(2n)!},$$ especially $p_1=\frac1n$ and $p_2=\frac2{n(2n-1)}$.

Answer 2

HINTS:

For the first question:

1. How many ways are there to pick $2$ of the $2n$ seats to be occupied by A and B?
2. How many ways are there to pick $2$ adjacent seats from the $2n$?

The possible pairs in (1) are the equally likely outcomes, and the ones in (2) are the successful outcomes; given those numbers, what’s the probability of success?

The second question is a bit harder, but the basic ideas are the same.

1. First you want to find the number of ways of choosing a pair of seats for each of the couples. There are $\binom{2n}2$ ways to choose a pair for A and B. Once that’s done, $2n-2$ seats remain free, so there are $\binom{2n-2}2$ ways to choose a pair for C and D. How do you combine those numbers to find the total number of ways to choose pairs of seats for both couples?

2. Now you want to find the number of ways to do it that seat each couple together. Suppose that A and B are sitting together in seats $k$ and $k+1$. There are $k-1$ seats to the left of this pair and $2n-k-1$ to the right of it. If $2\le k\le 2n-2$, each of these numbers is at least $1$, and there are $$\big((k-1)-1\big)+\big((2n-k-1)-1\big)=2n-4$$ pairs of adjacent seats open for C and D. If $k=1$ or $k=2n-1$, there are $2n-2$ seats open on one side of the pair, so there are $2n-3$ pairs of adjacent seats available for C and D. The total number of possibilities is therefore $$\begin{align*}\underbrace{(2n-3)}_{k=1}+\underbrace{\Big((2n-4)+\ldots+(2n-4)\Big)}_{2\le k\le 2n-2}+\underbrace{(2n-3)}_{k=2n-1}&=2(2n-3)+(2n-3)(2n-4)\\&=(2n-3)(2n-2)\;.\end{align*}$$