Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$N$ 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?

share|improve this question

2 Answers 2

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 $p_2=\frac2{n(2n-1)}$.

share|improve this answer

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 $$\begin{align*}\underbrace{(2n-3)}_{k=1}+\underbrace{\Big((2n-4)+\ldots+(2n-4)\Big)}_{2\le k\le 2n-2}+\underbrace{(2n-3)}_{k=2n-1}&=2(2n-3)+(2n-3)(2n-4)\\&=(2n-3)(2n-2)\;.\end{align*}$$

share|improve this answer
    
I got 1/n and 2/ 2n-1. is that correct? –  paul marcus Feb 10 '13 at 8:36
    
@paul: $1/n$ is right for the first question, but I get a rather different answer for the second. What were your intermediate calculations? –  Brian M. Scott Feb 10 '13 at 8:40
    
sorry, is it 2/ (2n-1)*n ? I used conditional probability –  paul marcus Feb 10 '13 at 8:45
    
If A and B already sit next to each other, then treat A and B as an entity. Insert D into 2n -1 places, the probability that D is next to C is 2 / 2n-1... What did you get? –  paul marcus Feb 10 '13 at 8:46
    
@paul: You can’t do it that way. If A and B are sitting in the first two seats, the probability that C and D sit together is $1/(n-1)$, as in the first problem. If A and B are sitting in 2nd and 3rd seats, the probability that C and D sit together is only $1/(n-3)$. Thus, you can’t treat all cases the same. –  Brian M. Scott Feb 10 '13 at 8:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.