Women and men around the table probability question

[twoflower]

Homework Statement

Let's have 2n persons, n men and n women. Suppose they sit randomly around a table with 2n chairs. What is the probability that no two persons of the same sex will sit next to each other?

The Attempt at a Solution

Here's my idea:

I will model this situation with classic probability. The set of all possible events (ie. the ways the people will sit around the table) is set of all strings consisting of 'W' and 'M' of length 2n. The only strings which satisfy our condition is

WMWMWM...WM
and
MWMWMW...MW

Number of all possibilities is $2^{2n}$, ie. the probability will be

$$\frac{1}{2^{2n-1}}$$

Is this solution correct?

 

[StatusX]

First, there are several different possible arrangements corresponding to a given string of M's and W's, so you should also show that every string corresponds to the same number of arrangements.

Assuming you've done this, you have still overcounted these strings, since you've included any string consisting of M's and W's, eg, MM...MM, WW...WW, etc. You need to limit to those with the same number of M's and W's.

 

[twoflower]

First, there are several different possible arrangements corresponding to a given string of M's and W's, so you should also show that every string corresponds to the same number of arrangements.

Yes, I understand. I don't know how to prove it rigorously, but I guess every string really correspond to the same number of arrangements..

Assuming you've done this, you have still overcounted these strings, since you've included any string consisting of M's and W's, eg, MM...MM, WW...WW, etc. You need to limit to those with the same number of M's and W's.

You're right, I didn't notice that. But what will be the total amount of strings then?

As I think about it, could it be

$$\left( \begin{array}{cc} 2n \\ n\end{array} \right)$$

? I mean, I have 2n positions and I draw n positions out of them for let's say men and these positions will uniquely determine the rest of the stiring.

Is it now correct?

 

[D H]

Let's have 2n persons, n men and n women. Suppose they sit randomly around a table with 2n chairs. What is the probability that no two persons of the same sex will sit next to each other?$$\frac{1}{2^{2n-1}}$$

Is this solution correct?

Designate the women $\lbrace W_1, W_2, \cdots, W_n \}$, the men $\{ M_1, M_2, \cdots, M_n\}$, and the chairs $\{ C_1, C_2, \cdots, C_{2n} \}$. A seating arrangement is a mapping of the union of the two gender sets to the set of chairs.

Consider the simplest possible case, one man and one woman. There are only two possible seating arrangements: $\{ M_1,W_1 \} \to \{ C_1,C_2 \}$ and $\{ W_1,M_1 \} \to \{ C_1,C_2 \}$. So in the case $n=1$, the probability that no two persons of the same sex will sit next to each other is zero.

 

[D H]

Another way to look at this problem is to forget half of the problem (i.e., forget the men or forget the women). If two men are sitting next to one another then so are two women. The problem is equivalent to asking, "what is the probability that n integers drawn without replacement from the set of integers from 1 to 2n are all odd or all even?"

 

[twoflower]

Another way to look at this problem is to forget half of the problem (i.e., forget the men or forget the women). If two men are sitting next to one another then so are two women. The problem is equivalent to asking, "what is the probability that n integers drawn without replacement from the set of integers from 1 to 2n are all odd or all even?"

Is the probability

$$\frac{2}{\left( \begin{array}{cc} 2n \\n\end{array} \right)}$$

?

 

[Dick]

Think of it this way. Pick a starting chair and seat people around the table. Compute two numbers i) the number of ways to pick people without restriction from a pool of 2*n people and ii) the number of ways to pick people alternately from two groups of n. It's actually pretty easy. Now it's your turn. Tell me what to do with those two numbers?

 

[twoflower]

Think of it this way. Pick a starting chair and seat people around the table. Compute two numbers i) the number of ways to pick people without restriction from a pool of 2*n people and ii) the number of ways to pick people alternately from two groups of n. It's actually pretty easy. Now it's your turn. Tell me what to do with those two numbers?

Ok Dick, for (i) I think it's $(2n)!$ and for (ii) I'd say it will be $n!.n!$. Is it ok?

But I don't know how to use or relate these two numbers, could you tell me a hint?

 

[Dick]

Almost! Just remember in the case of ii) you also have the choice of whether to start with men or women. Now think. i) is all possible seatings. ii) is all alternate seatings. How do I turn this is into a probability?

 

[twoflower]

Almost! Just remember in the case of ii) you also have the choice of whether to start with men or women. Now think. i) is all possible seatings. ii) is all alternate seatings. How do I turn this is into a probability?

So the final probability will be

$$\frac{2n!n!}{(2n)!} = \frac{2}{\left( \begin{array}{cc} 2n \\n\end{array} \right)}$$

?

 

[Dick]

That's my guess. I see you posted that before. I didn't recognize the correct answer in that form. Sorry!

 

[twoflower]

That's my guess. I see you posted that before. I didn't recognize the correct answer in that form. Sorry!

No problem, it's always better for me to see there are more ways how to look at the problem (in case it's the correct answer, of course )

Thank you very much for your help!