I am studying for my first actuary exam so these answers are for my review only and not for hw!

To start: I am working with n married couples sitting around a round table. I want to know the probability of the ith couple (i= 1,2,...,n) sitting together, given that the ordering is random except it must alternate men and women.

I went about this by first finding the possible overall orderings of all the 2n people, given that men have to alternate with women.

I have: n!n!*2

I got this by arranging the men in their seats (n! possible ways), arranging the women in their seats (n! possible ways) multiplying by 2 to account for the ordering to be woman-man and man-woman

Now, if I chose the ith couple, I then only have to order the other (n-1) couples (by the way I know the ith couple is not something I need to "choose"- this problem is actually a repeat in the textbook, just with the arrangement of men and women as an additional step, and the ith couple wasn't chosen in the last problem)

I have: (n-1)!(n-1)!*2

Thus, P(Ith couple sitting together) = [(n-1)!(n-1)!*2]/[n!n!*2] = 1/(n^2)

But this isn't right! The answer in the back of the book is 2/n

Where or where am I going wrong?! Any help would be wonderful!


The simple way is to seat the wife somewhere. There are $n$ seats available for men, and $2$ of them are next to the wife, so the chance the couple sits together is $\frac 2n$

In your approach, you neglected that there are $n$ different seats for the first of $i^{\text{th}}$ couple that you sit and $2$ seats for the second, then $(n-1)!^2$ for the rest of the people.

  • $\begingroup$ Ok... I totally get your reasoning for the top part (super simple and I like it!) But where I went wrong... what do you mean 2 seats for the second? Wouldn't that just be like multiplying by 2 for ordering M-W or W-M? Then you don't need to multiply by 2 again because the "ordering" is set? Or am I getting confused? $\endgroup$ – user159940 Jun 26 '14 at 1:29
  • $\begingroup$ Or is it more like this: lets chose to place 1 person from the ith couple- there are n possibilities. Then, let's place the other person from the ith couple to the right or the left- that's 2 possibilities. (Thus, multiplying by 2) Now, let's "place" the other (n-1)! couples, and then let's chose possible ordering M-W or W-M by multiplying by 2 again. I think I just talked myself through it lol thank you! $\endgroup$ – user159940 Jun 26 '14 at 1:35

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