# Probability of specfici couples around a table

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.