# How many ways are there to order $n$ women and $n$ men in circle

I have the following question :

How many ways are there to order $n$ women and $n$ men in circle so there is no man next to man and no woman next to man meaning the order is man,woman,man,woman...

This is how I started :

Lets sit down all the women in circle therefore we get $(n-1)!$ now we have $n$ men we know that each woman could have $n$ different men sitting next to her on one side and $n-1$ men at the other side and we know that sitting $n$ men in circle is $(n-1)!$, now I don't know to create a "law" to make the order man,woman,man,woman...

Any ideas how to approach this?

Thanks

Suppose one of the women is named “W” and one of the men is named “M”. Now you have already explained that there are $(n-1)!$ ways to arrange the women in a circle, and similarly $(n-1)!$ for the men. Any circular arrangement of men can be “interleaved” with any circular arrangement of women in $n$ different ways (“M” can be the first, second, third, ..., or $n$th man to the right of “W”, so there are $n\cdot(n-1)!\cdot(n-1)!$ arrangements.) These are all different arrangements, because either the circle of men is different, the circle of women is different, or the number of places between “W” and “M” are different.
• Thank you for your answer, just to making sure I got it correctly you placed all the women and men to sit meaning $(n-1)!*(n-1)!$ since each women as well as $"W"$ has $n$ different men that could sit near to her (or we could say that each men has $n$ different women that could sit near to him) then we multiply by $n$? – JaVaPG May 22 '16 at 21:03
• Yes, that is basically it. Once you know the circular orders separately of the men and of the women, you only need to specify one more thing: which of the $n$ men is at the immediate right of woman “W” to fully specify the arrangement. – Steve Kass May 22 '16 at 21:09
There are two scenarios here. Either the first person is a man or the first person is a woman. In each of the scenarios you now have to order two groups of $n$ objects into $n$ places. That means there are $n!$ different combinations and $n!$ for women. So there are $2(n!)^2$ ways to order them by those rules.