# Counting - seat men and women alternating at a table.

The problem:

Six men and six women are to be seated around a table, with men and women alternating. The chairs don't matter only who is next to whom, but right and left are different. How many seating arrangements are possible?

My initial approach is to think of it as pairs of boxes and find the way in which you can split the pairs and then multiply by 2 since left and right are not the same.

visually if $x$ is an empty spot to be taken with | denoting a boundary:

x x | x x | x x | x x | x x | x x

there are 11 gaps, if each boundary is put in between each x and you want to pick 2. Because chairs don't matter, I thought it might be:

$${11! \over {2!(11-2)!}} * 2$$

• Often I will point out what the mistaken number actually counts, but I'm at a loss for what your number describes in this scenario... my best would be that it counts the number of ways that you can line up twelve red marbles, a black marble, and a blue marble, such that the black marble and blue marble aren't adjacent and the furthest left and furthest right marbles are both red. (red marbles being your x's, the black and blue marble being your two gaps chosen between the x's, and the fact that black and blue marbles different colors accounts for your extra factor of 2). Seems far removed – JMoravitz Aug 5 '17 at 4:01

There are $5!$ ways to place the remaining men in the "man" seats.
Then there are $6!$ ways to seat the women in the "woman" seats.
So there are $6!5!$ total arrangements.