Permutation
posted by Rey .
In how many ways can 5 boys and 5 girls be seated around a table so that no 2 boys are adjacent?

Clearly they must alternate boygirlboygirl etc. So the question becomes: How many ways can that be done?
Consider chair #1. If a boy goes there, there are 5 possibilities. Then there are 5 for the next chair (any girl), then any of four boys, etc until you get a number of permutations of 5!*5! = 14,400
If chair #1 starts with a girl, then there are 14,400 more possible permutations. That is a total of 28,800
If you are only interested in the sequence of people and not who sits on chair #1, then you must divide by 10, giving 2880. 
Lets arrange the 5 girls first, fixing 1: (51)! = 4! = 24 .
Then arrange 5 boys = 5! = 120
so, total arrangements required: 120 x 24 = 2880