Posted by **Rey** on Sunday, February 27, 2011 at 5:25am.

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

- Permutation -
**drwls**, Sunday, February 27, 2011 at 6:09am
Clearly they must alternate boy-girl-boy-girl 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.

- Permutation -
**Sniper**, Sunday, February 27, 2011 at 6:22am
Lets arrange the 5 girls first, fixing 1: (5-1)! = 4! = 24 .

Then arrange 5 boys = 5! = 120

so, total arrangements required: 120 x 24 = 2880

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