n boys and n girls must be seated in 2n chairs. If every boy must only be seated next to a girl, how many possible seating arrangements are there?
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First count the ways the boys and girls can be seated if the boys are in the odd numbered chairs and the girls are in the even number chairs. There are n odd numbered chairs, and n boys can be seated in them in $$n!$$ ways. Similarly, $$n$$ girls can be seated in the $$n$$ even numbered chairs in $$n!$$ ways. They can both be seated in $$n!^2$$ ways. Then count the ways where the boys are in even chairs and girls in odd chairs; again, there are $$n!^2$$ ways. So in total, there are $$2(n!^2)$$ ways.