The number of combinations of n a's and n b's is clearly (2n)! / 2(n!), that is, the number of orderings of 2n objects, n alike of one kind and n of the other. But I am interested in the number of permutations of a's and b's such that, going from left to right, the number of b's never exceeds the number of a's. So with just 2 a's and 2 b's, aabb and abab are OK, but baab and abba break the rule. Empirically I am certain that the number of permutations is (2n)! / (n+1)! n!, but I have no idea how to prove that. Can anyone help?