For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].

- April 14th 2009, 09:52 PMqtpipin > 1, prove the number of even perm. of [n] equals the number of odd perm. of [n]
For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].

- April 17th 2009, 02:34 PMawkward
Let be a permutation of [n]. Define another permutation by switching and ; i.e., define

for .

Since and differ only by a transposition, they have opposite parity, i.e. one is odd and one is even. So we have established a bijection between the even and odd permutations, and there are equal numbers of each.