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

- Apr 14th 2009, 08:52 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].

- Apr 17th 2009, 01:34 PMawkward
Let $\displaystyle \pi$ be a permutation of [n]. Define another permutation $\displaystyle \sigma$ by switching $\displaystyle \pi(1)$ and $\displaystyle \pi(2)$; i.e., define

$\displaystyle \sigma(1) = \pi(2)$

$\displaystyle \sigma(2) = \pi(1)$

$\displaystyle \sigma(j) = \pi(j)$ for $\displaystyle j > 2$.

Since $\displaystyle \pi$ and $\displaystyle \sigma$ 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.