# Thread: n > 1, prove the number of even perm. of [n] equals the number of odd perm. of [n]

1. ## n > 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].

2. Originally Posted by qtpipi
For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].
Let $\pi$ be a permutation of [n]. Define another permutation $\sigma$ by switching $\pi(1)$ and $\pi(2)$; i.e., define

$\sigma(1) = \pi(2)$
$\sigma(2) = \pi(1)$
$\sigma(j) = \pi(j)$ for $j > 2$.

Since $\pi$ and $\sigma$ 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.