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

• Apr 14th 2009, 08:52 PM
qtpipi
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].
• Apr 17th 2009, 01:34 PM
awkward
Quote:

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 $\displaystyle \pi$ 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.