Cardinality of Even and Odd Permutations on Finite Set

Theorem
Let $n \in \N_{> 0}$ be a natural number greater than $0$.

Let $S$ be a set of cardinality $n$.

Let $S_n$ denote the symmetric group on $S$ of order $n$.

Let $R_e$ and $R_o$ denote the subsets of $S_n$ consisting of even permutations and odd permutations respectively.

Then the cardinality of both $R_e$ and $R_o$ is $\dfrac {n!} 2$.

Proof
From Order of Symmetric Group:
 * $\order {S_n} = n!$

where:
 * $\order {S_n}$ denotes the order of $S_n$
 * $n!$ denotes the factorial of $n$.

By definition:
 * $\card {R_e} + \card {R_o} = order {S_n}$

From Odd and Even Permutations of Set are Equivalent:
 * $\card {R_e} = \card {R_o}$

The result follows.