Number of Permutations of All Elements/Proof 3

Theorem

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

The number of permutations of $S$ is $n!$

Proof

From the definition, it can be seen that a bijection $f: S \to S$ is an $n$-permutation.

Hence, from Cardinality of Set of Bijections the number of $n$-permutations on a set of $n$ elements is:

${}^n P_n = \dfrac {n!} {\paren {n - n}!} = n!$

$\blacksquare$