Number of Permutations of All Elements/Proof 3
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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:
- ${}^n P_n = \dfrac {n!} {\paren {n - n}!} = n!$
$\blacksquare$