Order of Symmetric Group

Theorem
Let $S$ be a finite set of cardinality $n$

Let $\struct {\Gamma \paren S, \circ}$ be the symmetric group on $S$.

Then $\struct {\Gamma \paren S, \circ}$ has $n!$ elements (see factorial).

Proof
A direct application of Cardinality of Set of Bijections.

Example
Thus, when $S = \set {1, 2, 3}$, there are $3 \times 2 \times 1 = 6$ permutations:


 * $\begin{bmatrix}

1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix}$


 * $\begin{bmatrix}

1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{bmatrix}$