permutation

permutations: How many ways are there to arrange all elements of $A$?

$\pi :A\to A$, bijective permutation function
$f:x_1,x_2,\dots x_n\mapsto (x_1,\dots ,x_n),x_i=\pi (a_i)$
$f:S_A\to \{x\in A^n:\forall i\ne j:x_{i\ne x_j}\}$
$|S_A|=\#(x_1,\dots ,x_n):\forall i\ne j:x_i\ne x_j$ (exactly what we had in (ii) above)
$=n!\left(\genfrac{}{}{0ex}{}{n}{n}\right)=n!$
→ $|A!|=n!$

Link to original

Permutation notation

$M=\{1,2,3\}$

We can denote the permutation in three ways:
(1) Two rows – the first row is the original order, the second row is the new order:
$\left(\begin{array}{ccc}1& 2& 3\\ \pi (1)& \pi (2)& \pi (3)\end{array}\right)$, e.g.: $\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)⟺1\to 2,2\to 1,3\to 3$
(2) or we just take the second row: $213$,
(3) or we take the cycle notation: $(12)(3)$, i.e. $(1\to 2\to 1)(3\to 3)$.
Each group is its own cycle – no duplicates / loops (else it wouldn’t be bijective which is a requirement for a permutation).

If you think of a permutation as a function, the group operation is function composition.

Permutation composition

This example shows the composition of two permutations in cycle notation: $(1)(3)(24)\circ (1234)$.
The first permutation $(1234)$ creates a cyclic shift (1→2→3→4→1).
When composed with $(1)(3)(24)$, which swaps positions 2 and 4 while fixing 1 and 3, the result is $(14)(23)$ – mirroring the square horizontally, swapping 1 with 4 and 2 with 3.