A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality

Let $A$ be a finite set with $n:=\mid A\mid$ and write

$\mathrm{Bij}(A):=\{f:A\to A:f\text{ is a bijection}\}\mathrm{.}$

Then $\mathrm{Bij}(A)$ is finite and $\mid \mathrm{Bij}(A)\mid =n!$ (The factorial $n!$ and the falling factorial $n^{\underset{‾}{k}}$, defined by recursion in $\mathbb{N}$).

More generally, for finite sets $X$ and $Y$ write $\mathrm{Bij}(X,Y)$ for the set of bijections $X\to Y$. If $\mid X\mid =\mid Y\mid =n$ then $\mathrm{Bij}(X,Y)$ is finite with $n!$ elements, and if $\mid X\mid \mathrm{\ne }\mid Y\mid$ then $\mathrm{Bij}(X,Y)=\mathrm{\varnothing }$.

• No group vocabulary is used or needed. $\mathrm{Bij}(A)$ is written here as a set of bijections. Composition makes it a group, and that structure, together with the name symmetric group, is introduced in The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition ↗ later in the reading order; the pointer is orientation only and nothing above rests on it. The count $n!$ proved here is what a later page needs in order to say that the symmetric group on $n$ letters has $n!$ elements.

• Why this is on the main page and not among the examples. Later pages consume this count, and an examples page is a leaf that nothing else may depend on.

• The two-set form costs one line and is used immediately. The closed formula for $\left(\genfrac{}{}{0px}{}{n}{k}\right)$ counts the bijections between an initial segment and an arbitrary $k$-element subset, which is exactly $\mathrm{Bij}(X,Y)$ with $X\mathrm{\ne }Y$.