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

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

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

Then $\operatorname{Bij}(A)$ is finite and $\lvert\operatorname{Bij}(A)\rvert = n!$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$, defined by recursion in $\mathbb{N}$).

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

• No group vocabulary is used or needed. $\operatorname{Bij}(A)$ 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 $\operatorname{Sym}(X)$: 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 $\binom{n}{k}$ counts the bijections between an initial segment and an arbitrary $k$-element subset, which is exactly $\operatorname{Bij}(X,Y)$ with $X \ne Y$.