Number of Permutations/Proof 2

Theorem
Let $S$ be a set of $n$ elements.

Let $r \in \N: r \le n$.

Then the number of $r$-permutations of $S$ is:
 * ${}^r P_n = \dfrac {n!} {\left({n-r}\right)!}$

When $r = n$, this becomes:
 * ${}^n P_n = \dfrac {n!} {\left({n-n}\right)!} = n!$

Using the falling factorial symbol, this can also be expressed:
 * ${}^r P_n = n^{\underline r}$

Proof
From the definition, an $r$-permutations of $S$ is an ordered selection of $r$ elements of $S$.

It can be seen that an $r$-permutation is an injection from a subset of $S$ into $S$.

From Cardinality of Set of Injections‎, we see that the number of $r$-permutations ${}^r P_n$ on a set of $n$ elements is given by:
 * ${}^r P_n = \dfrac {n!} {\left({n-r}\right)!}$

From this definition, it can be seen that a bijection $f: S \to S$ is an $n$-permutation.

Hence the number of $n$-permutations on a set of $n$ elements is ${}^n P_n = \dfrac {n!} {\left({n-n}\right)!} = n!$.