Number of Permutations/Proof 2

Theorem

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

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

The number of $r$-permutations of $S$ is:

${}^n P_r = \dfrac {n!} {\paren {n - r}!}$

Using the falling factorial symbol, this can also be expressed:

${}^n P_r = n^{\underline r}$

Proof

From the definition, an $r$-permutation 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 ${}^n P_r$ on a set of $n$ elements is given by:

${}^n P_r = \dfrac {n!} {\paren {n - r}!}$

$\blacksquare$