Cardinality of Set of Subsets/Proof 2

Theorem
Let $S$ be a set such that $\left|{S}\right| = n$.

Let $m \le n$.

Then the number of subsets $T$ of $S$ such that $\left|{T}\right| = m$ is:
 * ${}^m C_n = \dfrac {n!} {m! \left({n - m}\right)!}$

Proof
Let ${}^m C_n$ be the number of subsets of $m$ elements of $S$.

From Number of Permutations, the number of $m$-permutations of $S$ is:
 * ${}^m P_n = \dfrac {n!} {\left({n - m}\right)!}$

Consider the way ${}^m P_n$ can be calculated.

First one makes the selection of which $m$ elements of $S$ are to be arranged.

This number is ${}^m C_n$.

Then for each selection, the number of different arrangements of these is $m!$, from Number of Permutations.

So: