Cardinality of Set of Subsets/Proof 4

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * ${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$

Basis for the Induction
$\map P 1$ is the case:
 * ${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * ${}^m C_k = \dfrac {k!} {m! \paren {k - m}!}$

from which it is to be shown that:
 * ${}^m C_{k + 1} = \dfrac {\paren {k + 1}!} {m! \paren {k + 1 - m}!}$

Induction Step
This is the induction step:

The number of ways to choose $m$ elements from $k + 1$ elements is:
 * the number of ways to choose $m$ elements elements from $k$ elements (deciding not to select the $k + 1$th element)

added to:
 * the number of ways to choose $m - 1$ elements elements from $k$ elements (after having selected the $k + 1$th element for the $n$th selection)

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{>0}: {}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$