# Cardinality of Set of Subsets/Proof 4

## Theorem

Let $S$ be a set such that $\card S = n$.

Let $m \le n$.

Then the number of subsets $T$ of $S$ such that $\card T = m$ is:

${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$

## 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:

${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$
 $\ds {}^m C_1$ $=$ $\ds \begin {cases} 1 & : m = 0 \text { or } m = 1 \\ 0 & : \text {otherwise} \end {cases}$ $\ds$ $=$ $\ds \dfrac {1!} {0! \paren {1 - 0}!}$ for $m = 0$ $\ds$ $=$ $\ds \dfrac {1!} {1! \paren {1 - 1}!}$ for $m = 1$

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)

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)

 $\ds {}^m C_{k + 1}$ $=$ $\ds {}^m C_k + {}^{m - 1} C_k$ $\ds$ $=$ $\ds \binom k m + \binom k {m - 1}$ Induction Hypothesis $\ds$ $=$ $\ds \binom {k + 1} m$ Pascal's Rule

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

Therefore:

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