Cardinality of Set of Subsets/Proof 4

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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)

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)


\(\ds {}^m C_{k + 1}\) \(=\) \(\ds {}^m C_k + {}^{m - 1} C_k\)
\(\ds \) \(=\) \(\ds \binom k m + \binom k {m - 1}\) 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:

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



Sources