Cardinality of Set of Subsets

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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 1

For each $X \subseteq \N_n$ and $Y \subseteq S$, let $B \left({X, Y}\right)$ be the set of all bijections from $X$ onto $Y$.

Let $\Bbb S$ be the set of all subsets of $S$ with $m$ elements.

By Cardinality of Power Set of Finite Set and Cardinality of Subset of Finite Set, $\Bbb S$ is finite, so let $s = \left|{\Bbb S}\right|$.

Let $\beta: B \left({\N_n, S}\right) \to \Bbb S$ be the mapping defined as $\forall f \in B \left({\N_n, S}\right): \beta \left({f}\right) = f \left({\N_m}\right)$.


For each $Y \in \Bbb S$, the mapping:

$\Phi_Y: \beta^{-1} \left({Y}\right) \to B \left({\N_m, Y}\right) \times B \left({\N_n - \N_m, S - Y}\right)$

defined as:

$\Phi_Y \left({f}\right) = \left({f_{\N_m}, f_{\N_n - \N_m}}\right)$

is also (clearly) a bijection.


By Cardinality of Set of Bijections:

$\left|{B \left({\N_m, Y}\right)}\right| = m!$

and:

$\left|{B \left({\N_n - \N_m, S - Y}\right)}\right| = \left({n - m}\right)!$

So by Cardinality of Cartesian Product:

$\left|{\beta^{-1} \left({Y}\right)}\right| = m! \left({n - m}\right)!$

It is clear that $\left\{{\beta^{-1} \left({Y}\right): Y \in \Bbb S}\right\}$ is a partition of $B \left({\N_n, S}\right)$.

Therefore by Number of Elements in Partition:

$\left|{B \left({\N_n, S}\right)}\right| = m! \left({n - m}\right)! s$

Consequently, as $\left|{B \left({\N_n, S}\right)}\right| = n!$ by Cardinality of Set of Bijections, it follows that:

$m! \left({n - m}\right)! s = n!$

and the result follows.

$\blacksquare$


Proof 2

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:

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

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:

\(\displaystyle m! \cdot {}^m C_n\) \(=\) \(\displaystyle {}^m P_n\) Product Rule for Counting
\(\displaystyle \) \(=\) \(\displaystyle \frac {n!} {\paren {n - m}!}\) Number of Permutations
\(\displaystyle \leadsto \ \ \) \(\displaystyle {}^m C_n\) \(=\) \(\displaystyle \frac {n!} {m! \paren {n - m}!}\) Product Rule for Counting

$\blacksquare$


Proof 3

Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.

Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.

Let $r \in \N: 0 < r \le n$.

Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:

$B_r := \set {S \subseteq \N_n: \card S = r}$


Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:

$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$

where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.

From Group Action of Symmetric Group on Subset it is established that $*$ is a group action.


The stabilizer of any $U \in B_r$ is the set of permutations on $\N_n$ that fix $U$.


Let $U = \set {1, 2, \ldots, r}$.

So:

$\tuple {a_1, a_2, \ldots, a_r}$ can be any one of the $r!$ permutations of $1, 2, \ldots, r$
$\tuple {a_{r + 1}, a_{r + 2}, \ldots, _n}$ can be any one of the $\paren {n - r}!$ permutations of $r + 1, r + 2, \ldots, n$.

Thus:

$\order {\Stab U} = r! \paren {n - r}!$

From Group Action of Symmetric Group on Subset is Transitive:

$B_r = \Orb U$

and so:

$\card {B_r} = \card {\Orb U}$

From the Orbit-Stabilizer Theorem:

$\card {\Orb U} = \dfrac {\order {S_n} } {\order {\Stab U} } = \dfrac {n!} {r! \paren {n - r}!}$

But $\card {B_r}$ is the number of subsets of $\N_n$ of cardinality $r$.

Hence the result.

$\blacksquare$


Proof 4

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:

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


\(\displaystyle {}^m C_{k + 1}\) \(=\) \(\displaystyle {}^m C_k + {}^{m - 1} C_k\)
\(\displaystyle \) \(=\) \(\displaystyle \binom k m + \binom k {m - 1}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \binom {k + 1} m\) Pascal's Rule

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


Therefore:

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


Also see


Sources

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