# Cardinality of Set of Subsets

## 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:

$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$

## Proof 1

For each $X \subseteq \N_n$ and $Y \subseteq S$, let $\map B {X, Y}$ 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 $s = \card {\Bbb S}$.

Let $\beta: \map B {\N_n, S} \to \Bbb S$ be the mapping defined as:

$\forall f \in \map B {\N_n, S}: \map \beta f = \map f {\N_m}$

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

$\Phi_Y: \map {\beta^{-1} } Y \to \map B {\N_m, Y} \times \map B {\N_n - \N_m, S - Y}$

defined as:

$\map {\Phi_Y} f = \tuple {f_{\N_m}, f_{\N_n - \N_m} }$

is also (clearly) a bijection.

$\card {\map B {\N_m, Y} } = m!$

and:

$\card {\map B {\N_n - \N_m, S - Y} } = \paren {n - m}!$
$\card {\map {\beta^{-1} } Y} = m! \paren {n - m}!$

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

Therefore by Number of Elements in Partition:

$\card {\map B {\N_n, S} } = m! \paren {n - m}! s$

Consequently, as $\card {\map B {\N_n, S} } = n!$ by Cardinality of Set of Bijections, it follows that:

$m! \paren {n - m}! s = n!$

and the result follows.

$\blacksquare$

## Proof 2

Let $\dbinom n m$ denote 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 $\dbinom n m$.

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

So:

 $\ds m! \cdot \dbinom n m$ $=$ $\ds {}^m P_n$ Product Rule for Counting $\ds$ $=$ $\ds \frac {n!} {\paren {n - m}!}$ Number of Permutations $\ds \leadsto \ \$ $\ds \dbinom n m$ $=$ $\ds \frac {n!} {m! \paren {n - m}!}$ 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:

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

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:

$\dbinom n m = \dfrac {n!} {m! \paren {n - m}!}$

### Basis for the Induction

$\map P 1$ is the case:

 $\ds \dbinom 1 m$ $=$ $\ds \dfrac {1!} {1! \paren {m - 1}!}$ $\ds$ $=$ $\ds \begin {cases} 1 & : m = 0 \text { or } m = 1 \\ 0 & : \text {otherwise} \end {cases}$ $\ds$ $=$ $\ds \dfrac {1!} {0! \paren {1 - 0}!}$ 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:

$\dbinom k m = \dfrac {k!} {m! \paren {k - m}!}$

from which it is to be shown that:

$\dbinom {k + 1} m = \dfrac {\paren {k + 1}!} {m! \paren {k + 1 - m}!}$

### 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)

 $\ds \dbinom {k + 1} m$ $=$ $\ds \binom k m + \binom k {m - 1}$ $\ds$ $=$ $\ds \dfrac {k!} {m! \paren {k - m}!} + \dfrac {k!} {\paren {m - 1}! \paren {k - m + 1}!}$ Induction Hypothesis $\ds$ $=$ $\ds \dfrac {\paren {k + 1}!} {m! \paren {k - m + 1}!}$ after algebra $\ds$ $=$ $\ds \binom {k + 1} m$

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

Therefore:

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

## Also denoted as

The number of subsets of cardinality $m$ that can be taken from a set of cardinality $n$ can also be seen as:

${}_n C_m$
${}^n C_m$
${C_n}^m$

In order to avoid any ambiguity, $\mathsf{Pr} \infty \mathsf{fWiki}$ will tend to use $\dbinom n m$.

## Also presented as

This result is also presented in the form:

The number of combinations of $n$ objects taken $m$ at a time

or:

The number of selections of $n$ objects taken $r$ at a time

or:

The number of ways of selecting $m$ objects out of $n$

or:

The number of ways of choosing $m$ objects from $n$ where order does not matter

or:

The number of ways of choosing a set of $m$ elements from the $n$ elements of the set $S$

Such wording is insufficiently precise for $\mathsf{Pr} \infty \mathsf{fWiki}$.