Cardinality of Set of Subsets
Contents
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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