Cardinality of Set of Subsets/Proof 2

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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:

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


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.


\(\ds m! \cdot \dbinom n m\) \(=\) \(\ds {}^m P_n\) 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