Cardinality of Set of Subsets/Proof 1

Proof
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.

By Cardinality of Set of Bijections:
 * $\card {\map B {\N_m, Y} } = m!$

and:
 * $\card {\map B {\N_n - \N_m, S - Y} } = \paren {n - m}!$

So by Cardinality of Cartesian Product:
 * $\card {\map {\beta^{-1} } Y} = m! \paren {n - m}!$

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.