Theorem: Cardinality of Set of Injections

Let $S$ and $T$ be finite sets.
The number of injections from $S$ to $T$, where $\card S = m, \card T = n$ is often denoted ${}^m P_n$, and is:
:${}^m P_n = \begin {cases} \dfrac {n!} {\paren {n - m}!} & : m \le n \\ 0 & : m > n \end {cases}$

Proof:
Let $m > n$.
By the Pigeonhole Principle, there can be no injection from $S$ to $T$ when $\card S > \card T$.
Once $\card T$ elements of $S$ have been used up, there is no element of $T$ left for the remaining elements of $S$ to be mapped to such that they all still map to different elements of $T$.
Let $m = 0$.
The only injection from $\O \to T$ is $\O \times T$ which is $\O$.
So if $m = 0$ there is $1 = n! / n!$ injection.
Let $0 < m \le n$.
As in the proof of Cardinality of Set of All Mappings, we can assume that $S = \N_m$ and $T = \N_n$.
For each $k \in \closedint 1 n$, let $\map {\mathbb H} {k, n}$ be the set of all injections from $\N_k$ to $\N_n$.
The proof now proceeds by induction.
Let:
:$\ds \mathbb S = \set {k \in \closedint 1 n: \card {\map {\mathbb H} {k, n} } = \frac {n!} {\paren {n - k}!} }$
=== Basis for the Induction ===
Let $k = 1$.
From Cardinality of Set of All Mappings, there are $n^1 = n$ different mappings from $S$ to $T$.
From Mapping from Singleton is Injection, each one of these $n$ mappings is an injection.
Thus:
:$\ds \card {\map {\mathbb H} {1, n} } = n = \frac {n!} {\paren {n - 1}!}$
and so it follows that:
:$1 \in \mathbb S$.
This is the basis for the induction.
=== Induction Hypothesis ===
We suppose that:
:$\ds \card {\map {\mathbb H} {k, n} } = \frac {n!} {\paren {n - k}!}$.
This is the induction hypothesis.
We need to show that:
:$\ds \card {\map {\mathbb H} {k + 1, n} } = \frac {n!} {\paren {n - \paren {k + 1} }!}$.
=== Induction Step ===
This is the induction step:
Let $k \in \mathbb S$ such that $k < n$.
Let $\rho: \map {\mathbb H} {k + 1, n} \to \map {\mathbb H} {k, n}$ be the mapping defined by:
:$\forall f \in \map {\mathbb H} {k + 1, n}: \map \rho f =$ the restriction of $f$ to $\N_k$
Given that $g \in \map {\mathbb H} {k, n}$ and $a \in \N_n - \map g {\N_k}$, let $g_a: \N_{k + 1} \to \N_n$ be the mapping defined as:
:$\map {g_a} x = \begin {cases} \map g x & : x \in \N_k \\ a & : x = k \end {cases}$
Now $g$ is an injection as $g \in \map {\mathbb H} {k, n}$, and as $\map {g_a} a \notin \map g {\N_k}$ it follows that $g_a$ is also an injection.
Hence:
:$g_a \in \map {\mathbb H} {k + 1, n}$
It follows from the definition of $\rho$ that:
:$\map {\rho^{-1} } {\set g} = \set {g_a: a \in \N_n - \map g {\N_k} }$
Since $g$ is an injection, $\map g {\N_k}$ has $k$ elements.
Therefore $\N_n - \map g {\N_k}$ has $n - k$ elements by Cardinality of Complement.
As $G: a \to g_a$ is clearly a bijection from $\N_n - \map g {\N_k}$ onto $\map {\rho^{-1} } {\set g}$, that set has $n - k$ elements.
Clearly:
:$\set {\map {\rho^{-1} } {\set g}: g \in \map {\mathbb H} {k, n} }$
is a partition of $\map {\mathbb H} {k + 1, n}$.
Therefore by Number of Elements in Partition:
:$\card {\map {\mathbb H} {k + 1, n} } = \paren {n - k} \dfrac {n!} {\paren {n - k}!} = \dfrac {n!} {\paren {\paren {n - k} - 1}!}$
as $k \in \mathbb S$.
But:
:$\paren {n - k} - 1 = n - \paren {k + 1}$
So:
:$k + 1 \in \mathbb S$
By induction:
:$\mathbb S = \closedint 1 n$
and in particular:
:$m \in \mathbb S$
{{qed}}