# Cardinality of Set of Injections

## Theorem

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:

- ${}^m P_n = \begin{cases} \dfrac {n!} {\paren {n - m}!} & : m \le n \\ 0 & : m > n \end{cases}$

### Corollary

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping.

Then $f$ cannot be an injection if:

- $\card S > \card T$

where $\card S$ denotes the cardinality of $S$.

## Informal Proof

This is the same question as determining how many permutations there are of $m$ objects out of $n$.

Informally, the thinking goes like this.

We pick the elements of $S$ in any arbitrary order, and assign them in turn to an element of $T$.

The first element of $S$ can be mapped to any element of $T$, so there are $n$ options for the first element.

The second element of $S$, once we've mapped the first, can be mapped to any of the remaining $n-1$ elements of $T$, so there are $n-1$ options for that one.

And so on, to the $m$th element of $S$, which has $n - \paren {m - 1}$ possible elements in $T$ that it can be mapped to.

Each mapping is independent of the choices made for all the other mappings, so the total number of mappings from $S$ to $T$ is:

- $n \paren {n - 1} \paren {n - 2} \cdots \paren {n - m + 1} = \dfrac {n!} {\paren {n - m}!}$

$\blacksquare$

## Formal Proof

Let $m > n$.

By the Pigeonhole Principle, there can be no injection from $S$ to $T$ when $\left|{S}\right| > \left|{T}\right|$.

Once $\left|{T}\right|$ 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 $\varnothing \to T$ is $\varnothing \times T$ which is $\varnothing$.

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 $k \in \left[{1 \,.\,.\, n}\right]$, let $\mathbb H \left({k, n}\right)$ be the set of all injections from $\N_k$ to $\N_n$.

The proof now proceeds by induction.

Let:

- $\displaystyle \mathbb S = \left\{{k \in \left[{1 \,.\,.\, n}\right]: \left|{\mathbb H \left({k, n}\right)}\right| = \frac {n!} {\left({n - k}\right)!}}\right\}$

### 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:

- $\displaystyle \left|{\mathbb H \left({1, n}\right)}\right| = n = \frac {n!} {\left({n - 1}\right)!}$

and so it follows that:

- $1 \in \mathbb S$.

This is the basis for the induction.

### Induction Hypothesis

We suppose that:

- $\displaystyle \left|{\mathbb H \left({k, n}\right)}\right| = \frac {n!} {\left({n - k}\right)!}$.

This is the induction hypothesis.

We need to show that:

- $\displaystyle \left|{\mathbb H \left({k+1, n}\right)}\right| = \frac {n!} {\left({n - \left({k+1}\right)}\right)!}$.

### Induction Step

This is the induction step:

Let $k \in \mathbb S$ such that $k < n$.

Let $\rho: \mathbb H \left({k + 1, n}\right) \to \mathbb H \left({k, n}\right)$ be the mapping defined by:

- $\forall f \in \mathbb H \left({k + 1, n}\right): \rho \left({f}\right) =$ the restriction of $f$ to $\N_k$

Given that $g \in \mathbb H \left({k, n}\right)$ and $a \in \N_n - g \left({\N_k}\right)$, let $g_a: \N_{k + 1} \to \N_n$ be the mapping defined as:

- $g_a \left({x}\right) = \begin{cases} g \left({x}\right) & : x \in \N_k \\ a & : x = k \end{cases}$

Now $g$ is an injection as $g \in \mathbb H \left({k, n}\right)$, and as $g_a \left({a}\right) \notin g \left({\N_k}\right)$ it follows that $g_a$ is also an injection.

Hence:

- $g_a \in \mathbb H \left({k + 1, n}\right)$

It follows from the definition of $\rho$ that:

- $\rho^{-1} \left({\left\{{g}\right\}}\right) = \left\{{g_a: a \in \N_n - g \left({\N_k}\right)}\right\}$

Since $g$ is an injection, $g \left({\N_k}\right)$ has $k$ elements.

Therefore $\N_n - g \left({\N_k}\right)$ has $n - k$ elements by Cardinality of Complement.

As $G: a \to g_a$ is clearly a bijection from $\N_n - g \left({\N_k}\right)$ onto $\rho^{-1} \left({\left\{{g}\right\}}\right)$, that set has $n - k$ elements.

Clearly:

- $\left\{{\rho^{-1} \left({\left\{{g}\right\}}\right): g \in \mathbb H \left({k, n}\right)}\right\}$

is a partition of $\mathbb H \left({k + 1, n}\right)$.

Therefore by Number of Elements in Partition:

- $\left|{\mathbb H \left({k + 1, n}\right)}\right| = \left({n - k}\right) \dfrac {n!} {\left({n - k}\right)!} = \dfrac {n!} {\left({\left({n - k}\right) - 1}\right)!}$

as $k \in \mathbb S$.

But:

- $\left({n - k}\right) - 1 = n - \left({k + 1}\right)$

So:

- $k + 1 \in \mathbb S$

By induction:

- $\mathbb S = \left[{1 \,.\,.\, n}\right]$

and in particular:

- $m \in \mathbb S$

$\blacksquare$

## Examples

### Set of Cardinality $2$ to Set of Cardinality $3$

Let $A = \set {a, b}$.

Let $B = \set {1, 2, 3}$.

Then the set of injections from $A$ to $B$, expressed in two-row notation, is:

- $\set {\dbinom {a \ b} {1 \ 2}, \dbinom {a \ b} {1 \ 3}, \dbinom {a \ b} {2 \ 1}, \dbinom {a \ b} {2 \ 3}, \dbinom {a \ b} {3 \ 1}, \dbinom {a \ b} {3 \ 2} }$

thus demonstrating that there are $\dfrac {3!} {\paren {3 - 2}!} = \dfrac {3!} {1!} = \dfrac 6 1 = 6$ injections from $A$ to $B$.

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $3$. Mappings: Exercise $1$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $5.5$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $4$: Mappings: Exercise $4 \ \text{(ii) (a)}$