Given two sets, is there a formula to know the quantity of injective function realationships? For example. 
\begin{align*}
&A\quad =\quad \{ 1,2\} \\ 
&B\quad =\quad \{ a,b\} \\
&\texttt{Then 2 injective functions from } A\rightarrow B \\ 
&\left\{ (1,a),(2,b) \right\} \quad \{ (1,b),(2,a)\}  \\ \\
&A\quad =\quad \{ 1,2\} \\ 
&B\quad =\quad \{ a,b,c\} \\
&\texttt{Then 6 injective functions from } A\rightarrow B \\ 
&\left\{ (1,a),(2,b) \right\} \quad \{ (1,a),(2,c)\} \quad \\ 
&\{ (1,b),(2,a)\} \quad \{ (1,b),(2,c)\} \\ 
&\{ (1,c),(2,a)\} \quad \{ (1,c),(2,b)\} \\ \\
&A\quad =\quad \{ 1,2,3\} \\ 
&B\quad =\quad \{ a,b,c\} \\
&\texttt{Then 12 injective functions from } A\rightarrow B \\ 
&\left\{ (1,a) (2,b) (3,a) \right\} 
\left\{ (1,a) (2,b) (3,c) \right\}\\ 
&\left\{ (1,a) (2,c) (3,a) \right\} 
\left\{ (1,a) (2,c) (3,b) \right\}\\ 
&\left\{ (1,b) (2,a) (3,b) \right\} 
\left\{ (1,b) (2,a) (3,c) \right\}\\ 
&\left\{ (1,b) (2,c) (3,a) \right\} 
\left\{ (1,b) (2,c) (3,b) \right\}\\ 
&\left\{ (1,c) (2,a) (3,b) \right\} 
\left\{ (1,c) (2,a) (3,c) \right\}\\ 
&\left\{ (1,c) (2,b) (3,a) \right\} 
\left\{ (1,c) (2,b) (3,c) \right\}\\ 
\end{align*}

Let $A$ and $B$ be nonempty, finite sets such that $A$ has at most as many elements as $B$. Let $m$ be the number of elements in $A$ and let $n$ be the number of elements in $B$. WLOG assume that $A = \{ 1, 2, \ldots, m \}$ and $B = \{1, 2, \ldots, n \}$. Then there are precisely $\binom{n}{m} \cdot m!$ many injections $A \to B$.
To see this, let $[B]^m$ be the set of all subsets of $B$ with $m$ elements. And let $S_m$ be the permutation group on $m$ elements. Let $\mathcal F$ the the set of all injections $A \to B$. Then
$$\pi \colon \mathcal F \to [B]^m \times S_m, f \mapsto (\operatorname{im}(f), f^*)$$
is a bijection, where $f^* \colon m \to m$ is the unique permutation such that for all $x,y \in \{1, \ldots, m\}$ we have $x < y$ iff $f \circ f^*(x) < f \circ f^*(y)$.
Since $[B]^m$ has $\binom{n}{m}$ many elements and $S_m$ has $m!$ many elements, the result follows.
