Let X be a finite set and Y be a countable set. Prove that the set of functions from X to Y is countable. Enumerate X = {$x_1,x_2,...,x_n$} with ${x_j}$ distinct elements since x is finite.
Define $f:X\rightarrow Y$ with $f(x_j) = y_k$ 
where $y_k \in $ Y = {$y_0,y_1,...$}.
So besides the definitions I do not know how to proceed
 A: You can identify each function with its image vector, that is, the function $f\colon X \to Y$ is identified with $(f(x_1), \dots, f(x_n))$. This maps the set of functions from $X$ to $Y$ to a subset of the cartesian product of $Y$, $|X|$ times, namely $Y\times \stackrel{|X|}\dots\times Y$.
Since the cartesian product of countable sets is countable, then so is the finite cartesian product of countable sets.
A: This set of functions is the set $Y^{X}$, which is a Cartesian product of finitely many (to be specific, of $|X|$) copies of the countable set $Y$, hence is countable.
A: The cardinality of $Y^X$ is $\mid Y^X\mid=\mid Y\mid^{\mid X\mid}=\aleph_0^n=\aleph_0 $.
For the proof of the last equality (the fact that a countable union of countable sets is countable) see Cantor's pairing function . 
A: Let's consider $A_1$, the set of functions $f:X\to Y$ such that one element of $X$ maps to $y_{1}$. Clearly $|A_1| = \binom{n}{1}$ where $n$ is the number of elements in $X$. 
Similarly let's consider $A_2$, the set of functions $f:X\to Y$ such that two elements of $X$ maps to two elements of $Y$, namely $y_{1}$ and $y_{2}$. Here we can also trivially conclude that $|A_2| = \binom{n}{2}$.
For $0\leq k\leq n$, we can follow a similar procedure to construct sets $A_1, \dots, A_{k}, \dots A_{n}$ that are clearly finite and so countable. 
For $k \geq n+1$, we have to do something slightly different. In particular, let's consider the set $A_{n+1}$, the set of functions $f:X\to Y$ such that all ($n$) elements of $X$ map to $n$ elements of $Y$ with indices less than or equal to $n+1$. For clarity, the range of the functions in $A_{n+1}$ can be described by the set $\{y\in Y:y_i=f_{{n+1}_{k}}(x_j) \text{ for some } x_j\in X \text{ and } 0\leq i\leq n+1\}$. 
The cardinality $|A_{n+1}| = \binom{n+1}{n}\cdot n!$ as there are $n+1$ spots for $n$ elements of $X$ and also $n!$ ways to choose those $n$ elements of $X$. With that in mind, we can also produce a formula for the cardinality of sets $A_k$ of functions $f:X\to Y$ where $k\geq n+1$. In particular, $|A_{k}| = \binom{k}{n}\cdot n!$.
Clearly, for all $k\in \mathbb{N}$, the sets $A_k$ are all finite and therefore countable. Because the countable union of countable sets is countable, the set of functions from $X \to Y$, which is the union of all $A_k$, is also countable.
A: Fix $p_1,\dots, p_n$ distinct prime numbers.
For a function $f:X\to Y$, say $f(x_j)=y_{k_j}$ for $j=1,\dots,n$.   Define $\varphi(f)=p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}$, this defines an injection $\varphi: Y^X\to \mathbb{N}$.
