How many Functions/ Bijections X → X do exist, if the set X has 4 elements? So I belive the amount of functinos is $4^4$. But I can't seem to find the right explaination to prove it.
My idea was, that 
$a\mapsto a,b,c,d$
$b \mapsto a,b,c,d$
...
...
are the possible ways the elements are related. 
... and if you change/exclude the diffrent elements you get to $4^4$. But I know that this is not a real prove and I can't seem to find one.
And the amount of bijections is $4!$. Because the defintions of bijection is that one element can't be related to a diffrent amount of elements, than exactly one element.
Would you have suggestions how you could do that
 A: $X=\{a,b,c,d\}$.
We search the number of functions $f: X\to X$. 
Note that every element of $X$ has exactly one value $f(x)$ under $f$.
For every $x\in X$ there are four possibilities to choose $f(x)$. Therefore there are $4\cdot 4\cdot 4\cdot 4=4^4$ different functions $f: X\to X$.
Now we want to obtain the number of bijective functions $f:X\to X$.
Note that every element in $X$ has to be mapped on exactly once.
For the first value $x_1$ we have 4 possibilities to choose the image $f(x_1)$.
For the second value $x_2$ we have 3 possibilities to choose the image $f(x_2)$, since $f$ has to bijectiv and we can not map onto the same value more then once.
We get $4!=4\cdot 3\cdot 2\cdot 1$ possible bijective functions.
A: Suppose $X = \{a,b,c,d\}$. Consider a bijection $f\colon X \rightarrow X$. You are looking for all the possible combinations of $f(a),f(b),f(c),f(d)$. Start with $f(a)$. You have $4$ possibilities for it, i.e. $a,b,c$ or $d$. Now $f(a)$ is fixed, so $f(b)$ will have to be different from $f(a)$ in order for $f$ to be injective. Then $f(b)$ can be chosen in $3$ ways. Similarly, $f(c)$ can be chosen in $2$ ways and $f(d)$ is then forced to be the only remaining element in $X$. 
To sum up, you got $4$ possibilities for $f(a)$, and for each of them $3$ possibilities for $f(b)$, then again $2$ for $f(c)$ and $1$ for $f(d)$. Hence $4 \cdot 3 \cdot 2 \cdot 1 = 4!$ possibilities.
