Finding $A,B,C,D \subseteq \{1,2,...n\}$ such that $A \cup B \cup C \cup D = \{1,2,...,n\}$ I have this combinatorial question:

Find the number of $(A,B,C,D)$ of sets $A,B,C,D \subseteq \{1,2,...,n\}$ such that $A \cup B \cup C \cup D = \{1,2,...,n\}$

I started by saying:

We choose the $k$ elements that are in $A$ which is $n \choose k$ and then we distribute all the $n-k$ elements  to the rest of the 3 sets.

Which is obviously wrong, because even if we choose $k$ to be in $A$, there still can be a set which has common elements of $A$.
I'll be happy if anyone could give me a direction.
EDIT:

Perhaps the answer is $4^n$? For every element in $N$ we choose if it is in $A, B, C$ or $D$

 A: For each element $k$ we can choose if it is in $A, B, C, D$. This amounts to $16$ possible choices for $k$.
For only one of these choices, we have that $k \notin A \cup B \cup C \cup D$.
Since we have such a consideration for each $1 \le k \le n$, the solution is $15^n$.

In general, when distributing elements over sets, if counting from the point of view of sets is hard, a good first reflex is to try and approach the problem from the point of view of the elements. Often this makes for an easy solution.
A: A simple way to verify your answer in mathematics is to substitute many small numbers in the formula. You can then verify your answer by hand calculation. 
In your case, what is the answer for $n=1$ if you list all the collection of sets? Does it match your 'guessed' formula for $n=1$? What about $n=2$? and so on.
Now coming back to your problem.
In Polya's words:"If you can't solve a problem, then there is an easier problem you can solve: find it."
Let's try a simpler problem: How many pairs $A,B$ can you find such that $ A \cup B=\{1,2,3 \cdots n\}$? 
In this case, clearly each of the $n$ elements have a choice of being in $A$, $B$ or both. In other words, each of the $n$ elements have $3$ choices independently. So the answer would be $3^n$.
Now can you solve your problem?
