The task is to find how many possibilities are for sets $A_1,A_2,A_3,A_4 \subseteq [n]$ so that $A_1\bigcap A_2\bigcap A_3\bigcap A_4 = \emptyset$ and $A_1\bigcup A_2\bigcup A_3\bigcup A_4 = [n]$
My attempt:
First i go on every element of $[n]$ and decide if it's in $A_1$ or not, which accumulates to $2^n$. then I do the same for $A_2$ and $A_3$. After that, I add all the elements that aren't picked for any of the first 3 groups to $A_4$.
All together is $2^n * 2^n * 2^n = 2^{3n}$.
However, I still think I count something double times so I'm not sure about that answer.
Notice that each element of $[n]$ must satisfy the following two things:
It is in at least one of $A_1,A_2,A_3,A_4$
It is not in at least one of $A_1,A_2,A_3,A_4$
This is to guarantee that the union of $A_1,A_2,A_3,A_4$ is the entirety of $[n]$ and the intersection of $A_1,A_2,A_3,A_4$ is empty.
This implies that each element of $[n]$ can be uniquely described by one of the $2^{4}-2$ length $4$ binary strings with at least one $0$ and at least one $1$ corresponding to which of $A_1,A_2,A_3,A_4$ the element is in.
We have then, every collection of $A_1,A_2,A_3,A_4$ can be described uniquely by a sequence of such length $4$ binary strings where each has at least one $0$ and at least one $1$.
By multiplication principle then we have as a final count of
$$(2^4-2)^n=14^n$$
possible collections of $A_1,A_2,A_3,A_4$