What I've been thinking is that there are $2^n$ ways of choosing $A$, then from the remaining elements of $\{1,…,n\}$ I could choose any subset (there would be $2^{n-|A|}$ ways to do that) and add that to A and form B, using this same procedure we can then construct C. The total ways to choose the three subsets would be:
$$ \sum_{k=0}^{n}\sum_{j=k}^{n} 2^{n}2^{n-k}2^{n-j}$$
I've tried this formula for the case n=1 and it does not work but I don't know where I went wrong.
Best Answer
We need to divide our set into $4$ pairwise disjoint parts, $A_1,A_2,A_3,A_4$ and set $A=A_1$, $B=A_1\cup A_2$, $C=A_1\cup A_2\cup A_3$, and $\{1,2,\dots,n\}=A_1\cup A_2\cup A_3\cup A_4$.
For every element $j$ of $\{1,2,\dots, n\}$, there are $4$ possible decisions: $j$ lands in $A_1$ or $A_2$ or $A_3$ or $A_4$. Thus the number of choices is $4^n$.
Remark: The number of choices for $B$ depends on the size of the set $A$. So an approach like yours should involve binomial coefficients. It can be pushed through. Ultimately you will end up calculating the sum of all the multinomial coefficients $\binom{n}{k_1,k_2,k_3,k_4}$. By the Multinomial Theorem this sum is $4^n$.