, 2017

Solution:

Firstly, there are $4^{2017}$ possible pairs of subsets, as each of the $2017$ elements can be in neither subset, in $A$ only, in $B$ only, or in both.

Now let us count the number of pairs of subsets for which $A$ is a subset of $B$. Under these conditions, each of the $2017$ elements could be in neither subset, in $B$ only, or in both $A$ and $B$. So there are $3^{2017}$ such pairs.

By symmetry, there are also $3^{2017}$ pairs of subsets where $B$ is a subset of $A$. But this overcounts the pairs in which $A$ is a subset of $B$ and $B$ is a subset of $A$, i.e. $A = B$. There are $2^{2017}$ such subsets.

Thus, in total, there are $4^{2017} - 2 \cdot 3^{2017} + 2^{2017}$ good pairs of subsets.