How many of these are topologies? 
Let $X$ be a set with $3$ elements. The set of subsets of the power
  set of $X$ is $2^{2^3}$ elements. How many of these are topologies?

Is there a trick to this problem, or is it just a "plug and chug" casework thing?
 A: Hint: a topology on a finite set is determined by its minimal basis (which amounts to choosing a minimal open set containing each point).
A: I think it's really just systematic listing, and trial and error.
http://en.wikipedia.org/wiki/Finite_topological_space#3_points
A: Let the elements be $a$, $b$, and $c$. We find the number of topologies.
Case 1.
$\mathcal{T}$ has no singleton sets.


*

*It could have no doubleton sets. Thus it is the trivial topology, so there is $1$ way to pick this.

*It could have $1$ doubleton. There are $3$ ways to pick this.

*It could have more than $1$. But then their intersection would be a singleton, which is impossible.


Case 2.
$\mathcal{T}$ has $1$ singleton set.


*

*It could have no doubleton. There are $3$ ways to pick the singleton.

*It could have $1$ doubleton. All of these work, so there are $3 \times 3 = 9$ ways $(3$ ways to pick single, $3$ double$)$.

*It could have $2$ doubletons. Then their intersection must be the singleton $($since $X$ has $3$ elements, the intersection of any $2$ element set is a $1$-element one$)$. Thus once we fix the singleton, we have to select the doubletons containing it. So there are $3$ ways to select.

*It could have all $3$ doubletons. But then it would have more than $1$ singleton, impossible.


Case 3.
$\mathcal{T}$ has $2$ singleton sets. Thus it must contain their union, a doubleton.


*

*It could have $1$ doubleton. This works, so there are $3$ ways.

*It could have $2$ doubletons. This also works. There are $3$ ways to pick singles, $2$ to pick extra double. So there are $3 \times 2 = 6$ ways.

*It could have $3$ doubletons. But this is impossible, since it would have all $3$ singletons.


Case 4.
$\mathcal{T}$ has $3$ singletons. This is the discrete topology, so there is $1$ way.

So the total is $$1 + 3 + 3 + 9 + 3 + 3 + 6 + 1 = 29 \text{ topologies}.$$
