The answer is 29.
How to get this answer?
There turn out to be 9 different topologies on a three point set {1,2,3} (for definiteness' sake),
and then some more topologies can be obtained from permuting the points.
1. The discrete topology; all sets are open. There is of course only 1.
2. The indiscrete topology; no sets are open except the empty set and the whole set.
In the following I will not include these sets any more...
There is of course only 1.
3. A topology with no isolated points (=open singletons) and 1 (non-trivial)
open set consisting of a doubleton. So {{0,1}} + the trivial ones, eg.
If these there are 3: "3 choose 2" ways of chosing the doubleton.
4. A topology with one isolated point, and no more open sets.
There are 3 ways of chosing the isolated point. So that makes 3 more.
5. A topology with one isolated point, and the two other points also
form an open set. Of this there are 3 as well.
6. A topology with one isolated point, and the other two
points are in its closure, but not in each other's closure, so
eg. {1}, {1,2}, {1,3} as non-trivial open sets. This topology is determined by
the choice of the isolated point, so there are 3 of them.
7. A topology with one isolated point, another point is in the closure of it,
but not in the closure of the third, while the third is in the closure of both the others.
eg.: {1}, {1,2} as non-trivial open sets. All three points have different
"roles" here, so there are 6 = 3! of these. (3 ways of picking the isolated point,
and then 2 to pick the point "inbetween".)
8. A topology with 2 isolated points as only non-trivial open sets.
So eg {1},{2}. There are 3 ways of picking the non-isolated point,
which fixes the type, so there are 3 of these.
9. A topology with 2 isolated points, and the third one is in the closure
of one of them. Eg.: {1},{2},{2,3} as non-trivial open sets.
There are 3 ways of fixing the non-isolated point, and then 2 to
fix the point in whose closure it is. So 6 of these in all.
That's all. Eg if there are 2 open sets of 2 points, their intersection
will be an isolated point, etc. A bit of thought shows that these
are all the types.
This gives: 2*1 + 5*3 + 2*6 = 12 + 15 + 2 = 29 topologies, divided
over 9 types.