Number of arrangements of paths connecting edges of a fixed hexagon Assume that a path must simply connect one edge to another. Then there could be up to 3 separate paths through a hexagon (they do not merge but they may cross).
I have 15 single paths, 33 non-crossing double paths,  15 crossing double paths, 2 non-crossing triples, and 6+6+1 triples with 1,2,3 crossings. The total is 78, but I'm not at all sure this is a good approach or the answer is correct. Perhaps there is a much simpler way to calculate/enumerate.

If rotations and/or reflections are allowed then the number of distinct tiles is much reduced. If paths are allowed to merge when they cross the number is quite a bit higher. Perhaps the question has 3 different answers, but the answer I'm most interested in is the distinct paths on fixed tiles.
[The application is to games like Tantrix, but really a hex maze.]

Edit: there are 30 non-crossing double paths, not 33. The total is 75.
 A: Since the hexagon edges are distinct, the question becomes

How many ways are there to pair up some or all of the edges using one to three indistinguishable paths?

Your answer of 78 is wrong and the correct answer is 75; crossings are irrelevant for the answer. The derivation is as follows:


*

*One path. There are $\binom62=15$ pairs that can be connected (choose two from the six edges available).

*Two paths. As before, there are 15 ways to choose the first pair. After this we have four remaining edges, so $\binom42=6$ ways to choose the second pair. However, since the two paths are indistinguishable, we must divide by $2!$ to count each choice once. So there are $\frac{15\cdot6}2=45$ ways to make two paths in the hexagon.

*Three paths. After inserting two paths, there are only two edges left, so the third path is forced (I like to consider it as $\binom22=1$ way for completeness, though). Now that there are three paths, we must divide by $3!$ instead of $2!$, leading to $\frac{15\cdot6\cdot1}6=15$ ways of making three paths.


Summing up, we get $15+45+15=75$ arrangements of paths on the fixed hexagon. We can generalise this to other polygons; the number of arrangements $a(n)$ on the fixed $n$-gon is given by A001189, which has $a(6)=75$. Based on how I calculated this specific term above, $a(n)$ has the following formula:
$$a(n)=\sum_{k=1}^{\lfloor n/2\rfloor}\frac1{k!}\prod_{j=1}^k\binom n{2j}$$
