Problem: Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).

Solution: We claim the following conditions:
Call the vertices $\omega_1, \omega_2, \omega_3, \omega_4, \omega_5$ . WLOG, let $\omega_1 \to \omega_2$ . Then, WLOG let $\omega_2 \to \omega_3$ . We have two ways to go from here:
If $\omega_3 \to \omega_1$ , we call it a "cycle".
We will prove (3):
Notice that once $\omega_4, \omega_5$ can reach one of the vertices on the "cycle", it can reach any of the other vertices on the same cycle simply by travelling along the arcs of the cycle. Additionally, one of $\omega_4, \omega_5$ must be able to travel to the cycle. Consider that it cannot, then $\omega_3 \to \omega_4$ and $\omega_5 \to \omega_4$ are forced. Now, $\omega_4 \to \omega_1, \omega_2, \omega_3$ which is on the cycle. Now, if $\omega_5 \to \omega_4$ , it will also be able to travel to the cycle.  Otherwise, the directed arc will directly travel to the cycle. Also, each of the vertices $\omega_1, \omega_2, \omega_3$ will travel to $\omega_4, \omega_5$
If our graph does not satisfy (1) and (2), $\omega_4, \omega_5$ will connect to the cycle on both directions. This can be shown below:
Clearly $\omega_4, \omega_5$ are connected. Assuming the graph does not satisfy (1) and (2), $\omega_4, \omega_5$ are connected to the cycle consisting of $\omega_1, \omega_2, \omega_3$ . WLOG, let $\omega_4 \to \omega_5$
Again, to avoid (2), $\omega_5 \to \omega_a$ where $a\in{1,2,3}$ . Now to avoid (1), one of $\omega_a \to \omega_4, a\in{1,2,3}$
Now it is clear that each vertex can go to any other vertex. This concludes our proof.
Say we try to avoid (3) so that WLOG $\omega_5 \to \omega_1$ . If we keep trying to avoid a "cycle", $\omega_3 \to \omega_4$ and $\omega_4 \to \omega_5$ which is a cycle.
Therefore, if there is any path $\omega_a \to \omega_b \to \omega_c$ , it will lead to a cycle and ultimately a valid path. The only way to avoid this is to have (1) or (2).
Consider (1):
WLOG, let $\omega_1$ have no arcs leading to it. Then all other vertices will have an arc directed toward them away from $\omega_1$ . Therefore, one path can have only 1 unique vertex that has no arcs leading to it. There are 5 ways to choose the vertex and $2^{\binom{5}{2}-4}=2^6$ ways for the remaining 6 arcs since they can be directed in any way. There are $2^6\cdot5=320$ paths for this condition.
Consider (2):
WLOG, let $\omega_1$ have all arcs leading to it. Similarly, all other vertices will have an arc that is directed away from them and points to $\omega_1$ . Therefore, only one unique vertex can have all arcs leading to it. There are 5 ways to choose the vertex and $2^{\binom{5}{2}-4}=2^6$ ways for the remaining arcs. There are also $2^6\cdot5=320$ paths for this condition.
However, we have overcounted paths that satisfy both (1) and (2). These paths will have one vertex that has all arcs leading to it and another that has all arcs directed away from it. There are $5\cdot4=20$ ways to choose these 2 vertices, and the remaining $\binom{3}{2}=3$ arcs each have 2 ways to be directed toward each other, so there are a total of $2^3\cdot 20=160$ overlapping paths.
By PIE, the number of invalid cases is $320+320-160=480$
The total number of cases is $2^{\binom{5}{2}}=2^{10}=1024$ since there are 10 edges and each can go 2 ways. By complementary counting, the number of valid paths is just $1024-480=\boxed{544}$

Answer:  544