The country of HMMTLand has 8 cities. Its government decides to construct several two-way roads between pairs of distinct cities. After they finish construction, it turns out that each city can reach exactly 3 other cities via a single road, and from any pair of distinct cities, either exactly 0 or 2 other cities can be reached from both cities by a single road. Compute the number of ways HMMTLand could have constructed the roads.
Solution: Let the cities be numbered $1,2,3,4,5,6,7,8$. WLOG, 1 is connected to 2,3 , and 4 . First suppose 2 and 3 are connected; then 3 and 1 share a second common neighbor, which must be 4 (as 1 is not connected to anything else). Likewise 2 and 4 are connected, and so 5, 6, 7, 8 are pairwise connected as well, so the graph consists of two disjoint copies of $K_{4}$ :\\ There are $\frac{1}{2}\left(\begin{array}{l}8 \\ 4\end{array}\right)=35$ ways to partition the 8 vertices into two groups of 4 , so there are 35 such graphs. Otherwise, none of 2, 3, 4 are connected to each other. Then 2 and 3 must share a common neighbor, as must 3 and 4 , and 2 and 4 . If these are the same neighbor, this vertex would share all three neighbors with 1 , so they must be pairwise distinct. The last vertex must then be connected to these three, creating a cube graph. A cube has 48 symmetries, so the number of such graphs is $\frac{8 !}{48}=840$. The total is $35+840=875$. $\fbox{875}$.