3c. In a state, there are 100 cities. It is required to connect some pairs of cities with flights in such a way that from any city it is possible to fly (possibly with layovers) to any other city, and for any four cities $A$, $B, C, D$, for which there are flights $A B, B C, C D$, there is also a flight $A D$. How many ways are there to do this?
Answer: $2^{99}$ ways.

Notice that from any city to any other city, one can fly with no more than one layover. Indeed, from the condition, it follows that any path with two or more layovers can be shortened. Consider an arbitrary city $A$. Cities that have flights from $A$ will be called nearby, and the rest of the cities will be called distant. Consider an arbitrary distant city $D$. As we have found, it is connected to one of the nearby cities $B_{1}$. For any other nearby city $B_{2}$, considering the route $B_{2}-A-B_{1}-D$, we see that $D$ and $B_{2}$ are also connected. Therefore, any distant city is connected to any nearby city. If we also call city $A$ distant, the specified property remains. Thus, all cities are divided into two non-empty groups, cities from different groups are connected pairwise. If there are no more flights, then such a system of flights fits, and it can be chosen in $2^{99}$ - 1 ways (if we fix city $A$, any of the other cities must be either distant or nearby, with the only restriction being that all cities cannot be distant). If, however, there is at least one flight between cities of the same group, then it is not hard to see that in this case any two cities are connected by a flight. Such a system of flights also fits. In total, $2^{99}-1+1=2^{99}$.

## Fourth Round