The cars can line up in \(20!\) different ways, too many variations for a simple parking space.

There are 2 arrangments of the cars that alternate: starting with US-made and starting with foreign-made. The total number of arrangments is \({ 20 \choose 10}\), since a specific arrangment is about how to mix foregin and US, which in turn is decided by how you place 10 US cars into 20 slots. Therefore, the probability of alternation is just \(\frac{2}{20 \choose 10}\).

An alternative solution incorporates permutations: \(\frac{2\cdot 10! \cdot 10!}{20!}\).