In how many orders can 4 red and 5 blue cars they be arranged if the colors must alternate? I see two cases. One starting with red and another starting with blue.
Case 1: Starting with red. $R_4\times B_5\times R_3\times B_4\times R_2\times B_3\times R_1\times B_2$. However, this doesn't seem possible since there are no reds available to multiply after $B_2$
Case 2: Starting with blue. $B_5\times R_4\times B_4\times R_3\times B_3\times R_2\times B_2\times R_1\times B_1$ . This means $5\times4\times4\times3\times3\times2\times2\times1\times1=2880$ ways.
As you've noted, in all odd positions, there must be a blue car, and in even positions, there must be a red car. As such, all that matters is which red car goes where, and likewise for blue cars. This is why the answer is $5!\cdot4!=\color{red}{2880}$