Posted by Andy on Friday, June 29, 2012 at 8:41pm.
This sounds a bit badly worded to me, since strictly speaking there is NO way that eight cars can fit into two spaces that between them can accommodate at most five cars. However I assume that the question is intended to be "In how many ways can exactly five out of eight cars be parked in two big spaces, one of which will take three and the other of which will take two?".
However... if that's right, then there's another question to answer first, which is "Do we distinguish between the six different configurations that any three cars can occupy in the big space, or between the two different configurations that any two cars can occupy in the small space?".
If we DON'T distinguish between them, then I think the answer would be 8!/(3! x 5!) x 5!/(2! x 3!) = 42 x 10 = 420. The reason for that is that we need to calculate the number of ways in which three cars out of eight can occupy the larger space {8!/(3! x 5!)}, and that needs to be multiplied by the number of ways in which two of the remaining five cars can occupy the smaller space (5!/(2! x 3!).
If we DO distinguish between them, then I think the answer would be (8!/3!) x (5!/2!) = (8.7.6.5.4).(5.4.3) = 403,200.
Having said that, I'm not actually certain about either answer, so please do check them.