MATH
posted by Jennifer .
A parking lot has 5 spots remaining and 5 different cars in line waiting to be parked. If one of the cars is too big to fit in the two outermost spots, in how many different ways can the five cars be arranged?

One car must go into any one of the three largeenough spots. The remaining four cars can be distributed among the other four slots 4! = 24 ways. The total number of arrangement possibilities is 3*24 = 72

4*3*2*1*3 there are 4 possibilities for the first space... so u only have 3 options for the fifth spot because 1 car has already parked... now since two cars have parked in the first and last spots you have 3 cars out of the five that can still park.. so 3 can park in the second spot and then 2 can park in the third spot and then 1 in the fourth space.