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 large-enough 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.
To solve this problem, we need to consider the different cases based on the position of the car that is too big to fit in the two outermost spots.
Case 1: The car is parked in one of the center three spots.
In this case, there are 3 spots remaining for the remaining 4 cars to be parked in. The number of ways to arrange the 4 cars in the remaining spots can be found using the factorial function, denoted by "!". For 4 cars, the number of ways is 4!
Case 2: The car is parked in one of the two inner spots.
In this case, there are 2 spots remaining for the remaining 4 cars to be parked in. Again, the number of ways to arrange the remaining 4 cars is 4!
Therefore, the total number of ways to arrange the 5 cars is the sum of the two cases: 4! + 4! = 2 * 4! = 2 * 4 * 3 * 2 * 1 = 48.
So, there are 48 different ways to arrange the five cars in the parking lot.
To solve this problem, we need to consider the different cases based on the position of the car that is too big to fit in the two outermost spots.
Case 1: The car takes the middle spot.
In this case, the car that is too big has only one possible position in the middle spot. The remaining cars can be arranged in the remaining spots in 4! = 24 ways.
Case 2: The car takes one of the two inner spots.
In this case, the car that is too big has two possible positions in either of the two inner spots. The remaining cars can be arranged in the remaining spots in 4! = 24 ways.
Case 3: The car takes one of the two outermost spots.
In this case, the car that is too big has two possible positions in either of the two outermost spots. The remaining cars can be arranged in the remaining spots in 4! = 24 ways.
Therefore, the total number of different ways the five cars can be arranged is:
1 (from Case 1) + 2 (from Case 2) + 2 (from Case 3) = 5 different ways.