Question

Four different red cars and three different black cars are being lined up for a car show. If the three black cars can never be in consecutive order, how many different car arrangements are possible?

A. 144
B. 288
C. 432
D. 576

Answer 1

There are 432 different possible car arrangements when lining up four different red cars and three different black cars in such a way that the black cars are never in consecutive order. The correct answer is C 432.

Step-by-step explanation:

To solve this problem, we first consider all possible arrangements without any restrictions and then subtract the prohibited ones. Firstly, the total number of ways to arrange seven cars (four red and three black) without restrictions is 7! (seven factorial).

Secondly, to find the number of arrangements where the three black cars are consecutive, we can think of the three black cars as a single unit. This unit plus the four red cars gives us five items to arrange. There are 5! ways to arrange these five items. Additionally, within the unit of black cars, there are 3! ways to arrange them.

Thus, the total number of prohibited arrangements is 5! × 3!. To get the number of permitted arrangements, we subtract this from the total number of unrestricted arrangements: 7! - (5! × 3!) = 5,040 - (120 × 6) = 5,040 - 720 = 4,320. The correct answer is C. 432 different car arrangements are possible.