ktrout2020 wrote:The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?
A. 24
B. 32
C. 48
D. 60
E. 192
Source: Manhattan Prep
Given that there are 2 models and each of them is available in 4 colors, we have 2*4 = 8 cars
Let's choose the first car.
The number of ways to choose the first car = 8 (Say this a black color car).
Now since we have already selected a black car, we cannot have another black car,; so for the second car, we have 6 cars to choose from.
The number of ways to choose the second car = 6 (Say this a blue color car).
Now since we have already selected a black and a blue car, we cannot have another black or blue car, so for the third car, we have 4 cars to choose from.
The number of ways to choose the third car = 4 (Say this a red color car).
Total no. of arrangements = 8*6*4 = 192
Note that 192 is the number of arrangements of three cars, but we are not interested in the arrangement; we are interested in the selection.
In 192 arrangements, there are arrangements such as (BlackA, BlueB, RedA) and ( RedA, BlackA, BlueB); however, from the selection point of view, both are the same. So, we must remove such arrangements.
The three selected cars can be arranged in 3! = 6 ways.
Thus, the number of different combinations of three cars the Carsons can select if all the cars are to be different colors = 192/6 = 32
The correct answer:
B
Hope this helps!
-Jay
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