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The Carson family will purchase three used cars. There are

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Intern
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Joined: 22 Apr 2017
Posts: 36

Kudos [?]: 19 [0], given: 40

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Re: The Carson family will purchase three used cars. There are [#permalink]

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New post 27 Sep 2017, 13:16
VeritasPrepKarishma wrote:
deya wrote:
Hello,
I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B------> C(4,2)* C(2,1) = 12 ----> C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them.

Secondly, we can select 1 car from A and 2 cars from B-----> C(4,1) * C(3,2) = 12

Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong.

Thank You in advance. :-D


What about selecting all three cars from A or all three cars from B?

All three cars can have the same model. The only constraint is that they need to be of different colors.

Select all from A - In 4C3 = 4 ways
Select all from B - In 4C3 = 4 ways

Total = 24+4+4 = 32 ways


Thanks a lot for the explanation VeritasPrepKarishma. I was stuck with 24 as solution. Now I realized the conceptual gap.

Kudos [?]: 19 [0], given: 40

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Joined: 14 Sep 2015
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Re: The Carson family will purchase three used cars. There are [#permalink]

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New post 29 Sep 2017, 12:48
enigma123 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


My method in solving this was a little different.

First, there are 8 cars to choose from and we need to pick 3 cars. This can be done in 8!/5!*3!=56 ways.
Next, let's choose all cars of the same color. We have 4 colors and 2 models. There is no way all 3 cars can be of the same color.
Next, let's choose 2 cars with the same color. let's assume we have chosen both the red cars, now we need only 1 car. the way we choose this 1 car is 6!/5!*1!=6.
but we could have just as easily chosen any of the other 3 colors for our first 2 cars, so we multiply 6*4=24.

now, since the problem asks us to find the combinations of cars that have a different color each, we need to subtract the number of cars with the same color from the total combinations. i.e. 56-24=32.

Therefore B.

Kudos [?]: 78 [0], given: 19

Re: The Carson family will purchase three used cars. There are   [#permalink] 29 Sep 2017, 12:48

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