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Re: The Carson family will purchase three used cars. There are
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27 Sep 2017, 12:16

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.

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.

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29 Sep 2017, 11:48

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.

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21 Apr 2018, 06:21

Top Contributor

1

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

Take the task of selecting cars and break it into stages.

Stage 1: Select 3 different colors. Since the order in which we select the colors does not matter, we can use combinations. We can select 3 colors from 4 colors in 4C3 ways (4 ways).

ASIDE: If anyone is interested, we have a video on calculating combinations (like 4C3) in your head (see below)

Stage 2: For one color, choose a model There are two models (A or B) so this stage can be accomplished in 2 ways.

Stage 3: For another color, choose a model There are two models (A or B) so this stage can be accomplished in 2 ways.

Stage 4: For the last remaining color, choose a model There are two models (A or B) so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus select the 3 cars) in (4)(2)(2)(2) ways (= 32 ways)

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

We need to select 3 colors out of 4 colors.There are two model of cars too!

So, [C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]+[C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)] -------Model A first,Model B 2nd-------- -------Model B first,Model A 2nd--------

Thus, on calculating we get: 28+28=56
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We need to select 3 colors out of 4 colors.There are two model of cars too!

So, [C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]+[C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)] -------Model A first,Model B 2nd-------- -------Model B first,Model A 2nd--------

Thus, on calculating we get: 28+28=56

I am not sure of the logic you have followed here. Using 4C3, you select 3 colours of the 4. Then why do you use 4C2 and 4C1? Also each colour is available in 2 different models so next set is to pick a model for each colour. Say you picked blue, black and red. For blue, you pick either model A or B so 2 options. For black you pick either model A or B so 2 options. For red, you pick either model A or B so 2 options.

In all, 4C3 * 2 * 2 *2 ways.

Also, it is just a selection. No arrangement involved so first second order is immaterial.
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[b]Karishma Veritas Prep GMAT Instructor

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We need to select 3 colors out of 4 colors.There are two model of cars too!

So, [C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)]+[C(4,3)+C(4,2)*C(2,1)+C(4,1)*C(3,2)] -------Model A first,Model B 2nd-------- -------Model B first,Model A 2nd--------

Thus, on calculating we get: 28+28=56

I am not sure of the logic you have followed here. Using 4C3, you select 3 colours of the 4. Then why do you use 4C2 and 4C1? Also each colour is available in 2 different models so next set is to pick a model for each colour. Say you picked blue, black and red. For blue, you pick either model A or B so 2 options. For black you pick either model A or B so 2 options. For red, you pick either model A or B so 2 options.

In all, 4C3 * 2 * 2 *2 ways.

Also, it is just a selection. No arrangement involved so first second order is immaterial.

Thanks Karishma!!
_________________

Thanks, Rnk

Please click on Kudos, if you think the post is helpful

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25 Aug 2018, 15:13

I understand that 1C8* 1C6 *1C4 = 8*6*4 is "trying to arrange the sequence btw 3 cars" so we need to "un-arrange them" by dividing 8*6*4 over 3!/1!2! (which equals to 3), so the whole equation is "8*6*4 then divide by 3"= 64. BUT, the official correct answer is 32.

Re: The Carson family will purchase three used cars. There are
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01 Sep 2018, 04:50

Its straightforward:-

Either all three of one model which means 4C3 ways for model A and 4C3 for model B; OR

2 could be selected from model A and one from model B and vice versa but then the precaution which one needs to take is that all must be of different colours.

Therefore it will be 4C2*2C1 for 2 of Model A AND one of model B OR 4C2*2C1 for 2 of Model B AND one of model A.

Therefore total ways is: 2*4C3 + 2*4C2*2C1 = 2*4 + 2*6*2 = 32 ways. (ANS)
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Re: The Carson family will purchase three used cars. There are
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01 Sep 2018, 05:21

We have 8 types of cars:Model A (B, B, R & G) + Model B (B,B, R & G). Let's select 3 cars of different colors from these 8 choices. 1st selection - 8 choices 2nd selection- 6 choices (As one color selection already done above, let's say Model A - Black. This put restrictions on model B - Black as well as we need different colors). 3rd selection - 4 choices (As 2 colors already selected above so we are left with only 2 colors of models A&B).

Using FCP, Number of combinations is 8X6X4/3! = 32. Choice B.

gmatclubot

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01 Sep 2018, 05:21