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Re: The Carson family will purchase three used cars. There are two models [#permalink]
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GreenlightTestPrep wrote:
GeminiHeat 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 GRE. So, be sure to learn it.


FCP is indeed a great method to solve such problems. Could you mention any situations in counting problems where FCP isn't appropriate?
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Re: The Carson family will purchase three used cars. There are two models [#permalink]
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computerbot wrote:
FCP is indeed a great method to solve such problems. Could you mention any situations in counting problems where FCP isn't appropriate?


The most common situation in which the FCP won't work is when the order in which we select objects doesn't matter (i.e., combination questions)
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Re: The Carson family will purchase three used cars. There are two models [#permalink]
I understand the approach of 4c3 x 2 x 2 x 2 (Here we are selecting three colors out of 4 and the 3 cars can be any of the A,)


What is wrong with my approach

Ways to select first car -> 4c1 * 2c1 = 8
Ways to select first car -> 3c1 * 2c1 = 6
Ways to select first car -> 2c1 * 2c1 = 4
8+6+4 = 18

Any help is appreciated
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Re: The Carson family will purchase three used cars. There are two models [#permalink]
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We are selecting 3 different color cars out of 4 possible colors. In how many ways it can be done? \(C^3_4=4\), selecting 3 out of 4.

Next, there are 2 models of each selected car of a certain color available, hence each selected car has 2 options: Model A or Model B. Since there are 3 selected cars then total ways is 2*2*2.

Grand total 4*2^3=32.
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Re: The Carson family will purchase three used cars. There are two models [#permalink]
Carcass wrote:
We are selecting 3 different color cars out of 4 possible colors. In how many ways it can be done? \(C^3_4=4\), selecting 3 out of 4.

Next, there are 2 models of each selected car of a certain color available, hence each selected car has 2 options: Model A or Model B. Since there are 3 selected cars then total ways is 2*2*2.

Grand total 4*2^3=32.


Could you please highlight the mistake in my approach
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The Carson family will purchase three used cars. There are two models [#permalink]
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\(C^3_4*2^3=4*8=32\), (\(C^3_4\) selecting 3 different colors from 4 and multiplying by 2*2*2=2^3 since each color car has two options: model A or model B).

but your mistake is common

quote

Quote:
The method of selecting 8 for the first choice, 6 for the second and 4 for the third is what we call 'basic counting principle'. When you do 8*6*4, you are effectively selecting and ARRANGING the cars: you say 'The FIRST car is selected in 8 ways, the SECOND car in 6 ways etc'. But you don't have a first second third car. You only have a group of 3 cars. So to un-arrange (so to say), you need to divide by 3!


Hope now is clear

see more here for the counting method https://gre.myprepclub.com/forum/gre-permu ... tml#p83002
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Re: The Carson family will purchase three used cars. There are two models [#permalink]
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GreenlightTestPrep wrote:
computerbot wrote:
FCP is indeed a great method to solve such problems. Could you mention any situations in counting problems where FCP isn't appropriate?


The most common situation in which the FCP won't work is when the order in which we select objects doesn't matter (i.e., combination questions)





But don't you think that even in this case the order didn't really matter?

We could've chosen any car at any position. We just had to choose 3 different cars.
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