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# The Carson family will purchase three used cars. There are

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

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22 Jun 2011, 18:57
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55% (hard)

Question Stats:

43% (00:22) correct 57% (00:52) wrong based on 130 sessions

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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

[Reveal] Spoiler: OE
This Combinations problem is asking for the number of ways to select 3 cars from 8 (each of the 2 models comes in 4 different colors for a total of 2 x 4 = 8 different types of cars) with the restriction that none of the selected cars be the same color.

We can treat the Carson family’s purchase as a sequence of decisions: the Carsons can initially purchase any one of the 8 cars, but once they have chosen the first car, their choice for the subsequent purchases is limited. This type of choice decision fits well into the Slot Method.
For the first choice, the family can choose from all 8 cars. After they have selected the first vehicle, they have fewer choices for the second pick because they cannot select another car of the same color. For example, if the family purchases a green Model A they cannot also purchase a green Model B. Therefore, we have eliminated both a green Model A and a green Model B from the second choice, leaving only 6 cars from which to choose.

A similar scenario occurs after the second choice, leaving only 4 cars for the final choice. Multiplying these choices together to get the total number of choices we have 8×6×4. (Don’t multiply this out yet! Save yourself some trouble by simplifying first.)

The order in which the purchases are made is not important so we must divide by the factorial of the number of choices to eliminate over-counting:

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-carson-family-will-purchase-three-used-cars-there-are-128876.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Jul 2013, 07:54, edited 1 time in total.
Renamed the topic and edited the question.

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23 Jun 2011, 04:28
This comes from the fact that the order does not matter. IF order doesn't matter you can divide by the factorial of the interchangeable elements, 3 interchangeable elements therefore 3!. After rereading your post again I realized I disregarded the restriction of different colors...

Last edited by Venchman on 23 Jun 2011, 22:20, edited 2 times in total.

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23 Jun 2011, 22:12
dividing by 3! is to avoid duplicates (when the order doesnt matter - meaning choosing A B or B A is the same)

let me change the question with two models and two colors for easy understanding.

let A , B be the models . X ,Y be the colors . so different color combinations can be chosen as follows

AX AY AY BX
[strike]AY AX[/strike] [strike]BX AY[/strike]

AX BY BX BY
[strike]BY AX[/strike] [strike]BY BX[/strike]

so if we go with 4 * 2 , we are getting 8 combinations , out of which only 4 are unique (duplicates are striked out)

Here we have divided by 2! ( two interchangeable positions) to avoid duplicates as Order Does not Matter.

= 4*2/(2!) = 4

Hope its clear now.

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21 Jul 2011, 07:34
Good explanation but i did not get why 3!

If we try to avoid the doubling why could we take just divide by 3 and not 3!

Spidy001 wrote:
dividing by 3! is to avoid duplicates (when the order doesnt matter - meaning choosing A B or B A is the same)

let me change the question with two models and two colors for easy understanding.

let A , B be the models . X ,Y be the colors . so different color combinations can be chosen as follows

AX AY AY BX
[strike]AY AX[/strike] [strike]BX AY[/strike]

AX BY BX BY
[strike]BY AX[/strike] [strike]BY BX[/strike]

so if we go with 4 * 2 , we are getting 8 combinations , out of which only 4 are unique (duplicates are striked out)

Here we have divided by 2! ( two interchangeable positions) to avoid duplicates as Order Does not Matter.

= 4*2/(2!) = 4

Hope its clear now.

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27 Aug 2011, 00:00
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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?

24

32

48

60

192

Choose 3 colors out of four AND choose 1 model for every color.

$$C^{4}_{3}*2*2*2$$

$$C^{4}_{3}$$: Ways to select 3 colors out of 4.
$$2*2*2$$: For every selected color, there are two options.

Ans: "B"
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29 Aug 2011, 10:01
fluke 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?

24

32

48

60

192

Choose 3 colors out of four AND choose 1 model for every color.

$$C^{4}_{3}*2*2*2$$

$$C^{4}_{3}$$: Ways to select 3 colors out of 4.
$$2*2*2$$: For every selected color, there are two options.

Ans: "B"

great explanation. i did it the other way, but i love the way u solve these. +1
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Re: combinations   [#permalink] 29 Aug 2011, 10:01
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