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B
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Difficulty:
65%
(hard)
Question Stats:
53% (01:36) correct
47%
(02:02)
wrong
based on 34
sessions
History
Date
Time
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Not Attempted Yet
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
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-carson-family-will-purchase-three-used-cars-there-are-128876.html
A) 24
B) 32
C) 48
D) 60
E) 192
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:
The correct answer is B
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:
The correct answer is B
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-carson-family-will-purchase-three-used-cars-there-are-128876.html
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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...
23 Jun 2011, 22:12
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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.
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.
