Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The Carson family will purchase three used cars. There are [#permalink]
22 Jun 2011, 17:57

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

52% (01:22) correct
48% (00:26) wrong based on 46 sessions

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?

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 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.

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, 21:20, edited 2 times in total.

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 _________________