The answer is A
There 3 different cars and 2 different model. We will have 6 combinations of 2 different models.
Then 6 combinations come with 4 colors, we will have 24 combinations of 3 used cars 2 different models and 4 colors

Or 3x2x4

minhchau1986 wrote:The answer is A
There 3 different cars and 2 different model. We will have 6 combinations of 2 different models.
Then 6 combinations come with 4 colors, we will have 24 combinations of 3 used cars 2 different models and 4 colors

Or 3x2x4

This is incorrect the answer is B.

[spoiler]4C3 and then each time you have two choice models... therefore 4C3 is 4 x 2 x 2 x 2 = 32[/spoiler]

OE:

[spoiler]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[/spoiler]

seal4913 wrote:OE:

[spoiler]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[/spoiler]

Thanks seal4913

. I got your process til 8x6x4 because we only have 8 choices for first, 6 for second and 4 for third cars. I got stuck at the factorial of the number of choices to eliminate over-counting

Is this because 3 cars can not have same colors out of 8x6x4 combinations? So we have 3x2=6 same color combinations.

Simplying by 6 to get 32

Alternate Method:
(by SUBTRACTION)

Required number of combinations = Total number of ways of selecting any 3 cars from the 8 cars - Total number of ways in which cars of same colour are chosen
= 8C3 - (Select 1 out of the 4 pairs of same colour AND Select 1 out of the remaining 6 cars)
=8C3 - 4C1*6C1
=56 - 24
=32