Problem Solving

Q: 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

OA: B

I understand the explanation partially. Could someone help me understand the bold faced part below. I am unable to understand why we did that step.

Explanation:
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:


(8*6*4)/3! = 8*4 = 32

The correct answer is B.

Thank you

krnverma wrote:Q: 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

OA: B

Color options:
Number of ways to choose 3 colors from 4 choices = 4C3 = 4.

Model options:
Each color comes in two models (A and B).
Number of model options for the first color = 2.
Number of model options for the second color = 2.
Number of model options for the third color = 2.
To combine these options, we multiply:
2*2*2 = 8.

Car options:
To combine our color options with our model options, we multiply the results above:
4*8 = 32.

The correct answer is B.

In the MGMAT solution, 8*6*4 = the number of ways the 3 cars could be ARRANGED:
Blue A, Red A, Green A
Blue A, Green A, Red A
Red A, Blue A, Green A
etc.

Each entry in the list above represents a different arrangement but the SAME combination of cars.
In order not to overcount the duplicate combinations, we must divide by the number of ways to arrange 3 elements (3!):
(8*6*4)/3! = 32.

Hi - consider this example (-also present in the problem solving forum) - in how Many ways John can select 2 squares from the chess board such that the 2 blocks are not present in the same column.
As we know a chess board consists of 64 square blocks.
So first one can be selected in 64 ways
For our next selection - the column from which we selected our first quare will be out of the selection process.
So the blocks left for selection are 56 ( 64-8)
Let's name the first block as A
And the second block as B
Now the pair AB is same as pair BA
So, if we multiply 64*56 we will get repetitive pairs
Therefore we must divide the product by 2
The same logic applies in your question
Green blue yellow
Is same as blue yellow green

The number of ways

Good Question