Hi anksm22,

This is a Combination Formula question with a twist and the "math" behind it can be done in a couple of different ways. Here's an approach that will allow you to focus on one car at a time:

We're told that there are 2 models of car and 4 colors that the cars can come in. This essentially means that there are 8 different cars available.

We're asked for the total number of combinations of 3 cars with DIFFERENT COLORS.

1st Car: Can be any of the 8 possible cars. Once that car is chosen, you must remove that color from the remaining options.

2nd Car: With the first color removed, there are now 6 possible cars. Once the second car is chosen, you must remove that second color from the remaining options.

3rd Car: With the first 2 colors removed, there are now 4 possible cars.

(8)(6)(4) = 192

We're not done though. Since we were asked for possible Combinations, the order of the cars does not matter, so we have to remove the duplicate entries.

If we call our 3 cars X, Y and Z, there are actually 6 different ways that those 3 cars could have been chosen:

XYZ
XZY
YXZ
YZX
ZXY
ZYX

We are NOT allowed to count this combination 6 times; we're only allowed to count it once. Thus, we must take our prior total and divide by 6:

192/6 = 32

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

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

Take the task of selecting cars and break it into stages.

Stage 1: Select 3 different colors.
Since the order in which we select the colors does not matter, we can use combinations.
We can select 3 colors from 4 colors in 4C3 ways (4 ways).

If anyone is interested, we have a free video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: For one color, choose a model
There are two models (A or B) so this stage can be accomplished in 2 ways.

Stage 3: For another color, choose a model
There are two models (A or B) so this stage can be accomplished in 2 ways.

Stage 4: For the last remaining color, choose a model
There are two models (A or B) so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus select the 3 cars) in (4)(2)(2)(2) ways ([spoiler]= 32 ways[/spoiler])

Answer: B

------------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent

gmatdriller wrote:I tried it the 1st time like this:

All model A's: 4C3 = 4 ways
All model B's: 4C3 = 4 ways

Mixing model A & B:
1A with 2B: 4C1*3C2 = 12
4C1= No of ways of selecting 1 of 4 colors from model A cars
3C2= Ways of selecting 2 cars to match 1 from A w/out matching colors

2A with 1B: 4C2*2C1 = 12
4C2: No of ways of selecting 2 colors from A;
2C1: ways of selecting 1 cars from B to match 2 from A w/out matching colors

Total options = 4+4+12+12 = 32

Could somebody pls criticize my method.

Thanks

Hi gmatdriller,

It's hard to critique your approach, because it's unclear what you're doing with each step.
For example, what do you mean by the following:

All model A's: 4C3 = 4 ways
All model B's: 4C3 = 4 ways

It's hard to tell what your rationale is for other steps as well.
I'm not saying that your approach is incorrect; I just can't tell what those numbers are representing.

Cheers,
Brent