## Permutations vs. Combinations

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### Permutations vs. Combinations

by krithika1993 » Mon Oct 17, 2016 8:38 am
Hello,

I got this question wrong when I tried an MGMAT practice exam.

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

*************************************************************************************************************************

After trying the problem again, I understand that the # of options available for each additional car after the 1st one is limited, but the MGMAT answer explanation for this question wants me to divide the # of options by 3!, in order to get the final correct answer.

I just don't understand why I should do that - when a question asks for total # of combinations, I was under the impression that this is the product of all options available (i.e. # of options available for car 1 x # of options available for car 2 x # of options available for option 3); MGMAT says that this would result in over-counting.

Could somebody please explain this to me?

Your help would be appreciated. Thank you!

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by GMATGuruNY » Mon Oct 17, 2016 8:47 am
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by [email protected] » Mon Oct 17, 2016 9:00 am

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
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).

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])

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here's a video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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by [email protected] » Mon Oct 17, 2016 9:02 am
krithika1993 wrote: After trying the problem again, I understand that the # of options available for each additional car after the 1st one is limited, but the MGMAT answer explanation for this question wants me to divide the # of options by 3!, in order to get the final correct answer.

I just don't understand why I should do that - when a question asks for total # of combinations, I was under the impression that this is the product of all options available (i.e. # of options available for car 1 x # of options available for car 2 x # of options available for option 3); MGMAT says that this would result in over-counting.

Could somebody please explain this to me?

Your help would be appreciated. Thank you!
You need not necessarily divide by 3! at the end.
See my solution above.

Cheers,
Brent
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by [email protected] » Mon Oct 17, 2016 10:27 am
Hi krithika1993,

The logic behind this prompt is 'thick', so you might want to approach the work 'one car at a time'

We have two Models of cars and 4 possible colors, so there are initially 8 different cars available.

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

For the 1st car, there are 8 options. Once we pick one, we've removed one of the colors....
For the 2nd car, there are now 6 options (each of the two Models in each of the remaining 3 colors). Once we pick one, we're removed another color...
For the 3rd car, there are now 4 options (each of the two Models in each of the remaining 2 colors).

So, at first glance, it might appear that there are (8)(6)(4) possibilities. HOWEVER, we've done a permutation so far, but the order of the cars DOES NOT MATTER, so we have to adjust the calculation...

If we called the specific cars X, Y and Z, then there are 6 different 'orders' that we could choose those 3 cars...
XYZ
XZY
YXZ
YZX
ZXY
ZYX

But those are NOT 6 different outcomes - it's the same outcome 6 different times. When we're asked for combinations, we're not allowed to count 'duplicates.' Thus, we have to divide our prior total by 6...

192/6 = 32