combinatorics problem

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combinatorics problem

by [email protected] » Mon Jul 22, 2013 1:50 pm
Can someone give me tips on how to crack this question.

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

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by [email protected] » Mon Jul 22, 2013 2:28 pm
[email protected] wrote:
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])

Cheers,
Brent

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

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by GMATGuruNY » Mon Jul 22, 2013 6:24 pm
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by [email protected] » Tue Jul 23, 2013 11:06 am
Thank you so much Brent and GMAT Guru. Makes much more sense now. Will apply to future problems.

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by smkhan » Sat Sep 06, 2014 11:42 am
Hi,

Does it matter how we setup the stages? I am a little confused when you say "For one color, choose a model ". Can it be solved in the manner below;

Stage 1: Possibilities for color, 4C3= 4
Stage 2: Select a model for 3 cars where we have 2 choices for each car. 2x2x2=8

Result: 8x4 =32

So instead of focusing on the color we focus on the choice of model available and the color combination can come from 4C3.

Thanks!

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by [email protected] » Sat Sep 06, 2014 2:29 pm
Hi smkhan,

Yes, you can "do the math" in the way that you described. GMAT Quant questions can often be solved in a variety of ways, so it pays to be flexible. Knowing more than one approach to questions will give you options (and could save you some significant time, if you can spot the "fast" way to answer a question among the various approaches).

Permutation, Combination and Probability questions are often just "shades" of the same concepts, so these questions can be solved in more than one way.

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by [email protected] » Sat Sep 06, 2014 6:03 pm
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by nikhilgmat31 » Wed Oct 07, 2015 12:34 am
this question statement itself asks for How many combinations. but everyone is counting permutations.

I also counted till 192 only using slot method. Please suggest how to decide whether to divide the answer by factorial of number of places.

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by GMATGuruNY » Wed Oct 07, 2015 2:41 am
nikhilgmat31 wrote: I also counted till 192 only using slot method. Please suggest how to decide whether to divide the answer by factorial of number of places.
I explain why we must divide by 3! here:
https://www.beatthegmat.com/pls-clear-my ... 85743.html
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by [email protected] » Wed Oct 07, 2015 9:21 am
Hi nikhilgmat31,

You can also approach the prompt '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

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by nikhilgmat31 » Thu Oct 08, 2015 12:09 am
Thanks Rich,

I got you, basically picking the RED car at the first place or the last place doesn't matter so it is combination of cars.

I clearly remember
nCr = n!/r!(n-r)!
nPr = n!/r!

But I am always confused if order of items matter then number of options of placement will be more or less. But now I got it that if order matters number of options will be more ... so as in permutations.

--> One more question.
Most of times we simply calculate the total by just using the number of options in first place & second place & so on... & then multiple all of them to get final total.

But some times we use nCr method for it. How to choose among a simple factorial or nCr or nPr.

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by nikhilgmat31 » Thu Oct 08, 2015 12:23 am
Also how to understand Combinations with Repetitions , It seems very complex to me.

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by [email protected] » Thu Oct 08, 2015 8:59 am
Hi nikhilgmat31,

The biggest 'clue' is usually in the way that the prompt is written. Certain words almost always 'point' to a certain type of math/logic ("combinations" implies the Combination Formula; "arrangements" implies Permutations, etc.). If you're not clear on what approach to take, then you have to think about the 'logic' behind the question that is asked. In this prompt, we're really just dealing with groups of 3 cars - nothing in the wording implies that the order of the cars matters, so this has to ultimately be about combinations.

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by nikhilgmat31 » Thu Oct 08, 2015 8:13 pm
Thanks Rich, It is crystal clear now.

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by Stuti567 » Wed May 04, 2016 11:55 pm
[email protected] wrote:
[email protected] wrote:
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])

Cheers,
Brent

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