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
combinatorics problem
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Take the task of selecting cars and break it into stages.[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
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/gmatcounting?id=789
Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmatcounting?id=775
 GMATGuruNY
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I posted solutions here:
https://www.beatthegmat.com/combinatoric ... 89780.html
https://www.beatthegmat.com/plsclearmy ... 85743.html
https://www.beatthegmat.com/combinatoric ... 89780.html
https://www.beatthegmat.com/plsclearmy ... 85743.html
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Thank you so much Brent and GMAT Guru. Makes much more sense now. Will apply to future problems.
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!
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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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.
GMAT assassins aren't born, they're made,
Rich
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.
GMAT assassins aren't born, they're made,
Rich
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 Posts: 15965
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Looks good, smkhan.
The FCP can be used to solve the MAJORITY of counting questions on the GMAT. Check out our free video (https://www.gmatprepnow.com/module/gmatcounting?id=775) and then solve the following questions:
EASY
 https://www.beatthegmat.com/whatshould ... 67256.html
 https://www.beatthegmat.com/countingpro ... 44302.html
 https://www.beatthegmat.com/pickinga5 ... 73110.html
 https://www.beatthegmat.com/permutation ... 57412.html
 https://www.beatthegmat.com/simpleonet270061.html
 https://www.beatthegmat.com/mousepelletst274303.html
MEDIUM
 https://www.beatthegmat.com/combinatoric ... 73194.html
 https://www.beatthegmat.com/arabianhors ... 50703.html
 https://www.beatthegmat.com/subsetspro ... 73337.html
 https://www.beatthegmat.com/combinatoric ... 73180.html
 https://www.beatthegmat.com/digitsnumberst270127.html
 https://www.beatthegmat.com/doubtonsep ... 71047.html
 https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
 https://www.beatthegmat.com/wonderfulp ... 71001.html
 https://www.beatthegmat.com/pscountingt273659.html
 https://www.beatthegmat.com/permutation ... 73915.html
 https://www.beatthegmat.com/pleasesolve ... 71499.html
 https://www.beatthegmat.com/notwoladie ... 75661.html
 https://www.beatthegmat.com/lanierasco ... 15764.html
Cheers,
Brent
The FCP can be used to solve the MAJORITY of counting questions on the GMAT. Check out our free video (https://www.gmatprepnow.com/module/gmatcounting?id=775) and then solve the following questions:
EASY
 https://www.beatthegmat.com/whatshould ... 67256.html
 https://www.beatthegmat.com/countingpro ... 44302.html
 https://www.beatthegmat.com/pickinga5 ... 73110.html
 https://www.beatthegmat.com/permutation ... 57412.html
 https://www.beatthegmat.com/simpleonet270061.html
 https://www.beatthegmat.com/mousepelletst274303.html
MEDIUM
 https://www.beatthegmat.com/combinatoric ... 73194.html
 https://www.beatthegmat.com/arabianhors ... 50703.html
 https://www.beatthegmat.com/subsetspro ... 73337.html
 https://www.beatthegmat.com/combinatoric ... 73180.html
 https://www.beatthegmat.com/digitsnumberst270127.html
 https://www.beatthegmat.com/doubtonsep ... 71047.html
 https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
 https://www.beatthegmat.com/wonderfulp ... 71001.html
 https://www.beatthegmat.com/pscountingt273659.html
 https://www.beatthegmat.com/permutation ... 73915.html
 https://www.beatthegmat.com/pleasesolve ... 71499.html
 https://www.beatthegmat.com/notwoladie ... 75661.html
 https://www.beatthegmat.com/lanierasco ... 15764.html
Cheers,
Brent

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this question statement itself asks for How many combinations. but everyone is counting permutations.GMATGuruNY wrote:I posted solutions here:
https://www.beatthegmat.com/combinatoric ... 89780.html
https://www.beatthegmat.com/plsclearmy ... 85743.html
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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I explain why we must divide by 3! here: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.
https://www.beatthegmat.com/plsclearmy ... 85743.html
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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
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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
Final Answer: B
GMAT assassins aren't born, they're made,
Rich

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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!(nr)!
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.
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!(nr)!
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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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.
GMAT assassins aren't born, they're made,
Rich
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.
GMAT assassins aren't born, they're made,
Rich

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[email protected] wrote:Take the task of selecting cars and break it into stages.[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
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/gmatcounting?id=789
Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmatcounting?id=775
Hi Brent,
Thanks a lot for your wonderful explanation.
However, I have difficulties in breaking a task into stages. For example, in this question, why can we first select the model and then the color for each model? I understand that the variety of models is less than the number of cars to be selected still I don't feel very confident about it. Is there any rule or trick to divide a task into stages?