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
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I posted solutions here:
https://www.beatthegmat.com/combinatoric ... 89780.html
https://www.beatthegmat.com/pls-clear-my ... 85743.html
https://www.beatthegmat.com/combinatoric ... 89780.html
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Take the task of selecting cars and break it into stages.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
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
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One last approach:
Our first car can be A or B, and any of four colors: this gives us 2 * 4 = 8 options.
Our second car can be A or B, and any of three colors: this gives us 2 * 3 = 6 options.
Our third car can be A or B, and either of two colors: this gives us 2 * 2 = 4 options.
So we have (8 * 6 * 4) options, but we don't care about the order in which we're arranging them, so we divide by the number of ways we can arrange three things (3 * 2 * 1).
This gives us (8*6*4)/(3*2*1) = 32 options.
Our first car can be A or B, and any of four colors: this gives us 2 * 4 = 8 options.
Our second car can be A or B, and any of three colors: this gives us 2 * 3 = 6 options.
Our third car can be A or B, and either of two colors: this gives us 2 * 2 = 4 options.
So we have (8 * 6 * 4) options, but we don't care about the order in which we're arranging them, so we divide by the number of ways we can arrange three things (3 * 2 * 1).
This gives us (8*6*4)/(3*2*1) = 32 options.
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Chicagosummer
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Total #of ways to choose 3 cars from 8 without any constraints = 8C3 = 56
Now ways to choose the cars with constraint = 1 x 6/7 x 4/6 = 24/42 = 8/14 ways from 56 = 8/14 x 56 = 32
Now ways to choose the cars with constraint = 1 x 6/7 x 4/6 = 24/42 = 8/14 ways from 56 = 8/14 x 56 = 32
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Here comes the last method to solve could be
We have four pairs (2 models in 4 colors)
Blue - A
Blue - B
Black - A
Black - B
Red - A
Red - B
Green - A
Green - B
Total ways to choose 3 cars randomly (favorable and unfavorable cases) = 8C3 = 56
Total unwanted cases (in which we have 2 cars of same color) = 4(ways to choose 2 cars of same color) x 6(third car of different color) = 24
Total favorable cases = Total cases - Unfavorable cases = 56 - 24 = 32
Answer: Option B
We have four pairs (2 models in 4 colors)
Blue - A
Blue - B
Black - A
Black - B
Red - A
Red - B
Green - A
Green - B
Total ways to choose 3 cars randomly (favorable and unfavorable cases) = 8C3 = 56
Total unwanted cases (in which we have 2 cars of same color) = 4(ways to choose 2 cars of same color) x 6(third car of different color) = 24
Total favorable cases = Total cases - Unfavorable cases = 56 - 24 = 32
Answer: Option B
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Alternatively:
Imagine a 3-digit integer: PQR
P has 4 values, Q has 4 values, R has 4 values
There are 4^3 = 48 combinations
If we dismiss P=Q=R, that removes 4 combinations
If we dismiss P=Q, Q=R and P=R, that removes 3 x 4 = 12 combinations
So, 48 - 4 - 12 = 32 combinations
Imagine a 3-digit integer: PQR
P has 4 values, Q has 4 values, R has 4 values
There are 4^3 = 48 combinations
If we dismiss P=Q=R, that removes 4 combinations
If we dismiss P=Q, Q=R and P=R, that removes 3 x 4 = 12 combinations
So, 48 - 4 - 12 = 32 combinations
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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
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
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Hi gmatdriller,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
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:
It's hard to tell what your rationale is for other steps as well.All model A's: 4C3 = 4 ways
All model B's: 4C3 = 4 ways
I'm not saying that your approach is incorrect; I just can't tell what those numbers are representing.
Cheers,
Brent
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The way I approach this problem is to start with 8 cars (four colors of A and four colors of B).
We have a constraint that we cannot chose 2 of the same color.
1st choice: we can chose any of the 8 cars. Once we pick one car we remove that car and other car with the same color - now there are 6 to chose from.
2nd choice: we can chose any of the 6 cars remaining (using the same principle above we are now down to 4 cars)
3rd choice: we can chose any of the 4 cars remaining
permutations = 8*6*4
as we want combinations we need to divide by 3! (as order does not matter).
Answer = 8*6*4/3! = 32
We have a constraint that we cannot chose 2 of the same color.
1st choice: we can chose any of the 8 cars. Once we pick one car we remove that car and other car with the same color - now there are 6 to chose from.
2nd choice: we can chose any of the 6 cars remaining (using the same principle above we are now down to 4 cars)
3rd choice: we can chose any of the 4 cars remaining
permutations = 8*6*4
as we want combinations we need to divide by 3! (as order does not matter).
Answer = 8*6*4/3! = 32
