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The Carson family will purchase three used cars. There are

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The Carson family will purchase three used cars. There are [#permalink] New post 11 Mar 2012, 01:03
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A
B
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Difficulty:

  45% (medium)

Question Stats:

40% (02:10) correct 59% (01:09) wrong based on 77 sessions
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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 13 Jul 2013, 06:51, edited 1 time in total.
Added the OA.
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 11 Mar 2012, 01:18
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enigma123 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?

A. 24
B. 32
C. 48
D. 60
E. 192

Any idea how to solve this guys? I don't have an OA unfortunately.


C^3_4*2^3=4*8=32, C^3_4 selecting 3 different colors from 4 and multiplying by 2*2*2=2^3 since each color car has two options: model A or model B.

Answer: B.

Similar question to practice:
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
if-a-committee-of-3-people-is-to-be-selected-from-among-98533.html

Hope it helps.
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 26 Mar 2012, 19:31
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 26 Mar 2012, 23:41
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calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


We are selecting 3 different color cars out of 4 possible colors. In how many ways it can be done? C^3_4=4, selecting 3 out of 4.

Next, there are 2 models of each selected car of a certain color available, hence each selected car has 2 options: Model A or Model B. Since there are 3 selected cars then total ways is 2*2*2.

Grand total 4*2^3=32.

Check the links in my previous post for similar questions.

Hope it helps.
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 07 May 2012, 20:46
Bunnel,

I tried solving it through 8C3 = 56, as there are 8 options available Model A with 4 colors, and Model B with 4 colors, and we have to select 3 out of those..Can you please explain what is wrong in this approach ?

Thanks!
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 07 May 2012, 23:44
Expert's post
gmihir wrote:
Bunnel,

I tried solving it through 8C3 = 56, as there are 8 options available Model A with 4 colors, and Model B with 4 colors, and we have to select 3 out of those..Can you please explain what is wrong in this approach ?

Thanks!


8C3 gives total # of ways to select 3 cars out of 8. But in the question we have a restriction saying that "all the cars (selected) are to be different colors", naturally restriction will reduce the number of combinations possible, so 56 cannot be a correct answer.

Hope it's clear.
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 09 May 2012, 07:21
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 09 May 2012, 07:29
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ashish8 wrote:
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?


Check this problem: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html and this post there: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html#p775925
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 09 May 2012, 08:26
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ashish8 wrote:
calreg11 wrote:
Bunuel, I don't understand the logic behind why you grouped it like that.
in MGMAT they said to use the slot method, first car has 8 choices, second has 6, third has 6. to account for overcounting we divide by the factorial 3!. same number but i understand that logic a little better but i am inclined to use your method on the test.
Can you explain?


Bunuel, Calgreg, if I follow the MGMAT method I don't understand what over counting they are talking about. I understand the 8 for the first choice, 6 for the second, and 4 for the third, but why do the division afterwards?


The method of selecting 8 for the first choice, 6 for the second and 4 for the third is what we call 'basic counting principle'. When you do 8*6*4, you are effectively selecting and ARRANGING the cars: you say 'The FIRST car is selected in 8 ways, the SECOND car in 6 ways etc'. But you don't have a first second third car. You only have a group of 3 cars. So to un-arrange (so to say), you need to divide by 3!

I have explained this concept in this post: http://www.veritasprep.com/blog/2011/11 ... binations/
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 26 Dec 2012, 13:28
Bunuel, Karishma,

You gotta help me out here, m getting shaky on all my Combinatorics concepts.

I approached the problem as

8C8 * 7C6 * 4C1

What exactly m I doing wrong?
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 26 Dec 2012, 23:33
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eaakbari wrote:
Bunuel, Karishma,

You gotta help me out here, m getting shaky on all my Combinatorics concepts.

I approached the problem as

8C8 * 7C6 * 4C1

What exactly m I doing wrong?


I have no idea how you got 8C8, 7C6 and 4C1.

I could have understood 8C1*6C1*4C1 (I have explained why this doesn't work in the post above)

The best way to solve it is by first selecting 3 colors out of the given 4 in 4C3 ways. (say, you got black, red and green)
Now for each color, you have 2 choices - model A or B
So you select a model in 2 ways.
No of ways of selecting the 3 cars = 4C3 * 2 * 2 * 2 = 32
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 26 Dec 2012, 23:51
VeritasPrepKarishma wrote:
eaakbari wrote:
Bunuel, Karishma,

You gotta help me out here, m getting shaky on all my Combinatorics concepts.

I approached the problem as

8C8 * 7C6 * 4C1

What exactly m I doing wrong?


I have no idea how you got 8C8, 7C6 and 4C1.

I could have understood 8C1*6C1*4C1 (I have explained why this doesn't work in the post above)

The best way to solve it is by first selecting 3 colors out of the given 4 in 4C3 ways. (say, you got black, red and green)
Now for each color, you have 2 choices - model A or B
So you select a model in 2 ways.
No of ways of selecting the 3 cars = 4C3 * 2 * 2 * 2 = 32


Well, my logic was

The number of ways of selecting r objects from n different objects is nCr.

We have 8 options and can select either of the 8 as the first - 8C8

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4

I reckon that my understanding of Combinatorics logic is fundamentally flawed. Appreciate if you could tell me where and what
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 27 Dec 2012, 00:28
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eaakbari wrote:

Well, my logic was

The number of ways of selecting r objects from n different objects is nCr.

We have 8 options and can select either of the 8 as the first - 8C8

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4

I reckon that my understanding of Combinatorics logic is fundamentally flawed. Appreciate if you could tell me where and what


Yes, I think you need to go through a PnC book from scratch. You can check out the Veritas Combinatorics and Probability book at amazon.

"The number of ways of selecting r objects from n different objects is nCr" - This is correct. You need to apply it correctly now.

As for this question,

We have 8 options and can select either of the 8 as the first - 8C8
--- We have 8 options and we need to select 1 out of those 8. So n = 8 and r = 1. This gives us 8C1 ways. (Say you selected BlackA)

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6
--- Now one car is gone and we cannot select another one. Now we have 6 cars to choose from and we again have to select 1. So n = 6 and r = 1. We get 6C1 ways (Say you selected GreenB)

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4
--- Now 2 cars are gone and 2 are forbidden. You can choose out of only 4 cars and you have to select only 1. Again, n = 4 and r = 1 so we get 4C1 (Say you selected RedA)

You select 3 cars in 8C1 * 6C1 * 4C1 ways but there is a flaw here. You selected BlackA, GreenB and RedA.
In another case, you could have selected RedA first, then BlackA and then GreenB. This combination is the same as the previous one but we are counting it separately. The point is we have arranged the cars as first, second and third which we should not do since it is a group.

As I said before, you might want to start with fundamentals from a book.
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 27 Dec 2012, 00:35
VeritasPrepKarishma wrote:
eaakbari wrote:

Well, my logic was

The number of ways of selecting r objects from n different objects is nCr.

We have 8 options and can select either of the 8 as the first - 8C8

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4

I reckon that my understanding of Combinatorics logic is fundamentally flawed. Appreciate if you could tell me where and what


Yes, I think you need to go through a PnC book from scratch. You can check out the Veritas Combinatorics and Probability book at amazon.

"The number of ways of selecting r objects from n different objects is nCr" - This is correct. You need to apply it correctly now.

As for this question,

We have 8 options and can select either of the 8 as the first - 8C8
--- We have 8 options and we need to select 1 out of those 8. So n = 8 and r = 1. This gives us 8C1 ways. (Say you selected BlackA)

One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6
--- Now one car is gone and we cannot select another one. Now we have 6 cars to choose from and we again have to select 1. So n = 6 and r = 1. We get 6C1 ways (Say you selected GreenB)

Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4
--- Now 2 cars are gone and 2 are forbidden. You can choose out of only 4 cars and you have to select only 1. Again, n = 4 and r = 1 so we get 4C1 (Say you selected RedA)

You select 3 cars in 8C1 * 6C1 * 4C1 ways but there is a flaw here. You selected BlackA, GreenB and RedA.
In another case, you could have selected RedA first, then BlackA and then GreenB. This combination is the same as the previous one but we are counting it separately. The point is we have arranged the cars as first, second and third which we should not do since it is a group.

As I said before, you might want to start with fundamentals from a book.


Thanks Karishma,

That was very helpful indeed.

I will take your advice on the PnC book too.

Thanks
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Re: The Carson family will purchase three used cars. There are [#permalink] New post 27 Dec 2012, 20:43
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enigma123 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?

A. 24
B. 32
C. 48
D. 60
E. 192

Any idea how to solve this guys? I don't have an OA unfortunately.


How many ways to select either Model A or Model B for three cars? 2*2*2 = 8
How many ways to select distinct colors for 3 cars? 4!/3!1! (NOTE: Order doesn't matter) = 4

4*8 = 32

Answer: B
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The Carson family will purchase three used cars. [#permalink] New post 18 Oct 2013, 08:40
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


I don't get this, I study for months, then I take practice exams and it's like I've forgotten everything I've learned.
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Re: The Carson family will purchase three used cars. [#permalink] New post 18 Oct 2013, 08:41
I get an answer of 12. 3C2 (3 cars out of 2)=3, 4C3 (4 colors, 3 cars)=4, 4*3=12. I think the provided answers are all wrong, or maybe I' missign something. I've been drilling this type of program for 6 weeks, and either I haven't learned a thing, or the answers are incorrect.
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Re: The Carson family will purchase three used cars. [#permalink] New post 18 Oct 2013, 08:46
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AccipiterQ 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?


A) 24
B) 32
C) 48
D) 60
E) 192


I don't get this, I study for months, then I take practice exams and it's like I've forgotten everything I've learned.


Merging similar topics. Please refer to the solutions above.
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Re: The Carson family will purchase three used cars.   [#permalink] 18 Oct 2013, 08:46
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