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The Carson family will purchase three used cars. There are [#permalink]
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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
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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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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]
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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 ?
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Re: The Carson family will purchase three used cars. There are [#permalink]
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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]
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09 May 2012, 07:29
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: iftherearefourdistinctpairsofbrothersandsisters99992.html and this post there: iftherearefourdistinctpairsofbrothersandsisters99992.html#p775925
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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 unarrange (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]
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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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26 Dec 2012, 23:33
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]
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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]
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27 Dec 2012, 00:28
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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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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Re: The Carson family will purchase three used cars. [#permalink]
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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. There are [#permalink]
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17 May 2014, 12:43
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. Hi Karishma, I made a similar error in the beginning and looking back at it, it might have been because I might have tried to mix in "probability" into the combinatorics problem. If you treat it as a probability, wouldn't the above posters approach be correct?



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Re: The Carson family will purchase three used cars. There are [#permalink]
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18 May 2014, 20:33
russ9 wrote: 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. Hi Karishma, I made a similar error in the beginning and looking back at it, it might have been because I might have tried to mix in "probability" into the combinatorics problem. If you treat it as a probability, wouldn't the above posters approach be correct? The approach of the poster here is quite wrong. You just need the number of cases here so it is not a probability question. If you do want to figure out the correct method in case it were a probability question, give the question you want to discuss and the solution you will use. Then I can tell you whether you are correct.
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Re: The Carson family will purchase three used cars. There are [#permalink]
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03 Jun 2014, 20:21
VeritasPrepKarishma wrote: 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
Quote: My doubt was
Can you please explain why we have multiplied 2*2*2 with 4C3.
what i did to solve this was 4C3 * 2C1 where 2C1 are for selection of model . Can you please let me know what i have missed here.
Thanks
What you did is correct but incomplete. You select 3 colors in 4C3 ways is correct. Say you select blue, black and red. Now you have 3 different colors but for each color you have 2 models available. So you must select a model for EACH color. You do that in 2C1 ways for EACH color. Say model A for blue, model B for black and model A for red. Total selection can be made in 4C3 * 2C1 * 2C1 * 2C1 ways
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27 Jun 2014, 11:51
Hello, I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B> C(4,2)* C(2,1) = 12 > C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them. Secondly, we can select 1 car from A and 2 cars from B> C(4,1) * C(3,2) = 12 Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong. Thank You in advance.



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Re: The Carson family will purchase three used cars. There are [#permalink]
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27 Jun 2014, 22:12
deya wrote: Hello, I have solved the problem in the following way. From the question I came to know that there are two categories of cars of 4 different colors each. So, in total there are 8 cars. We need to select 3 cars of different colors. So, we can do it in two ways. We can select 2 cars from category A and one from B> C(4,2)* C(2,1) = 12 > C(2,1) for B as we have already selected two colors from A so we are left with the rest 2 colors and we have selected 1 from them. Secondly, we can select 1 car from A and 2 cars from B> C(4,1) * C(3,2) = 12 Therefore, total ways of combinations = 12+12 = 24. But, unfortunately this is not the correct answer. Kindly tell me where I went wrong. Thank You in advance. What about selecting all three cars from A or all three cars from B? All three cars can have the same model. The only constraint is that they need to be of different colors. Select all from A  In 4C3 = 4 ways Select all from B  In 4C3 = 4 ways Total = 24+4+4 = 32 ways
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