Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Nov 2015, 18:17

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The Carson family will purchase three used cars. There are

Author Message
TAGS:
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
GMAT 1: 730 Q49 V40
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 44

Kudos [?]: 1622 [2] , given: 217

The Carson family will purchase three used cars. There are [#permalink]  11 Mar 2012, 01:03
2
KUDOS
10
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

42% (02:17) correct 58% (01:13) wrong based on 415 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

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Last edited by Bunuel on 13 Jul 2013, 06:51, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 30376
Followers: 5088

Kudos [?]: 57269 [6] , given: 8811

Re: The Carson family will purchase three used cars. There are [#permalink]  11 Mar 2012, 01:18
6
KUDOS
Expert's post
8
This post was
BOOKMARKED
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.

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.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6062
Location: Pune, India
Followers: 1596

Kudos [?]: 8936 [6] , given: 195

Re: The Carson family will purchase three used cars. There are [#permalink]  09 May 2012, 08:26
6
KUDOS
Expert's post
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/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 30376 Followers: 5088 Kudos [?]: 57269 [2] , given: 8811 Re: The Carson family will purchase three used cars. There are [#permalink] 26 Mar 2012, 23:41 2 This post received KUDOS Expert's post 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. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1596 Kudos [?]: 8936 [2] , given: 195 Re: The Carson family will purchase three used cars. There are [#permalink] 26 Oct 2015, 21:34 2 This post received KUDOS Expert's post davesinger786 wrote: VeritasPrepKarishma wrote: 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 Hi Karishma, Need some help here. I tried the method given above as well.But I'm confused regarding this.If I'm selecting 2 cars of Model A in 4c2 ways and correspondingly I select I car from Model B in 2c1 ways(2 colors removed because already selected in A) then why are not taking the cases other way around i.e 2 cars of Model B in 4c2 and 1 car of B in 2c1.After all,they are all different cars.Won't that be added to the combinations as well. In that case, the number of combinations would become (12+12+4)*2=56 ways..Kindly put your thoughts here. Right, so 4C2 * 2C1 = 12 (2 cars of model A and 1 car of model B) From where do we get the other 12? It represents the case where we select 2 cars of model B and 1 car of model A. So these are already accounted for. Next, you select all from A - In 4C3 = 4 ways and then select all from B - In 4C3 = 4 ways Total = 12+12+4+4 = 32 ways _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Math Expert
Joined: 02 Sep 2009
Posts: 30376
Followers: 5088

Kudos [?]: 57269 [1] , given: 8811

Re: The Carson family will purchase three used cars. There are [#permalink]  09 May 2012, 07:29
1
KUDOS
Expert's post
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
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6062
Location: Pune, India
Followers: 1596

Kudos [?]: 8936 [1] , given: 195

Re: The Carson family will purchase three used cars. There are [#permalink]  26 Dec 2012, 23:33
1
KUDOS
Expert's post
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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1596 Kudos [?]: 8936 [1] , given: 195 Re: The Carson family will purchase three used cars. There are [#permalink] 27 Dec 2012, 00:28 1 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 18

Kudos [?]: 309 [1] , given: 11

Re: The Carson family will purchase three used cars. There are [#permalink]  27 Dec 2012, 20:43
1
KUDOS
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

_________________

Impossible is nothing to God.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6062
Location: Pune, India
Followers: 1596

Kudos [?]: 8936 [1] , given: 195

Re: The Carson family will purchase three used cars. There are [#permalink]  03 Jun 2014, 20:21
1
KUDOS
Expert's post
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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 13 Jun 2013 Posts: 279 Followers: 13 Kudos [?]: 227 [1] , given: 13 Re: The Carson family will purchase three used cars. There are two models [#permalink] 08 Nov 2014, 09:27 1 This post received KUDOS Avisek47 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 Case 1) all the three selected cars are from model A =4C3=4 Case 2) all the three selected cars are from model B= 4C3=4 Case 3) 1 car from model A and 2 car from model B = 4C1*3C2(because all three cars must have different colors)=12 Case 4) 1 car from model B and 2 car from model A = 4C1*3C2=12 thus total number of combinations= 4+4+12+12=32 Manager Joined: 27 Oct 2011 Posts: 191 Location: United States Concentration: Finance, Strategy GMAT 1: Q V GPA: 3.7 WE: Account Management (Consumer Products) Followers: 4 Kudos [?]: 90 [0], given: 4 Re: The Carson family will purchase three used cars. There are [#permalink] 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? _________________ DETERMINED TO BREAK 700!!! Intern Joined: 04 Mar 2012 Posts: 46 Followers: 0 Kudos [?]: 117 [0], given: 10 Re: The Carson family will purchase three used cars. There are [#permalink] 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! Math Expert Joined: 02 Sep 2009 Posts: 30376 Followers: 5088 Kudos [?]: 57269 [0], given: 8811 Re: The Carson family will purchase three used cars. There are [#permalink] 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. _________________ Manager Joined: 28 Sep 2011 Posts: 70 Location: United States GMAT 1: 520 Q34 V27 GMAT 2: 550 Q V GMAT 3: 690 Q47 V38 GPA: 3.01 WE: Information Technology (Commercial Banking) Followers: 1 Kudos [?]: 21 [0], given: 10 Re: The Carson family will purchase three used cars. There are [#permalink] 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? Manager Joined: 24 Mar 2010 Posts: 81 Followers: 0 Kudos [?]: 35 [0], given: 134 Re: The Carson family will purchase three used cars. There are [#permalink] 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? _________________ - Stay Hungry, stay Foolish - Manager Joined: 24 Mar 2010 Posts: 81 Followers: 0 Kudos [?]: 35 [0], given: 134 Re: The Carson family will purchase three used cars. There are [#permalink] 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 _________________ - Stay Hungry, stay Foolish - Current Student Joined: 26 Sep 2013 Posts: 221 Concentration: Finance, Economics GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 Followers: 3 Kudos [?]: 82 [0], given: 40 Re: The Carson family will purchase three used cars. [#permalink] 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. Senior Manager Joined: 15 Aug 2013 Posts: 328 Followers: 0 Kudos [?]: 28 [0], given: 23 Re: The Carson family will purchase three used cars. There are [#permalink] 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1596 Kudos [?]: 8936 [0], given: 195 Re: The Carson family will purchase three used cars. There are [#permalink] 18 May 2014, 20:33 Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Re: The Carson family will purchase three used cars. There are   [#permalink] 18 May 2014, 20:33

Go to page    1   2    Next  [ 32 posts ]

Similar topics Replies Last post
Similar
Topics:
1 Among the members of the Malmo family, there are three times 2 24 Mar 2014, 03:44
6 The Carson family will purchase three used cars. There are 6 22 Jun 2011, 17:57
6 What is the probability for a family with three children to 6 06 Feb 2011, 01:20
11 In May, Xiang sold 15 used cars. For these 15 cars, the 3 17 Jul 2010, 10:18
37 A used car dealer sold one car at a profit of 25 percent of 12 23 Jul 2008, 21:15
Display posts from previous: Sort by