|
Author |
Message |
|
TAGS:
|
|
|
Director
Status: Preparing for the 4th time -:(
Joined: 25 Jun 2011
Posts: 558
Location: United Kingdom
Concentration: International Business, Strategy
GMAT Date: 06-22-2012
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 8
Kudos [?]:
63
[0], given: 212
|
The Carson family will purchase three used cars. There are [#permalink]
 11 Mar 2012, 02:03
Question Stats:
75% (01:39) correct
25% (01:50) wrong based on 0 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 Any idea how to solve this guys? I don't have an OA unfortunately.
_________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610 
|
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11571
Followers: 1797
Kudos [?]:
9581
[2] , given: 826
|
Re: The Carson family will purchase three used cars. There are [#permalink]
11 Mar 2012, 02:18
2
This post received
KUDOS
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11571
Followers: 1797
Kudos [?]:
9581
[1] , given: 826
|
Re: The Carson family will purchase three used cars. There are [#permalink]
27 Mar 2012, 00:41
1
This post received
KUDOS
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.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3111
Location: Pune, India
Followers: 572
Kudos [?]:
2011
[1] , given: 92
|
Re: The Carson family will purchase three used cars. There are [#permalink]
27 Dec 2012, 01:28
1
This post received
KUDOS
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
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Senior Manager
Joined: 13 Aug 2012
Posts: 468
Followers: 12
Kudos [?]:
75
[1] , given: 11
|
Re: The Carson family will purchase three used cars. There are [#permalink]
27 Dec 2012, 21:43
1
This post received
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 Answer: B
|
|
|
|
|
|
Manager
Joined: 27 Oct 2011
Posts: 196
Location: United States
Concentration: Finance, Strategy
GMAT 1: Q V
GPA: 3.7
WE: Account Management (Consumer Products)
Followers: 1
Kudos [?]:
25
[0], given: 4
|
Re: The Carson family will purchase three used cars. There are [#permalink]
26 Mar 2012, 20:31
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: 48
Followers: 0
Kudos [?]:
6
[0], given: 10
|
Re: The Carson family will purchase three used cars. There are [#permalink]
07 May 2012, 21:46
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!
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11571
Followers: 1797
Kudos [?]:
9581
[0], given: 826
|
Re: The Carson family will purchase three used cars. There are [#permalink]
08 May 2012, 00:44
|
|
|
|
|
|
Manager
Joined: 28 Sep 2011
Posts: 64
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 [?]:
10
[0], given: 5
|
Re: The Carson family will purchase three used cars. There are [#permalink]
09 May 2012, 08:21
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?
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11571
Followers: 1797
Kudos [?]:
9581
[0], given: 826
|
Re: The Carson family will purchase three used cars. There are [#permalink]
09 May 2012, 08:29
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3111
Location: Pune, India
Followers: 572
Kudos [?]:
2011
[0], given: 92
|
Re: The Carson family will purchase three used cars. There are [#permalink]
09 May 2012, 09:26
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
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Manager
Joined: 24 Mar 2010
Posts: 82
Followers: 0
Kudos [?]:
4
[0], given: 133
|
Re: The Carson family will purchase three used cars. There are [#permalink]
26 Dec 2012, 14:28
Bunuel, Karishma, You gotta help me out here, m getting shaky on all my Combinatorics concepts. I approached the problem as 8C8 * 7C6 * 4C1What exactly m I doing wrong?
_________________
- Stay Hungry, stay Foolish -
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3111
Location: Pune, India
Followers: 572
Kudos [?]:
2011
[0], given: 92
|
Re: The Carson family will purchase three used cars. There are [#permalink]
27 Dec 2012, 00:33
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
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Manager
Joined: 24 Mar 2010
Posts: 82
Followers: 0
Kudos [?]:
4
[0], given: 133
|
Re: The Carson family will purchase three used cars. There are [#permalink]
27 Dec 2012, 00:51
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 -
|
|
|
|
|
|
Manager
Joined: 24 Mar 2010
Posts: 82
Followers: 0
Kudos [?]:
4
[0], given: 133
|
Re: The Carson family will purchase three used cars. There are [#permalink]
27 Dec 2012, 01:35
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
_________________
- Stay Hungry, stay Foolish -
|
|
|
|
|
|
|
Re: The Carson family will purchase three used cars. There are
[#permalink]
27 Dec 2012, 01:35
|
|
|
|
|
|
|
|
|
|
|
close
