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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
m04q29 - color coding [#permalink]
10 Sep 2008, 17:53
4
This post was BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
Re: m04 - color coding [#permalink]
10 Sep 2008, 19:01
sarzan wrote:
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
* 24 * 12 * 7 * 6 * 5
start from the least number:
5 colors can represent 5 single colored codes + 5C2=10 two colored codes = 15 color codes which should be suff for 12 clients.
Re: m04 - color coding [#permalink]
11 Sep 2008, 17:47
1
This post received KUDOS
sarzan wrote:
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
* 24 * 12 * 7 * 6 * 5
5c2+5c1 >12 => andmin value IMO E _________________
Re: m04q29 - color coding [#permalink]
20 Oct 2010, 05:15
Answer should be E.
Let say we have 5 color which is A,B,C,D,E
We can code 12 clients as A,B,C,D,E,AB,AC,AD,AE,BC,BD,BE.
so atleast 5 color required. Even the 4 color will not be able to code all 12 client.oNLY 10 CLIENT CAN BE CODED USNING 4 COLOR. A,B,C,D,AB,AC,AD,BC,BD,CD
so minimum 5 color required. Friends, Let me know if I am wrong.
Re: m04 - color coding [#permalink]
21 Oct 2010, 08:47
alpha_plus_gamma wrote:
scthakur wrote:
[quote="alpha_plus_gamma 5 colors can represent 5 single colored codes + 5C3=10 two colored codes = 15 color codes which should be suff for 12 clients.
E should be the answer
Did you mean 5C2 here? Although, the value will remain the same, but just wanted to check.
Yes. sorry for the typo![/quote]
I can get the problem drawing out the color codes but I was wondering if there was a better way. I see that you have 5C2 + 5 = 15. What does the 5C2 stand for? 5 color 2? Is there a specific equation for problems like these so I don't spend so much time drawing it out? _________________
I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!
Re: m04 - color coding [#permalink]
21 Oct 2010, 16:35
sonnco wrote:
alpha_plus_gamma wrote:
I can get the problem drawing out the color codes but I was wondering if there was a better way. I see that you have 5C2 + 5 = 15. What does the 5C2 stand for? 5 color 2? Is there a specific equation for problems like these so I don't spend so much time drawing it out?
5C2 is a "combinations" representation - denoting the number of unique combinations of 2 items you can get from a group of 5 items. The generic format is nCr = n!/(r!*(n-r)!)
Re: m04q29 - color coding [#permalink]
20 Jan 2012, 15:16
I've used my Veritas GMAT prep to understand this new concept but the company is seriously lacking in defining this new concept. Can anyone explain in more detail to someone new to this concept on how to solve this problem. I tried using the combinatoric formula or simple counting method but was lost by the time I tried to solve it.Thanks
Re: m04q29 - color coding [#permalink]
20 Jan 2012, 15:31
3
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
AzWildcat1 wrote:
I've used my Veritas GMAT prep to understand this new concept but the company is seriously lacking in defining this new concept. Can anyone explain in more detail to someone new to this concept on how to solve this problem. I tried using the combinatoric formula or simple counting method but was lost by the time I tried to solve it.Thanks
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes. A. 24 B. 12 C. 7 D. 6 E. 5
The concept is not that hard. We can use combination or trial and error approach.
Combination approach: Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).
Trial and error approach: If the minimum number of colors needed is 4 then there are 4 single color codes possible PLUS \(C^2_4=6\) two-color codes --> 4+6=10<12 --> not enough for 12 codes;
If the minimum number of colors needed is 5 then there are 5 single color codes possible PLUS \(C^2_5=10\) two-color codes --> 5+10=15>12 --> more than enough for 12 codes.
Actually as the least answer choice is 5 then if you tried it first you'd get the correct answer right away.
Re: m04q29 - color coding [#permalink]
24 Oct 2012, 04:40
1
This post received KUDOS
sarzan wrote:
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
If n colors are used, there are n codes with 1 color and n(n-1)/2 codes with 2 colors. First color - n options, second color n - 1 options, which gives n(n - 1) options, then divide by 2 because order in which we choose the two colors doesn't matter. It is obvious that the correct answer should be less than 12. Already for n = 5, we get 5 + 5x4/2 = 15 which is greater than 12.
Therefore, answer E. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: m04q29 - color coding [#permalink]
22 Oct 2013, 05:28
1
This post received KUDOS
E.
I tested answer choices. Started with the smallest answer (E).
5 different colors + _ _ (Slot method for counting....since order matters...you divide by the factorial number of slots where order matters, in this case 2)
= 5 + (5 choices of colors x 4 choices of colors divided by 2!) = 15.
15 different combinations is sufficient for 12 people.
I stopped there as the other choices were all larger.
gmatclubot
Re: m04q29 - color coding
[#permalink]
22 Oct 2013, 05:28