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m04q29 - color coding : Retired Discussions [Locked]

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m04q29 - color coding [#permalink]
10 Sep 2008, 17:53

3

This post was BOOKMARKED

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

1

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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

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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