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# m04q29 - color coding

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10 Sep 2008, 17:53
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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

[Reveal] Spoiler: OA
E

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Re: m04 - color coding [#permalink]

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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.

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Re: m04 - color coding [#permalink]

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11 Sep 2008, 11:23
Agree with e

5+5C2 = 15 .. >12
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11 Sep 2008, 17:04
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we need xC2 + x > 12 => x(x+1)/2 + x > 12 => x*(x+1) > 24

so x=5.

IMO E
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Re: m04 - color coding [#permalink]

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11 Sep 2008, 17:47
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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
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Re: m04q29 - color coding [#permalink]

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20 Oct 2010, 05:15

Let say we have 5 color which is A,B,C,D,E

We can code 12 clients as

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.

so minimum 5 color required.
Friends, Let me know if I am wrong.
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Re: m04q29 - color coding [#permalink]

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20 Oct 2010, 09:39
Ans should be E...
here's how I did it :

Lets say the colors are A,B,C,D,E and the clients are numbered from 1-12.

Therefore:
1-A, 2-B,....E-5,
10-BC,...,11-BE,
12-CD...and so on.
Actually 5 colors would do for 15 clients...

My method is a bit rudimentary, I've got to better my PnC PS.

Cheers,
R J
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Re: m04 - color coding [#permalink]

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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.

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?
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Re: m04 - color coding [#permalink]

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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)!)
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Re: m04q29 - color coding [#permalink]

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21 Oct 2010, 23:37
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Given:

Since, order of colors doesn't matter here ...use Combinations.

If, NC1 + NC2 >= 12 then what is N?
$$\frac{N!}{(N-1)!}$$ + $$\frac{N!}{(N-2)! * 2}$$ >= 12
==> $$N^2$$ >= 24
==> N >= 5

Cheers!
Ravi
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Re: m04q29 - color coding [#permalink]

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12 Dec 2010, 19:09
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I started by taking minimum value as 5 (from the answer key)

5 = 5 colors for five clients with single color

Since arrangement doest matter so it will be 5C2 and not 5P2

5C2 = 10

so 5 + 10 = 15 > 12

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Re: m04q29 - color coding [#permalink]

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24 Oct 2011, 07:46
let us take 5 single color codes

then it will be

5+5C2= 15

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Re: m04q29 - color coding [#permalink]

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01 Nov 2011, 05:09
I chose E.

5 single codes
5c2 = 10 pair codes

total 10+5 = 15, which is more than 12.
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Re: m04q29 - color coding [#permalink]

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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
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Re: m04q29 - color coding [#permalink]

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20 Jan 2012, 15:31
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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.

Hope it's clear.
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Re: m04q29 - color coding [#permalink]

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24 Oct 2012, 04:40
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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.

(A) 24
(B) 12
(C) 7
(D) 6
(E) 5

[Reveal] Spoiler: OA
E

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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.

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Re: m04q29 - color coding [#permalink]

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24 Oct 2012, 07:16
1 2 3 4 5
12 13 14 15
23 24 ...........so its enough..Answer 5..
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Re: m04q29 - color coding [#permalink]

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28 Oct 2012, 04:06
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Quick trial and error approach:
3 single colored codes + 3C2 doubled colored codes = 6 //not enough
Next 4:
4 single colored codes + 4C2 doubled colored codes = 10 //not enough
Next 5:
5 single colored codes + 5C2 doubled colored codes = 15 >12 //enough

(E) 5
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Re: m04q29 - color coding [#permalink]

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22 Oct 2013, 04:23
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Re: m04q29 - color coding [#permalink]

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22 Oct 2013, 05:28
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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.
Re: m04q29 - color coding   [#permalink] 22 Oct 2013, 05:28

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