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# John has 12 clients and he wants to use color coding to iden

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John has 12 clients and he wants to use color coding to iden [#permalink]

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03 Jan 2011, 18:02
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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
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04 Jan 2011, 01:28
I started with the smallest answer of all and worked my way up or so I thought

5 different colours = 5 single colours = 5 different clients

There are 7 clients to go

The use of the factorial is the easiest way to solve this kind of problem for me.
A B C D E (different colour)
C C N N N (2 colour coding + 3 colour non chosen)

5!/2!3!= 10

15 codings can be done with 5 different colours

Ans E
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04 Jan 2011, 03:16
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rtaha2412 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

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) $$n\geq{5}$$ --> $$n_{min}=5$$.

Hope it's clear.
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Re: John has 12 clients and he wants to use color coding to iden [#permalink]

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28 Nov 2014, 21:29
Bunuel,how did you calculate n(n+1)>=24. Is there any other way than putting in values and validating the equation to get the correct answer?
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Re: John has 12 clients and he wants to use color coding to iden [#permalink]

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01 Dec 2014, 05:52
Ralphcuisak wrote:
Bunuel,how did you calculate n(n+1)>=24. Is there any other way than putting in values and validating the equation to get the correct answer?

Since n must be an integer solving by number plugging is the best approach for n(n+1)>=24.

Check Constructing Numbers, Codes and Passwords problems for practice.

Hope it helps.
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01 Aug 2015, 08:52
Is there any other way to solve this question?
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John has 12 clients and he wants to use color coding to iden [#permalink]

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01 Aug 2015, 09:00
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SQUINGEL wrote:
Is there any other way to solve this question?

Well 'counting' is the only method applicable for this question.

As you are being asked to find the 'minimum' value, use the options to guide you.

Options A and B are out because of obvious reasons.

Start with E, Lets say you have 5 colors. Out of these 5 colors, look at how many 2 color combinations you can create = 5C2 = 10 and remaining 2 can be single colors. So there you go, you have your answer. An answer that will give you possible number of combinations $$\geq$$ 12 will be the answer as you need to cover all 12 clients uniquely.

If lets say the total number of clients would have been = 23, then with n = 5, you could at most have = 5C2 + 5 = 15 (<23) different ways, with n =6 you could have 6C2 + 6 = 21 different ways (<23). Thus n = 7 would have been the minimum number of colors.

Hope this helps.
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Re: John has 12 clients and he wants to use color coding to iden [#permalink]

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05 Aug 2015, 01:34
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.

We can backsolve using the answer choices and the formula n!/(r! (n-r)!) and add the number of colors we are using (to account for the single color codes) to get the total number of possible codes. As we're asked for the minimum, we can start with 5.

n=5 (five colors) and r=2 (because we're making paired color codes)
(5!)/(2! (5-2)!) = 10 possible paired color codes
10 paired possible color codes + 5 single codes for each color used = 15 possible codes. This is the minimum.
A. 24
B.12
C. 7
D. 6
E. 5
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Re: John has 12 clients and he wants to use color coding to iden   [#permalink] 05 Aug 2015, 01:34
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