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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] New post 03 Jan 2011, 17:02
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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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Re: combinations [#permalink] New post 04 Jan 2011, 00:28
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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Re: combinations [#permalink] New post 04 Jan 2011, 02:16
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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.

Answer: E.

Hope it's clear.
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Re: combinations   [#permalink] 04 Jan 2011, 02:16
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John has 12 clients and he wants to use color coding to iden

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