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

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

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26 Aug 2008, 17:27
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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.

* 24
* 12
* 7
* 6
* 5
Senior Manager
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Re: Color coding - PS [#permalink]

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26 Aug 2008, 17:39
5 :

let a,b,c,d,e

a ab b c ac bc d ad bd cd e ea ...............
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Re: Color coding - PS [#permalink]

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26 Aug 2008, 18:18
chan4312 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

I think 4 colors would be sufficient
With 2 colors, he can code 2P1 + 2P2 = 4 clients. If a,b are colors then a, b, ab, ba.
With 3 colors 3P1 + 3P2 = 9 clients
With 4 colors 4P1 + 4P2 = 16 clients.

So 4 should be the answer.

If order of color does not matter (which is nonsensical for color code) then 5 colors needed
with 2 colors 2C1 + 2C2 = 2
with 3 colors 3C1 + 3C2 = 6
with 4 colors 4C1 + 4C2 = 10
with 5 colors 5C1 + 5C2 = 15
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Re: Color coding - PS [#permalink]

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26 Aug 2008, 18:32
chan4312 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

You are looking for the number of combinations of one or two colors that sums to at least 12. The giveaway is the "either/or" clause "either a single color or a pair of two" in the stem.

That would be xC1 + xC2 > 12, and x=5 fits the bill.

5C1 + 5C2 > 12 --> 5 + 10 > 12.

cP
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Manager
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Re: Color coding - PS [#permalink]

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26 Aug 2008, 21:38
Thanks buddies
OA is E
Re: Color coding - PS   [#permalink] 26 Aug 2008, 21:38
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