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combination problem

Problem Solving — algebra and arithmetic
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by sukh » Mon Aug 08, 2011 6:15 am
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
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by Tani » Mon Aug 08, 2011 9:33 am
Since order doesn't matter you will use combinations. The simplest way to allow for a single color is to think of a single color designation as a color plus no color. Then you can consider "no color" as an additional color.

You can then analyze the number of possibilities with 5 colors by using the combination formula for 6 (5 colors plus no color)

6!/(2!*4!) = 15. Enough - the answer is 5 colors.

Looked at another way, combinations of two colors starting with 5 will give you 5!/(2! * 3!) = 10. That's ten combinations of two colors. You can also use any one of the five colors by itself - giving you again 15 possibilities.
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by sukh » Tue Aug 09, 2011 5:05 am
answer is 5 , i didnt get it - calculated 15 , how does it get reduced to 5
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by Tani » Tue Aug 09, 2011 7:55 am
5 colors will give you 15 alternatives. That is the smallest number you need to have at least 12 combinations available.
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