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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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?

Is there any other way to solve this question?

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

We need to be able to create AT LEAST 12 codes (to represent the 12 clients).

Let's test the options, starting with the smallest value....answer choice E

So, can we get 12 or more color codes with 5 colors?
Let's see . . .
1-color codes = 5 (since there are 5 colors)
2-color codes = We need to choose 2 colors from 5. This can be accomplished in 5C2 ways (using combinations). 5C2 = 10

So, using 5 colors, the total number of color codes we can create = 5 + 10 = 15
Perfect!

The answer is 5 (E)

RELATED VIDEO (calculating combinations, like 5C2, in your head)

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