Official Guide Explanation:
Problem Solving #132

Background

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Solution and Metadata

Question: 132
Page: 79
Difficulty: 5 (Moderate)
Category 1: Word Problems > Sets >
Category 2: Arithmetic > Counting Methods > Combinations

Explanation: When you see phrasing such as "the order of the colors in a pair does not matter," recognzie that you're dealing with a combinations problem. To find the number of possible pairs given a certain number of colors, use the combinations formula: ((n!)/(k!(n - k)!)), where n is the total number of colors and k is the size of the subgroup--in this case 2. Since centers can be coded with either a single color or a pair of colors, the number of possible codes that can be generated from a group of n colors is n + ((n!)/(k!(n - k)!)), which includes the number of single - color codes and two - color codes.

It would be extremely arduous to solve that problem by setting that expression equal to 12; much better to use the likely - looking answer choices. 12 and 24 are far too big -- if there are 12 colors, there'd be 12 single - color codes, not to mention all of the two - color codes. So, start in the middle, with n = 5:

n + ((n!)/(k!(n - k)!)) = 5 + ((5!)/(2!(5 - 2)!)) = 5 + ((5(4)(3!))/(2(3!))) = 5 + (20/2) = 5 + 10 = 15. This is possible; the only concern is that n = 4 might also offer enough codes, so we'll have to check n = 4:

n + ((n!)/(k!(n - k)!)) = 4 + ((4!)/(2!(4 - 2)!)) = 4 + ((4(3)(2!))/(2(2!))) = 4 + (12/2) = 4 + 6 = 10. Not big enough, so n = 5 is the smallest possible number. Choice (B) is correct.

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