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## Official Guide Explanation:

Problem Solving #132

**Background**

This is just one of hundreds of free explanations I've created to the quantitative questions in The Official Guide for GMAT Quantitative Review (2nd ed.). Click the links on the question number, difficulty level, and categories to find explanations for other problems.

These are the same explanations that are featured in my "Guides to the Official Guide" PDF booklets. However, because of the limitations of HTML and cross-browser compatibility, some mathematical concepts, such as fractions and roots, do not display as clearly online.

Click here for an example of the PDF booklets. Click here to purchase a PDF copy.

**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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