Official Guide Explanation:
Problem Solving #D11




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Solution and Metadata

Question: D11
Page: 22
Difficulty: 6 (Moderately Difficult)
Category 1: Arithmetic > Counting Methods > Permutations

Explanation: Since the question is limited to three - digit integers, we're looking at numbers between 701 and 999, inclusive. To make things simpler, focus on one set of 100 at a time.

In the 800s, there are three ways to create a number in which two digits are equal but the third is not:

a.    8x8. There are nine such numbers, from 808 to 898, not including 888.

b.    88x. There are, again, nine such numbers, from 880 to 889, not including 888.

c.    8xx. Nine such numbers, from 800 to 899, not including 888.

That gives us 27 possibilities in the 800s. The same logic applies to the 900s, which has 27 as well.

There are 27 in the 700s, as well, but remember that the question specifies "greater than 700." Since 700 is one of the numbers, that reduces the number to 26. Our total, then, is:

26 + 27 + 27 = 80, choice (C).

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