2009/10 AHSMC Part 1 Problem 7

Problem 7 of 16

The number of arrangements of the letters AABBCC in a row such that no two identical letters are adjacent is

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Solution

Assume that the first A appears before the first B, and the first B before the first C. Then we must start with AB and continue with A or C. If we continue with A, the last three letters must be CBC. If we start with ABC, we must continue with A or B. In either case, either of the last two letters can appear before the other. So the total is $1 + 2 \times 2 = 5$ . Relaxing the order of appearance, the total becomes $5 \times 3! = 30$ . The answer is (A).

University of Alberta

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