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2014/15 AHSMC Part 1 Problem 10

Problem 10 of 16

The number of five-digit positive integers such that the digits are alternately odd and even (either odd-even-odd-even-odd or even-odd-even-odd-even) is

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Solution

For the odd-even-odd-even-odd pattern, we have 555^5 such numbers. For the even-odd-even-odd-even pattern, we have 4×544 \times 5^4 such numbers. The total is 9×549 \times 5^4. The answer is (D).

University of Alberta