AIME 1993 P1

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Problem Statement

How many even integers between 4000 and 7000 have four different digits?

Problem Link

Solution

$$\text{Case 1: Number starts with 5}$$ $$\text{There are 5 options for the last digit, a number from {0, 2, 4, 6, 8}}$$ $$\text{There are then P(8, 2) options for the last two digits}$$ $$\therefore 5 \times 8 \times 7 = 280 \text{ possible numbers in this case.}$$ $$\text{Case 2: Number starts with 4}$$ $$\text{There are 4 options for the last digit, a number from {0, 2, 6, 8}}$$ $$\text{There are then P(8, 2) options for the last two digits}$$ $$\therefore 4 \times 8 \times 7 = 224 \text{ possible numbers in this case.}$$ $$\text{Case 3: Number starts with 6}$$ $$\text{There are 4 options for the last digit, a number from {0, 2, 4, 8}}$$ $$\text{There are then P(8, 2) options for the last two digits}$$ $$\therefore 4 \times 8 \times 7 = 224 \text{ possible numbers in this case.}$$ $$\therefore \text{ there are } 280 + 224 + 224 \text{ even integers between 4000 and 7000 that have four different digits.}$$