How many even, three-digit positive integers have the property that two of the digits are equal, but not all three are equal?
To solve this problem, we can break it down into cases:
Case 1: The two equal digits are the first two digits (XX_):
In this case, we have 9 options for the first digit (1 to 9, excluding 0) and 10 options for the third digit since it can be any digit from 0 to 9. So, there are a total of 9 * 10 = 90 possibilities.
Case 2: The two equal digits are the last two digits (_XX):
Similar to Case 1, we have 10 options for the first digit and 9 options for the second digit. Therefore, there are a total of 10 * 9 = 90 possibilities.
Case 3: The two equal digits are the first and last digits (X_X):
In this case, we have 9 options for the first and last digits (1 to 9, excluding 0) and 10 options for the second digit. So, there are a total of 9 * 10 = 90 possibilities.
Adding up the possibilities from all three cases, we get:
90 + 90 + 90 = 270
Therefore, there are 270 three-digit positive integers that have the property of having two equal digits, but not all three equal.