Hi swerve,

We're told that an employee identification code consists of a vowel followed by 3-digit number greater than 200 and EXACTLY 2 of the 3 digits in the code should be IDENTICAL. We're asked for the number of different codes that are possible.

To start, there are 5 vowels in the English language (A, E, I, O and U), so the answer must be a multiple of 5.

Next, we have to determine the number of values that will fit the rules the rules that the number be greater than 200 AND have 2 digits that are the same. As an example, let's consider just the numbers in the 900s. There are 3 possible outcomes that would fit what we're looking for:
-1st and 2nd digits the same; 3rd is different (re: 99X) - there are 9 possible numbers that fit this pattern (990, 991, 992.... but NOT 999)
-1st and 3rd digits the same; 2nd is different (re: 9X9) - there are 9 possible numbers that fit this pattern (909, 919, 929.... but NOT 999)
-2nd and 3rd digits the same; 1st is different (re: 9XX) - there are 9 possible numbers that fit this pattern (900, 911, 922.... but NOT 999)
Total numbers that start with '9' = 9 + 9 + 9 = 27

This pattern repeats for the numbers that start with 8, 7, 6, 5, 4 and 3. With the number 2 though, the option "200" does NOT fit (remember that the number has to be GREATER than 200), so there are 27 - 1 = 26 possible numbers that start with '2'

Total numbers = 27(7) + 26 = 189 + 26 = 215

Total possible codes = (5)(215) = 1075

Final Answer: C

GMAT assassins aren't born, they're made,
Rich