An employee identification code consists of a vowel followed

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An employee identification code consists of a vowel followed by 3-digit number greater than 200. Exactly 2 of the 3 digits in the code should be identical. How many different codes is it possible to form?

A. 211
B. 216
C. 1075
D. 1080
E. 2160

The OA is C.

Please, can someone explain this PS question? I can't get the correct answer. I need help. Thanks.

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by [email protected] » Mon Jun 04, 2018 9:45 am
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

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