Casework Counting...again

Here are the permutations of 3-digit numbers beginning with "1". You have a total of 100 permutations and it appears that 10 EVEN combinations have a repeated digit. Since, there: 9 x 10 x 5 =450 EVEN 3-digit numbers between 100 - 999, and since there are 9 numbers beginning with "1", then I would generalize that there are: 10 x 9 = 90 such EVEN numbers with repeated digits.

{1, 0, 0} | {1, 0, 1} | {1, 0, 2} | {1, 0, 3} | {1, 0, 4} | {1, 0, 5} | {1, 0, 6} | {1, 0, 7} | {1, 0, 8} | {1, 0, 9} | {1, 1, 0} | {1, 1, 1} | {1, 1, 2} | {1, 1, 3} | {1, 1, 4} | {1, 1, 5} | {1, 1, 6} | {1, 1, 7} | {1, 1, 8} | {1, 1, 9} | {1, 2, 0} | {1, 2, 1} | {1, 2, 2} | {1, 2, 3} | {1, 2, 4} | {1, 2, 5} | {1, 2, 6} | {1, 2, 7} | {1, 2, 8} | {1, 2, 9} | {1, 3, 0} | {1, 3, 1} | {1, 3, 2} | {1, 3, 3} | {1, 3, 4} | {1, 3, 5} | {1, 3, 6} | {1, 3, 7} | {1, 3, 8} | {1, 3, 9} | {1, 4, 0} | {1, 4, 1} | {1, 4, 2} | {1, 4, 3} | {1, 4, 4} | {1, 4, 5} | {1, 4, 6} | {1, 4, 7} | {1, 4, 8} | {1, 4, 9} | {1, 5, 0} | {1, 5, 1} | {1, 5, 2} | {1, 5, 3} | {1, 5, 4} | {1, 5, 5} | {1, 5, 6} | {1, 5, 7} | {1, 5, 8} | {1, 5, 9} | {1, 6, 0} | {1, 6, 1} | {1, 6, 2} | {1, 6, 3} | {1, 6, 4} | {1, 6, 5} | {1, 6, 6} | {1, 6, 7} | {1, 6, 8} | {1, 6, 9} | {1, 7, 0} | {1, 7, 1} | {1, 7, 2} | {1, 7, 3} | {1, 7, 4} | {1, 7, 5} | {1, 7, 6} | {1, 7, 7} | {1, 7, 8} | {1, 7, 9} | {1, 8, 0} | {1, 8, 1} | {1, 8, 2} | {1, 8, 3} | {1, 8, 4} | {1, 8, 5} | {1, 8, 6} | {1, 8, 7} | {1, 8, 8} | {1, 8, 9} | {1, 9, 0} | {1, 9, 1} | {1, 9, 2} | {1, 9, 3} | {1, 9, 4} | {1, 9, 5} | {1, 9, 6} | {1, 9, 7} | {1, 9, 8} | {1, 9, 9}.

Sorry, but your answer was incorrect...

Luckily, I used a more efficient version of your method to get an answer of 118...

If the units digit is a 0, then the only two cases are that the 0 is the repeated digit, in which case the tens digit must be 0 and the hundreds digit can be 1 through 9, or the first two digits are the repeated ones, of which there are also 9 choices. Now, if the units digit is any even digit of the set \(\{ 2, 4, 6, 8 \}\), then the situation is different. If the repeated digit is to be the units digit, it can be repeated in the hundreds digit, in which case there are 9 choices of tens digit, or it can be repeated in the tens digit, in which case there are 8 choices of hundreds digit. Or, if the first two digits are the repeated ones, they can be any of 1 through 9 excluding the even units digit, for a total of 8. So, for each of 2, 4, 6 and 8, there are \(9 + 8 + 8 = 25\). So, the total number is \(2 \cdot 9 + 4 \cdot 25 = \boxed{118}\).

Thanks for your effort though!

Awe, thanks GingerAle, I certainly tried my best on that solution.

Good Job!!!!.