Let the 3-digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.
As the number has to even, the digits possible at C are 2 or 4 or 6. That is number of possible digits at C is 3.
Now, as the repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6.
Therefore, The total number possible 3 digit numbers = 6 × 6 × 3 = 108.
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