This problem involves finding the total count of three-digit numbers that meet specific criteria: they must be even, and they must be formed using the digits {1, 2, 3, 4, 5, 6}. Crucially, the digits can be repeated.
We need to form a three-digit number. Let's represent the positions as Hundreds, Tens, and Units.
We need to determine the number of possible choices for each of the three positions (hundreds, tens, and units) based on the given constraints.
The number must be even. The available digits are {1, 2, 3, 4, 5, 6}. The even digits in this set are {2, 4, 6}. Therefore, there are 3 possible choices for the units place.
Since repetition is allowed, any of the available 6 digits ({1, 2, 3, 4, 5, 6}) can be placed in the hundreds position. Thus, there are 6 possible choices.
Similarly, because repetition is allowed, any of the 6 available digits ({1, 2, 3, 4, 5, 6}) can be used for the tens place. This gives us 6 possible choices.
To find the total number of three-digit even numbers that can be formed, we apply the fundamental principle of counting (multiplication principle). We multiply the number of choices for each position:
Total Numbers = (Choices for Hundreds Place) $\times$ (Choices for Tens Place) $\times$ (Choices for Units Place)
Using LaTeX for the calculation:
$$ \text{Total Numbers} = 6 \times 6 \times 3 $$
Performing the multiplication:
$$ 6 \times 6 = 36 $$
$$ 36 \times 3 = 108 $$
Therefore, there are 108 possible three-digit even numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 when repetition of digits is allowed.
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