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To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
As you describe the problem, the condition is you can select any of the 6 numbers with replacement since you have 3 sets of numbers for each value. Per the related link, you have N = 6 and n = 3. The number of combinations is Nn, or 63, or 216. If it is easier to think of the problem in this way, here is another way to look at it. You have 6 choices for the first digit, 6 choices for the second digit, and 6 choices for the third digit. So, you have 6*6*6 = 216 combinations.
Assuming that leading zeros are not permitted (that is the lowest ten digit number is 1 000 000 000), then there are 4 500 000 000 possible even numbers
If order is important and repetition is allowed, then : 000 through 999 (1000 possible). If no leading zeros, then start at 100 through 999 (900 possible). See related link, if you need to limit to no repetition, or order doesn't matter (like if 123 is the same as 231).