By a 3 digit number, we will assume that "zero" can appear as the first digit.
So, 012 would be acceptable correct?
If that's the case, the first place digit can be chosen from 10 digits.
The next digit can be anything except the first digit so that gives us 9 choices.
The thirds digit gives us 8 choices.
So ALL possible combinations equals
10 * 9 * 8 = 720
The probability that all 3 digits are different = .720