You can think of the number this way:
Hundreds digit = {8 or 9}
Tens digit = {anything}
Units digit = {1, 3, 5, 7, or 9}
Probably the easiest way to build it is the way described above, where you pick a hundreds digit, then work with the other numbers accordingly.
For instance, suppose the 100s digit is 8. We have FIVE choices for the 1s digit {1, 3, 5, 7, 9}. For simplicity's sake, suppose we pick 3. Now our 10s digit can be anything OTHER THAN 8 or 3; in other words, we're choosing from {0, 1, 2, 4, 5, 6, 7, 9}, for a total of EIGHT choices.
Since we have ONE choice for the 100s, EIGHT for the 10s, and FIVE for the 1s, we have 1 * 8 * 5 = 40.
Now we'll have our 100s digit be 9. We have FOUR choices for the 1s digit {1, 3, 5, 7}, since we can't use 9 as the 100s and the 1s digit. As before, we have EIGHT choices for the 10s digit, so we now have 1 * 8 * 4 = 32 choices.
In total this gives us 72.
Another way would be to calculate ALL the possibilities, then to eliminate those that have repeating digits.
Hundreds digit = TWO choices {8,9}
Tens digit = TEN choices {any digit}
Units digit = FIVE choices {any odd}
So we start with 2 * 10 * 5 = 100 possibilities.
Now we eliminate the following cases:
Any number that goes 88Odd (e.g. 881, 883, etc.)
Any number that goes 99Odd (e.g. 991, 993, etc.)
Any number that ends in the same two odds {e.g. 811, 911, etc.}
Any number of the form 9_9 {e.g. 909, 919, etc.)
There will be FIVE 88Odds, FIVE 99Odds, TWO numbers ending 11, TWO numbers ending 33, TWO numbers ending 55, TWO numbers ending 77, and ONE number ending 99 (only 899, since we already counted 999 in the 99Odd). There will be NINE numbers of the form 9_9 (again, we don't count 999, since we already counted it.)
So we subtract 5+5+2+2+2+2+1+9 = 28 from our total, giving us 100 - 28 = 72.
