GRE Probability question How many 3-digit positive integers are odd and do not contain the digit 5 ?
My attempt: 100-999 3 -digit integers, 900/2=450 odd numbers. Now how to calculate odd numbers which do not contain digit 5 in it.
 A: Units digit should contain {1,3,7,9} = 4;
Tens digits can be {0,1,2,3,4,6,7,8,9} = 9 
Hundreds digit can be {1,2,3,4,6,7,8,9} = 8.
So the answer is 8$*$9$*$4 = 288
A: The important aspect in these kinds of permutations and combinations questioins is the direction with which you are approaching the problem.
For some questions, starting from the left most digit and then moving towards right may be a good strategy. Ex-- How many numbers are greater than 500 or less than 800 kind of questions, where hundred's digit has a lot of importance.
But for these kind of questions, where units digit is important, it is a better idea to start from the right-most digit.
Note: When repetition is not allowed, the above two strategies play an important role.
As repetition is allowed in this problem, it is a straight forward one.
First units digit : odd-1,3,5,7,9     No five : 4 possibilities
Middle digit      : all 10 values     No five : 9 possibilities
Hundred's digit   : zero not possible- 9 values    8 possibilities
Total = 4 * 9 * 8 = 288
