Question #ce2b6

3-digit numbers
They MUST start with 5 or 6 to be greater than 500.
This leaves 3 to choose from for the second digit.
And 2 to choose from for the third digit.
Number of possibilities: #2*3*2=12#

4-digit numbers
We choose from 4 for the first digit
From 3 for the second digit
From 2 for the third digit
And then there is only one left for the last.
Number of possibilities: #4*3*2*1=24#

Total number of possibilities (because it's EITHER 3-digit OR 4-digit) you may add the above:

#12+24=36# possibilities.

There are 2 cases to consider - 3 digit numbers and 4 digit
numbers.

3 digit numbers must start with 5 or 6 in order to be bigger than 500.
If they start with 5, then the next digit must be a 6 if we intend to use the 6, since the number cannot end on a 6 else it won't be odd. The last digit can then be 2 choices. So the total number by the multiplication principle is #1xx1xx2=2#
However, we need not use the 6 and can simply use the 1 and 3 instead so this leaves an additional #2# ways.
If the 3 digit number starts with a 6, then there are 3 choices for the next digit, and hence 2 choices for the last digit since digits may not be repeated. Hence #1xx3xx2=6# possibilities.

4 digit numbers may start with any digit, but may not end on a 6 else the number will not be odd.
Hence there are only 3 options for the last digit, and for each of these 3 options, there are 3 options for the second last, then 2 and then 1 option for the first digit, hence by the multiplication principle, a total of #1xx2xx3xx3=18# possible 4 digit numbers satisfying the constraints.

Now putting everything together with the addition principle since the different cases considered are mutually exclusive, the overall total possible numbers satisfying all the given constraints is
#2+2+6+18=28#