How many choices do we have for the digit a? We have 99 choices (since the first digit cannot be 00). Now for each possible choice of a we have 99 choices for b (since we want b to be a different digit to and we now allow 00).
So for choice of the ab part we have 9∗9=819∗9=81 possibilities.
Now for each of these we have 88 choices for c (to avoid c being the same as either a or b). And for each of these we have 77 choices for d (to avoid d being the same as either a,b or c).
So in total there are 9∗9∗8∗7=45369∗9∗8∗7=4536 possible numbers.
