For the first part of the ID number (consisting of digits) we have the following number of permutations
The first part 9 takes into consideration that the first digit is non-zero, so we only have 1-9 to choose from, while the other succeeding 5 digits we have 10 to choose from (0 - 9).
The second part of the ID number (two letter part) will have the number of permutations
First multiplied by 5, and then 4, since we cannot repeat the previous letter.
Therefore the number of different student numbers is solved as
18,000,000 possible student ID numbers.