For the first part of the ID number (consisting of digits) we have the following number of permutations
![9\cdot10\cdot10\cdot10\cdot10\cdot10](https://img.qammunity.org/2023/formulas/mathematics/college/eqr0i8hm6l6hd7f5vpplrv8qw7zyf0pqeu.png)
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
![5\cdot4](https://img.qammunity.org/2023/formulas/mathematics/college/ga32u50bbyhqgo8t975l91bfrizrbrpwk0.png)
First multiplied by 5, and then 4, since we cannot repeat the previous letter.
Therefore the number of different student numbers is solved as
![9\cdot10\cdot10\cdot10\cdot10\cdot10\cdot5\cdot4=18,000,000](https://img.qammunity.org/2023/formulas/mathematics/college/qbqpn8pvalk02fdpla8kzboel233xc8txg.png)
18,000,000 possible student ID numbers.