Solution:
(i) We know that the initial digit of the license plates cannot be zero. It is also forbidden to repeat digits.
Let’s say there are five boxes. The first box can be filled with any of the nine digits, thus 9C1 is an option.
Similarly, the second box can be filled with any of the nine digits available, so 9C1 is an option.
The third box can be filled with any of the eight digits available, therefore 8C1 is a possibility.
The fourth box can be filled with any of the seven digits available, therefore 7C1 is a possibility.
The fifth box can be filled with any of the six digits available, therefore 6C1 is an option.
As a result, the total number of outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 = 9 × 9 × 8 × 7 × 6 = 27,216
(ii) We know that zero cannot be the first digit of the license plates. And the repetition of digits is allowed.
Let us assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is 9C1
Similarly, the second box can be filled with one of the ten available digits, so the possibility is 10C1
The third box can be filled with one of the ten available digits, so the possibility is 10C1
The fourth box can be filled with one of the ten available digits, so the possibility is 10C1
The fifth box can be filled with one of the ten available digits, so the possibility is 10C1
Hence, the total number of possible outcomes is 9C1 × 10C1 × 10C1 × 10C1 × 10C1 = 9 × 10 × 10 × 10 × 10 = 90,000