Solution:
Let us assume we have three boxes.
The first box can be filled with any one of the nine digits (zero not allowed at first position)
So, possibilities are 9C1
The second box can be filled with any one of the ten digits
So the available possibilities are 10C1
Third box can be filled with any one of the ten digits
So the available possibilities are 10C1
Hence, the total number of possible outcomes are 9C1 × 10C1 × 10C1 = 9 × 10 × 10 = 900.