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.