Question -
Answer -
Let us assume we havethree boxes.
The first box can befilled with any one of the nine digits (zero not allowed at first position)
So, possibilitiesare 9C1
The second box can befilled with any one of the ten digits
So the availablepossibilities are 10C1
Third box can befilled with any one of the ten digits
So the availablepossibilities are 10C1
Hence, the totalnumber of possible outcomes are 9C1 × 10C1 × 10C1 =9 × 10 × 10 = 900.