Question -
Answer -
Let us assume we havethree boxes.
The first box can befilled with any one of the nine digits (0 not allowed at first place).
So, the possibilitiesare 9C1
The second box can befilled with any one of the nine digits
So the availablepossibilities are 9C1
The third box can befilled with any one of the eight digits
So the availablepossibilities are 8C1
Hence, the totalnumber of possible outcomes are 9C1 × 9C1 × 8C1 =9 × 9 × 8 = 648.