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┬а9C1
The second box can befilled with any one of the nine digits
So the availablepossibilities are┬а9C1
The third box can befilled with any one of the eight digits
So the availablepossibilities are┬а8C1
Hence, the totalnumber of possible outcomes are┬а9C1┬а├Ч┬а9C1┬а├Ч┬а8C1┬а=9 ├Ч 9 ├Ч 8 = 648.