Question -
Answer -
We know that in oddnumbers, the last digit consists of (1, 3, 5, 7, 9).
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 the possibilitiesare 9C1
The second box can befilled with any one of the ten digits
So the availablepossibilities are 10C1
The third box can befilled with any one of the five digits (1,3,5,7,9)
So the availablepossibilities are 5C1
Hence, the totalnumber of possible outcomes are 9C1 × 10C1 × 5C1 =9 × 10 × 5 = 450