Let us assume we have three boxes.
The first box can be filled with any one of the nine digits (0 not allowed at first place).
So, the possibilities are 9C1
The second box can be filled with any one of the nine digits
So the available possibilities are 9C1
The third box can be filled with any one of the eight digits
So the available possibilities are 8C1
Hence, the total number of possible outcomes are 9C1 × 9C1 × 8C1 = 9 × 9 × 8 = 648.Answered by Sakshi | 1 year ago
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