1.

How many three-digit numbers are there?

Answer»

Let us assume that we have three boxes.

First box can be filled with any one of the nine digits (zero not allowed at first position)

Therefore, possibilities are 9C1

The second box can be filled with any one of the ten digits

Therefore the available possibilities are 10C1

Third box can be filled with any one of the ten digits

Therefore the available possibilities are 10C1

Thus, the total number of possible outcomes are 9C1 × 10C1 × 10C1 = 9 × 10 × 10 = 900.



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