How many three-digit numbers are there with no digit repeated?

Asked by Aaryan | 1 year ago |  38

##### Solution :-

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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