# How many three-digit odd numbers are there?

In odd numbers, the last digit consists of (1, 3, 5, 7, 9)

Assume 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 available possibilities are 10C1, third box can be filled with any one of the five digits (1,3,5,7,9) 5C1

Hence, number of total outcomes possible are 9C1 × 10C1 × 5C1 = 9 × 10 × 5 = 450

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