How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?

Given: Odd number less than 1000 is required.

In order to make the number odd, the last digit has to either of (3, 5, 7)

Assume three boxes, in the first either of the three digits (3,5,7) can be placed, so the possibility is ³C₁

Case 1: Middle digit is zero

If the middle digit is zero, number of ways of placing odd numbers on the second box = 2

Hence, the total number of ways = 3 × 2 = 6 ways

Case 2: Middle digit is an odd number

Number of ways of filling middle box = 2

Number of ways of filling third box = 1

Hence, the total number of ways = 3 × 3 = 9 ways

Hence total number of outcomes possible = 6 + 9 = 15 ways