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?

#### Solution

Since the number is less than 1000, it could be a three-digit, two-digit or single-digit number.__Case I__: Three-digit number:

Now, the hundred's place cannot be zero. Thus, it can be filled with **three **digits, i.e. 3, 5 and 7.

Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the hundred's place.

Thus, the unit's place can be filled by only **2** digits.

Number of ways of filling the ten's digit = **2** (as repetition is not allowed)

Total three-digit numbers that can be formed = `3xx2xx2=12`

__Case II:__ Two-digit number:

Now, the ten's place cannot be zero. Thus, it can be filled with **three **digits, i.e. 3, 5 and 7.

Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the ten's place,

Thus, the unit's place can be filled by only **2** digits.

Total two-digit numbers that can be formed = `3xx2=6`

__Case III: __Single-digit number:

It could be 3, 5 and 7.

Total single-digit numbers that can be formed = 3

Hence, required number = 12 + 6 + 3 = 21