How many odd numbers less than 1000 can be formed by using the digits 0, 2, 5, 7 when the repetition of digits is allowed?

Explanation

Since each number is less than 1000, required numbers are the 1-digit, 2-digit and 3-digit numbers.
One-digit numbers: Clearly, there are two one -digit odd numbers, namely 5 and 7, formed of the given digits.
Two-digit numbers: Since we are to form 2- digit odd numbers, we may put 5 or 7 at the unit's place. So, there are 2 ways of filling the unit's place.
Now, we cannot use 0 at the ten's place and the repetition of digits is allowed. So, we may fill up the ten's place by any of the digits 2, 5, 7. Thus, there are 3 ways of filling the ten's place.
Hence, the required type of 2-digit numbers `=(2xx3)=6.`
Three-digit numbers: To have an odd 3-digit number, we may put 5 or 7 at the unit's place. So, there are 2 ways of filling the unit's place.
We may fill up the ten's place by any of the digits 0, 2, 5, 7. So, there are 4 ways of filling the ten's place.
We cannot put 0 at the hundred's place. So, the hundred's place can be filled by any of the digits 2, 5, 7 and so it can be done in 3 ways.
`therefore " the required number of 3-digit numbers"= (2xx4xx3) =24.`
Hence, the total number of required type of numbers `=(2+6+24)= 32.`