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 ?
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 ?
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 going 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 = (2×3)=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 units 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.
∴the required number of 3-digit number = (2×4×3)=24.
Hence, the total number of required type of numbers = (2+6+24)=32.
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 going 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 = (2×3)=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 units 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.
∴the required number of 3-digit number = (2×4×3)=24.
Hence, the total number of required type of numbers = (2+6+24)=32.
