Question -
Answer -
Any number less than1000 may be any of a number from one-digit number, two-digit number and three-digitnumber.
Case 1: One-digit oddnumber
In order to make thenumber odd, the last digit has to be either of (3, 5, 7)
In the first boxeither of the three digits (3,5,7) can be placed, so the possibility is 3C1 =3 possible ways.
Case 2: two-digit oddnumber
Tens place can befilled up by 3 ways (using any of the digit among 3, 5 and 7) and then the onesplace can be filled in any of the remaining 2 digits.
So, there are3 × 2 = 6 such 2-digit numbers
Case 3: three-digitodd number
Ignore zero at one’splace for some instance.
Hundreds place can befilled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tensplace in 3 ways by using remaining 3 digits (after using a digit, there will bethree digits) and then the ones place in 2 ways.
So, there are a totalof 3 *3 * 2 = 18 numbers of 3-digit numbers which includes bothodd and even numbers (ones place digit are zero). In order to get the oddnumbers, it is required to ignore the even numbers i.e. numbers ending with zero.
To obtain the even3-digit numbers, one’s place can be filled up in 1 way (only 0 to be filled),hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tensplace in 2 ways (using remaining 2 digits after filling up hundreds place).
So, there are a totalof 1 * 3 * 2 = 6 even 3-digit numbers using the digits 0,3, 5 and 7 (repetition not allowed)
Then, number ofthree-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed)= 18 – 6 = 12.
∴ The odd numbers lessthan 1000 that can be formed by using the digits 0, 3, 5, 7 when repetition ofdigits is not allowed are 3 + 6 + 12 = 21.