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?

Asked by Aaryan | 1 year ago |  22

##### Solution :-

Any number less than 1000 may be any of a number from one-digit number, two-digit number and three-digit number.

Case 1: One-digit odd number

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

In the first box either of the three digits (3,5,7) can be placed, so the possibility is 3C1 = 3 possible ways.

Case 2: two-digit odd number

Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.

So, there are 3 × 2 = 6 such 2-digit numbers

Case 3: three-digit odd number

Ignore zero at one’s place for some instance.

Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways.

So, there are a total of 3 *3 * 2 = 18 numbers of 3-digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero.

To obtain the even 3-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 tens place in 2 ways (using remaining 2 digits after filling up hundreds place).

So, there are a total of 1 * 3 * 2 = 6 even 3-digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed)

Then, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 –  6 = 12.

The odd numbers less than 1000 that can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.

Answered by Sakshi | 1 year ago

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