Q. How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?

Explanation : 

3-digit odd number divisible by 5

A 3-digit natural number is divisible by 5 if its last digit (the units place) is either 0 or 5. Since the number should only have odd digits and 0 is not an odd digit, the last digit must be 5.

We are given that each digit is odd and there is no repetition of digits. The odd digits are 1, 3, 5, 7, and 9. We have already used 5 as the last digit, so we are left with 1, 3, 7, and 9 to fill the other two positions (the hundreds and tens place).

There are 4 choices for the hundreds place (1, 3, 7, or 9) and 3 choices for the tens place (the remaining 3 odd digits after selecting the digit for the hundreds place). So, there are a total of 4 x 3 = 12 different numbers.