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 ?

The correct option is 12.

Let us break down the problem step-by-step to understand why this is the correct answer.

We need to find the number of 3-digit natural numbers that satisfy the following conditions:
1. The digits must not repeat (without repetition).
2. Each digit must be odd.
3. The number must be divisible by 5.

Let the 3-digit number be represented as ABC, where A, B, and C are its digits (Hundreds, Tens, and Units place respectively).

First, let's identify the set of odd digits available in our number system:
Odd digits = {1, 3, 5, 7, 9}

Next, we apply the divisibility rule for 5. A number is divisible by 5 if and only if its units digit (C) is either 0 or 5. Since all digits of our number must be odd, and 0 is an even digit, the units digit C must be 5.
Therefore, C = 5 (1 choice).

Now, we need to fill the hundreds place (A) and tens place (B) using the remaining odd digits. Since repetition of digits is not allowed and 5 is already used for the units place, the remaining available odd digits are:
Remaining digits = {1, 3, 7, 9} (4 available digits).

Let's choose the digit for the hundreds place (A):
Since A can be any of the remaining 4 odd digits, we have 4 choices for A.

Finally, let's choose the digit for the tens place (B):
Since B cannot be equal to C (which is 5) or A (which is already chosen), we have 3 choices left for B.

Using the fundamental counting principle, the total number of such 3-digit numbers is given by the product of the number of choices for each position:

$$\text{Total numbers}=\text{Choices for A}\times \text{Choices for B}\times \text{Choices for C}$$

$$\text{Total numbers}=4\times 3\times 1=12$$

Thus, there are exactly 12 such 3-digit natural numbers.