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?

Correct Answer: (B) 12

The available odd digits are $\{1, 3, 5, 7, 9\}$. For a number to be divisible by 5 and have all odd digits, its unit digit must be 5. With 5 fixed in the units place, the hundreds place can be filled by any of the 4 remaining odd digits $\{1, 3, 7, 9\}$, and the tens place can then be filled by any of the remaining 3 odd digits without repetition. Thus, there are $4$$\times$ 3 = 12 such numbers.