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 Option (B)

The problem requires forming 3-digit natural numbers using only odd digits, without repetition, such that the number is divisible by 5.

The set of odd digits available is {1, 3, 5, 7, 9}.

For a number to be divisible by 5, its unit digit must be 0 or 5. Since only odd digits are permitted, the unit digit of the 3-digit number must be 5.

With the unit digit fixed as 5, the remaining odd digits available for the hundreds and tens places are {1, 3, 7, 9}. There are 4 distinct digits remaining.

We need to select and arrange 2 distinct digits from these 4 for the remaining two positions (hundreds and tens place). This is a permutation problem, calculated as P(n, k) = n! / (n-k)!.

• Number of available digits (n) = 4
• Number of positions to fill (k) = 2
• Number of ways = P(4, 2) = 4! / (4-2)! = 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 4 × 3 = 12.

Therefore, there are 12 such 3-digit natural numbers that satisfy all the given conditions.

Incorrect Options:

Options A (8), C (16), and D (24) are incorrect. The precise calculation, considering the constraints of using only odd digits, no repetition, and divisibility by 5, yields 12 such numbers, as demonstrated above.