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 ?

(a) 8
(b) 12 ✓ UPSC's answer
(c) 16
(d) 24

Why the answer is (b)

• The number must be divisible by 5, so the units digit must be 5 (since 0 is not an odd digit and repetition is not allowed, but 0 is even anyway, so only 5 is possible for the units place among odd digits).

• The available odd digits are {1, 3, 5, 7, 9}. Since 5 is used in the units place, the remaining available digits for the hundreds and tens places are {1, 3, 7, 9}.

• The hundreds digit can be any of the 4 remaining digits (1, 3, 7, or 9).

• The tens digit can be any of the 3 remaining digits left after choosing the hundreds digit.

• Therefore, the total number of such 3-digit numbers is 4 (choices for hundreds) × 3 (choices for tens) × 1 (choice for units) = 12.

Why the other options are wrong

(a) 8: 8 is incorrect because it undercounts the permutations; there are 4 choices for the hundreds place and 3 for the tens place, yielding 12, not 8.
(c) 16: 16 is incorrect because it likely assumes 4 choices for both the hundreds and tens places, ignoring that the digits cannot be repeated.
(d) 24: 24 is incorrect because it likely assumes 4 choices for the hundreds place and 3 for the tens place but multiplies by 2 for the units place, forgetting that 0 is not an odd digit and thus only 5 is valid for the units place.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.