MAH MCA CET 2023
A five-digit number divisible by 3 is to be made using the digits 0, 1, 2, 3, 4, 5 without repetition. Find the number of such numbers.
Answer (Detailed Solution Below)View answer ▾Hide ▴
Answer: Option A — 216
Detailed Solution
. A five-digit number uses 5 of these 6 digits, so exactly one digit is left out.
Digit sum . This is divisible by 3 only if the left-out digit is 0 or 3.
Step 3 (leave out 0): digits 1, 2, 3, 4, 5 → numbers.
Step 4 (leave out 3): digits 0, 1, 2, 4, 5 → (remove numbers starting with 0).
Total
⚠ Common trap
240 (= 120 + 120) forgets that a number cannot start with 0.Quick tip
Divisible by 3 digit sum divisible by 3; subtract for leading zero
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Permutations & Combinations — Preparation
This question tests your understanding of Permutations & Combinations, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.
Permutations count arrangements where order matters, while combinations count selections where order does not matter. This topic covers the fundamental principle of counting, factorials, nPr and nCr, arrangements of letters in a word (including repeated letters), circular arrangements, and selecting committees or teams with conditions. The hardest part is deciding whether a question needs a permutation or a combination. Ask yourself whether swapping two chosen items gives a new outcome. If it does, order matters. Conditions such as "two people must sit together" are handled by treating them as a single unit, and "at least one" questions are usually easiest to solve as total minus none. Practise the permutation and combination questions below, each with a step-by-step solution showing how to set up the count.