NIMCET 2018
Let n be the number of different 5-digit numbers, divisible by 4, that can be formed with the digits 1, 2, 3, 4, 5 and 6, with no digit being repeated. What is the value of n?
Answer (Detailed Solution Below)View answer ā¾Hide ā“
Answer: Option C ā 192
Detailed Solution
A number is divisible by 4 when its last two digits are.
Valid endings from 1 to 6 without repetition: 12, 16, 24, 32, 36, 52, 56, 64, i.e. 8 endings.
The first three places use 3 of the remaining 4 digits: 4 x 3 x 2 = 24 ways. Total = 8 x 24 = 192.
ā Check
44 is ruled out (repeated digit) and 20, 40, 60 need the digit 0, which is not available.ā Common trap
(B) 168 = 7 x 24 comes from missing one of the eight endings.Quick tip
8 valid endings
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Permutations & Combinations ā 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.