NIMCET 2014
There are 8 students appearing in an examination of which 3 have to appear in Mathematics paper and the remaining 5 in different subjects. Then, the number of ways they can be made to sit in a row, if the candidates in Mathematics cannot sit next to each other is
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Answer: Option D — 14400
Detailed Solution
Seat the 5 other candidates: ways.
This leaves 6 gaps; place the 3 Mathematics candidates in different gaps: ways.
Total .
✓ Check
Gaps: 5 people in a row create 4 inner gaps and 2 ends, 6 in all.⚠ Common trap
(A) 2400 forgets to arrange the 3 Mathematics candidates among themselves.Quick tip
Gap method:
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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.