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S5 Mathematics: Permutations and Combinations Practice Paper 4

S5 Mathematics: Permutations and Combinations Practice Paper 4

  • 2026-11-03

S5 Mathematics: Permutations and Combinations Practice Paper 4

Set four closes the chapter with new numbers: seven in a line, two of eleven, two posts among nine and four letters into three boxes, then twelve choose three, three people together and at least one boy, then five couples adjacent and the letters of MATHEMATICS, ending with the error of ignoring which post is which.

Quick answer 5 040, 55, 72, 81, 220, 144, 74, 3 840, 4 989 600 and 20.

01 Paper details (check answers afterwards)

Item Detail
Questions 10 questions
Suggested time 25 minutes
Difficulty mix Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter Chapter: S5 Mathematics: Permutations and Combinations – Tutor Lesson Plan
Level S5

02 Questions (10 items)

  1. FoundationSeven students stand in a line. How many different orders are possible?
  2. FoundationTwo class representatives are chosen from 11 students with no separate posts. How many ways?
  3. FoundationFrom 9 students, choose a chairperson and a vice-chairperson who must be different people. How many outcomes?
  4. FoundationFour different letters are posted into 3 different boxes, any number per box. How many ways?
  5. IntermediateHow many 3-person groups can be formed from 12 students?
  6. IntermediateSix people queue up with A, B and C standing together. How many orders are possible?
  7. IntermediateFrom 4 boys and 5 girls, choose 3 people including at least one boy. How many combinations?
  8. ChallengeFive married couples stand in a line with each couple adjacent. How many orders?
  9. ChallengeHow many distinct arrangements are there of all 11 letters in MATHEMATICS?
  10. Error redoTwo of five people are chosen as leader and recorder. A student answers 10. What is correct?

03 Answers and one-line notes

# Answer One-line note
1 5 040 orders Seven factorial, that is 7 x 6 x 5 x 4 x 3 x 2 x 1, gives 5 040.
2 55 ways Order does not matter, so C(11,2) is 11 x 10 divided by 2, giving 55.
3 72 outcomes The two posts are distinct, so 9 x 8 gives 72.
4 81 ways Each letter has 3 choices, so 3 to the power 4 gives 81.
5 220 groups C(12,3) is 12 x 11 x 10 divided by 6, giving 220.
6 144 orders Bundle the three together: 3! x 4! is 6 x 24 = 144.
7 74 combinations Take all C(9,3) = 84 and subtract the all-girl C(5,3) = 10, leaving 74.
8 3 840 orders Each couple has 2 internal orders, so 2 to the power 5 times 5! is 32 x 120 = 3 840.
9 4 989 600 arrangements M, A and T each appear twice, so divide 11! by 2! x 2! x 2!, giving 4 989 600.
10 20 outcomes The posts differ so use P: 5 x 4 = 20, whereas 10 is C(5,2) and ignores which post is which.

04 What to do after marking

  • After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
  • Repeat the same item two days later; move to the next paper only when both attempts are correct.
  • Parents only need the answer and note columns; there is no need to explain every item.

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