When do I use P and when C?

Use P when posts or order differ, and C when you only pick a group with no order.

Why subtract for non-adjacent?

Counting non-adjacent cases directly misses many, so take all orders and subtract the blocked ones.

Why divide by the factorial of the group count?

When groups are unlabelled the same split is counted several times, so divide by 4!.

S5 Mathematics: Permutations and Combinations Practice Paper 3

2026-11-03

Tutor Prep Guide · Secondary Mathematics

Set three uses fresh numbers: six in a line, four of nine, two posts and three of five letters to fix P against C, then five of twelve, mixed groups and seven in a line with two people apart, then four groups of two and the letters of BOOKKEEPER, ending with the error of counting the chairperson twice.

Quick answer　720, 126, 20, 60, 792, 15, 3 600, 105, 1 663 200 and 60.

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)

FoundationSix students stand in a line. How many different orders are possible?
FoundationHow many ways are there to choose 4 books from 9 different books?
FoundationFrom 5 people, choose a chairperson and a vice-chairperson, who must be different. How many outcomes?
FoundationUsing three of the five letters A, B, C, D and E, how many three-letter strings can be formed?
IntermediateHow many 5-person teams can be formed from 12 students?
IntermediateFrom 5 boys and 3 girls, choose 1 boy and 2 girls. How many combinations?
IntermediateSeven people queue up with A and B not standing next to each other. How many orders are possible?
ChallengeEight people are split into four groups of 2, with the groups unlabelled. How many ways?
ChallengeHow many distinct arrangements are there of all 11 letters in BOOKKEEPER?
Error redoFrom 6 people, choose a committee of 3, then choose a chairperson from those 3. How many outcomes?

03　Answers and one-line notes

#	Answer	One-line note
1	720 orders	Six factorial, that is 6 x 5 x 4 x 3 x 2 x 1, gives 720.
2	126 ways	Order does not matter, so C(9,4) is 9 x 8 x 7 x 6 divided by 24, giving 126.
3	20 outcomes	The two posts differ, so 5 x 4 gives 20.
4	60 strings	Letters are not repeated and order matters, so P(5,3) is 5 x 4 x 3, giving 60.
5	792 teams	C(12,5) is 12 x 11 x 10 x 9 x 8 divided by 120, giving 792.
6	15 combinations	Take C(5,1) times C(3,2), which is 5 x 3 = 15.
7	3 600 orders	Subtract the 2 x 6! orders with A and B together from 7!, giving 5 040 minus 1 440, that is 3 600.
8	105 ways	Divide 8! by 2! to the fourth power times 4!, that is 40 320 divided by 384, giving 105.
9	1 663 200 arrangements	With 2 O, 2 K and 3 E, divide 11! by 2! x 2! x 3!, giving 1 663 200.
10	60 outcomes	Choose the committee then the chair: C(6,3) x 3 is 20 x 3 = 60, whereas 6 x 5 x 4 = 120 counts the post twice.

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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