Permutation and Combination

Q: Are the permutation and combination formulae given in the SPM exam?

Yes. Both nPr=n!(n−r)!{}^{n}P_{r} = \dfrac{n!}{(n-r)!}nPr=(n−r)!n! and nCr=n!(n−r)! r!{}^{n}C_{r} = \dfrac{n!}{(n-r)!\,r!}nCr=(n−r)!r!n! are printed in the formula list at the front of the paper. What you still have to supply is the judgement of which one a question needs, and the factorial and circular-arrangement facts, which are not on the list.

Q: Do I need to memorise anything if the formulae are given?

Yes. The given list only has nPr{}^{n}P_{r}nPr and nCr{}^{n}C_{r}nCr. You must memorise n!n!n!, 0!=10! = 10!=1, the circular-arrangement rule (n−1)!(n-1)!(n−1)!, and the identical-objects rule n!p! q!⋯\dfrac{n!}{p!\,q!\cdots}p!q!⋯n!, plus how to handle conditions.

What this chapter is

Permutation and Combination sits in the Statistics learning area of Form 5 Add Math (KSSM 3472). It teaches you to count the number of ways something can happen without writing out every possibility.

Instead of listing arrangements one by one, you learn a small set of counting tools, the multiplication rule, factorials, permutations and combinations, and choose the right one for the situation.

The single most important skill in this chapter is deciding whether order matters. When the order of the objects changes the outcome (a first, second and third prize; the seats around a table; a PIN code), you are counting permutations.

When order does not change the outcome (a committee of three people; a hand of cards; which subjects you sit), you are counting combinations. Everything else builds on that one decision.

The chapter is short but high-value. The methods are quick to state, the calculations are light, and the two key formulae are printed for you in the exam.

What separates full marks from lost marks is careful reading of the conditions in each question,'must sit together', 'cannot be adjacent', 'at least two girls', 'in a circle', and translating those words into the correct multiplication of cases.

This chapter also feeds straight into the next one, Probability Distribution. The combination formula ${}^{n}C_{r}$ reappears inside the binomial probability formula, so time spent here pays off twice.

Add Math is an elective, and this chapter assumes only a light background: comfortable arithmetic and the idea of listing outcomes that you met earlier in school. If you can count the outcomes of a small example by hand, you already have the intuition.

The chapter simply gives you formulae so you no longer have to list large cases one at a time, and it trains you to spot which of a handful of standard shapes a question really is.

Content standards

The DSKP splits this chapter into two content standards. Here is what each one covers.

Code	Content standard	What you learn
4.1	Permutation	The multiplication rule, factorial notation $n!$ , and ${}^{n}P_{r}$ for arranging objects, including arrangements with identical objects and around a circle, under one condition.
4.2	Combination	Comparing permutation with combination, finding ${}^{n}C_{r}$ when order does not matter, and solving selection problems that carry a condition.

Within these, the DSKP lists six learning standards:

4.1.1 Investigate and generalise the multiplication rule of counting.
4.1.2 Find the number of permutations of $n$ different objects, of $n$ objects taken $r$ at a time, and of $n$ objects that include identical objects.
4.1.3 Solve permutation problems with one condition, including identical objects or arrangements in a circle.
4.2.1 Compare and contrast permutation and combination.
4.2.2 Find the number of combinations of $r$ objects chosen from $n$ different objects.
4.2.3 Solve combination problems that carry a condition.

Key ideas

The multiplication rule

If one task can be done in $m$ ways and a second, independent task in $n$ ways, the two together can be done in $m \times n$ ways. This is the engine behind the whole chapter: break a counting problem into stages, count the choices at each stage, then multiply.

Multiplication ruleMust memorise

m \times n \times p \times \cdots

Multiply the number of choices at each independent stage.

Factorial notation $n!$

$n!$ (read 'n factorial') means the product of all whole numbers from $n$ down to 1, and by definition $0!= 1$ . It counts the number of ways to arrange $n$ different objects in a row.

FactorialMust memorise

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

Permutations: when order matters

A permutation is an ordered arrangement. The number of ways to arrange $r$ objects chosen from $n$ different objects, where order matters, is given below.

This formula is printed in the exam formula list, so you do not need to memorise it, but you must know when to use it.

PermutationsGiven in the exam

{}^{n}P_{r} = \dfrac{n!}{(n-r)!}

Given in the SPM formula list.

Combinations: when order does not matter

A combination is a selection where order is irrelevant. Because each unordered group of $r$ objects can itself be arranged in $r!$

ways, the combination count is the permutation count divided by $r!$ . This formula is also supplied in the exam.

CombinationsGiven in the exam

{}^{n}C_{r} = \dfrac{n!}{(n-r)!\,r!}

Given in the SPM formula list.

How permutation and combination connect

The two are linked by ${}^{n}P_{r} = {}^{n}C_{r} \times r!$ . Reading it aloud helps: to build an ordered arrangement, first choose the $r$ objects ( ${}^{n}C_{r}$ ways), then arrange them ( $r!$

ways). Two useful facts fall out: ${}^{n}C_{0} = {}^{n}C_{n} = 1$ and ${}^{n}C_{r} = {}^{n}C_{n-r}$ .

Arrangements with identical objects

When some objects are identical, swapping them does not create a new arrangement, so you divide by the factorial of each repeated group. The number of distinct arrangements of $n$ objects, where one kind repeats $p$ times and another repeats $q$ times, is given below.

You must memorise this one.

Arrangements with repeatsMust memorise

\dfrac{n!}{p!\,q!\,\cdots}

Divide n! by the factorial of each repeated group.

Arrangements in a circle

Around a circular table there is no fixed starting seat, so we fix one person and arrange the rest. The number of distinct circular arrangements of $n$ different objects is $(n-1)!$ .

The DSKP limits circular questions to one condition, such as two people who must sit together.

Circular permutationsMust memorise

(n-1)!

Turning conditions into cases

Most marks come from conditions. 'Must be together', glue the group into one block, arrange the blocks, then arrange within the block.

'Cannot be together', count everything, then subtract the arrangements where they are together. 'At least', either add up the allowed cases or subtract the unwanted ones from the total.

How it is examined

Additional Mathematics is assessed by two written papers, both answered with a non-programmable scientific calculator and marked by analytic scoring, so method marks are awarded step by step even when the final answer slips.

Paper 1 (3472/1) runs for 2 hours and carries 80 marks. Section A has 12 questions that you all answer (64 marks); Section B has 3 questions from which you answer 2 (16 marks).

Counting questions from this chapter fit Paper 1 well because they are short and self-contained.

Paper 2 (3472/2) runs for 2 hours 30 minutes and carries 100 marks. It has three sections: Section A with 7 questions to answer all (50 marks), Section B with 4 questions of which you answer 3 (30 marks), and Section C with 4 questions of which you answer 2 (20 marks).

Here a permutation or combination idea may appear as one part of a longer, multi-step question.

Across the papers, items are limited-response subjective and structured, and the overall difficulty spread is Low : Medium : High = 5 : 3 : 2. We do not know in advance which chapter carries which marks, and no one can predict the exact questions, so prepare the method thoroughly rather than chasing specific items.

Two worked snapshots

These two short examples show the two moves you will make most often, recognising a permutation, then a conditional combination. Both are written in the style of the exam, not taken from any paper.

Q1[3 marks]

Five students sit in a row of five chairs. In how many ways can they be seated if two particular students insist on sitting next to each other?

Show worked solution

Order matters in a row, so this is a permutation. Treat the two friends as a single block, giving four items, the block plus three other students, to arrange: $4!= 24$ ways. The two friends can swap inside the block in $2!= 2$ ways. Total $= 4!\times 2! = 24 \times 2 = 48$ ways.

Q2[4 marks]

A committee of three is chosen from five teachers and four parents. (a) How many committees are possible with no restriction?

(b) How many committees contain at least two teachers?

Show worked solution

Order does not matter in a committee, so use combinations. (a) Choose 3 from 9: ${}^{9}C_{3} = 84$ .

(b) 'At least two teachers' means two teachers with one parent, or three teachers with no parent: ${}^{5}C_{2}\times{}^{4}C_{1} + {}^{5}C_{3}\times{}^{4}C_{0} = (10)(4) + (10)(1) = 40 + 10 = 50$ committees.

In the exam

The permutation formula ${}^{n}P_{r}$ and the combination formula ${}^{n}C_{r}$ are both printed in the formula list at the front of the paper. You still have to decide which one the question needs, the list will not tell you whether order matters.

Common mistakes

These are the slips we see most often, and how to avoid each one.

Mixing up permutation and combination

This is the number-one error. If you can swap two chosen objects and get a genuinely different outcome, order matters and you want ${}^{n}P_{r}$ ; if swapping changes nothing, use ${}^{n}C_{r}$ .

Fix: before touching the calculator, write one word next to the question,'order?', and answer yes or no.

Treating $0!$ as 0

Students often write $0! = 0$ , which breaks formulae like ${}^{n}C_{n} = \dfrac{n!}{0!\,n!}$ . By definition $0!= 1$ . Fix: memorise $0!= 1$ and ${}^{n}C_{0} = {}^{n}C_{n} = 1$ as fixed facts.

Using $n!$ for a circle

In a circle there is no first seat, so $n!$ overcounts by rotations.

The correct count is $(n-1)!$ . Fix: fix one person's seat first, then arrange the remaining $n-1$ people relative to them.

Forgetting to divide by repeats

When arranging the letters of a word with repeated letters, say a five-letter word with two identical letters, students use $5!$ and overcount.

Fix: divide $n!$ by the factorial of each repeated group, giving $\dfrac{n!}{p!\,q!\cdots}$ .

Misreading 'at least' and 'at most'

'At least two' is not the same as 'exactly two'. Fix: list every case the condition allows and add them, or count the total and subtract the cases that break the condition, whichever is shorter.

Adding instead of multiplying stages

When a task happens in stages (choose a chairperson, then a secretary), the stage counts multiply, not add. Add only when you are combining separate, mutually exclusive cases.

Fix: ask 'and then' (multiply) versus 'or' (add).

How to study this chapter

Here is the order our teachers use with students, from first principles to exam speed.

1
Lock in the vocabulary
Make sure you can define permutation, combination, factorial and the multiplication rule in your own words, and state $0! = 1$ .
The key-terms page collects these.
2
Decide 'does order matter?' every time
Practise reading short questions and labelling each one permutation or combination before doing any arithmetic. This single habit prevents most lost marks.
3
Drill the four standard forms
${}^{n}P_{r}$ , ${}^{n}C_{r}$ , circular $(n-1)!$ , and arrangements with identical objects. Do several of each until the setup is automatic.
4
Layer in the conditions
Work through 'together', 'not together', and 'at least / at most' questions using the block, subtraction and case methods.
5
Time yourself on exam-style items
Move to the worked examples and practice questions under time, then review with the common-mistakes page.

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Frequently asked questions

Are the permutation and combination formulae given in the SPM exam?

Yes. Both ${}^{n}P_{r} = \dfrac{n!}{(n-r)!}$ and ${}^{n}C_{r} = \dfrac{n!}{(n-r)!\,r!}$ are printed in the formula list at the front of the paper. What you still have to supply is the judgement of which one a question needs, and the factorial and circular-arrangement facts, which are not on the list.

How do I know whether to use permutation or combination?

Ask whether order changes the outcome. Prizes ranked first, second and third, seats in a row, or digits in a code depend on order, so use permutation.

A committee, a team, or a hand of cards does not depend on order, so use combination. When in doubt, swap two of the chosen items and see if you get a genuinely different result.

Is this a hard chapter?

The formulae are short and the calculations are light, so the mechanics are among the friendlier parts of Form 5 Add Math. The challenge is reading conditions carefully,'must sit together', 'at least two', 'in a circle', and translating them into the right multiplication of cases.

With steady practice on conditions, most students find it a reliable place to score.

Do I need to memorise anything if the formulae are given?

Yes. The given list only has ${}^{n}P_{r}$ and ${}^{n}C_{r}$ .

You must memorise $n!$ , $0! = 1$ , the circular-arrangement rule $(n-1)!$ , and the identical-objects rule $\dfrac{n!}{p!\,q!\cdots}$ , plus how to handle conditions.

Can I use my calculator's nPr and nCr buttons?

Yes. A non-programmable scientific calculator has nPr and nCr keys, and it is allowed.

The keys speed up the arithmetic, but the marks are for choosing the right method and showing your setup, so always write the expression, for example ${}^{9}C_{3}$ , before pressing the key. Because scoring is analytic, a clear setup can still earn method marks even if you slip on the final number.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026