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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9709Paper 5 · Probability & Statistics 1Permutations and combinationsArrangements and factorials

Permutations and combinations · Subtopic 1 of 3

Arrangements and Factorials

How many ways can things be put in order? Multiply the number of choices for each place, which gives $n!$ for $n$ different objects, then divide out the orderings you cannot tell apart.

Cambridge 9709 Paper 5≈ 30 minRevision 5 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Use the multiplication principle to count outcomes.
Arrange $n$ different objects in $n!$ ways, and choose and arrange $r$ of them in $^nP_r$ ways.
Arrange objects with repeats by dividing by the factorial of each repeat.
Know that $0!=1$ .

Pause and think

Three friends, Ali, Bilal and Chen, sit on a bench. List the orders they could sit in. How many are there, and why is it $3\times2\times1$ ?

Section 2 of 14

Core idea

Start hereHow many passwords?

Arrange the letters A, B, C: ABC, ACB, BAC, BCA, CAB, CBA. Six ways. With 10 different letters there are $3\,628\,800$ ways: listing is hopeless. Counting principles get the answer in seconds.

The big idea

Multiplication principle: $a$ ways then $b$ ways gives $ab$ ways. So $n$ different objects can be arranged in $n!$ ways. With repeats, divide by the factorial of each repeat count. Arranging $r$ of $n$ : $\,{}^nP_r=\frac{n!}{(n-r)!}$ .

1. Factorials

Every way to arrange A, B and C

3\times2\times1=6=3!

arrangements. Each level of the tree has one fewer choice than the level before.

Why

n!

$\text{1st place: }n\text{ choices}$
Any object.
$\text{2nd: }n-1,\ \text{3rd: }n-2,\ \ldots,\ \text{last: }1$
One fewer each time.
$n(n-1)(n-2)\cdots1=n!$
Multiplication principle.

By convention

0!=1

(one way to arrange nothing).

2. Repeated objects

Arrangements of BANANA

$\text{label: }B\,A_1\,N_1\,A_2\,N_2\,A_3:\ 6!=720$
Pretend all letters differ.
$\text{each real arrangement counted }3!\times2!=12\text{ times}$
Swapping the A's (3!) or N's (2!) gives the same word.
$\frac{720}{12}=60$
Divide out the overcount.

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

for repeats of sizes

p,q,\ldots

Worked exampleSTATISTICS

How many arrangements of the letters of STATISTICS?

Goal: Count.
Strategy: Count each letter; divide.

10 letters: S×3, T×3, I×2, A×1, C×1.
$\frac{10!}{3!\,3!\,2!}=\frac{3628800}{72}=50400$
Formula.

Answer

50\,400

✓ Check

6\times6\times2=72

✓.

3. Arranging some of the objects

Worked exampleCodes

How many 4-letter arrangements from 7 different letters (no repeats)?

Goal: Count.
Strategy: Multiplication principle.

$7\times6\times5\times4=840$
Four places.
$={}^7P_4=\frac{7!}{3!}$
Formula.

Answer

840

✓ Check

\frac{5040}{6}=840

✓.

Pause and think

3-digit codes from digits 1–9, no repeats?

Summary

$n!$ ; divide by repeats.
$\,{}^nP_r$ for part arrangements.

Permutation: An ordered arrangement.
Factorial: $n!=n(n-1)\cdots1$ .

HistoryCounting in twelfth-century India. The Indian mathematician Bhāskara II, in his book Līlāvatī (1150), gave rules for counting arrangements, including arrangements of digits with repeats, and the number of ways to hold Vishnu's four attributes in his four hands:

24

. The exclamation-mark notation

n!

came much later, from the French mathematician Christian Kramp in 1808.

Fun factYour shuffle is (almost certainly) unique. A deck of

52

cards can be arranged in

52!\approx8\times10^{67}

ways. That is so many that a well-shuffled deck is almost certainly in an order that has never existed before in the history of the universe, and never will again.

Section 3 of 14

Key formulae

n different objects in a line

n!=n(n-1)(n-2)\cdots1,\qquad0!=1

r objects chosen in order from n

^nP_r=\frac{n!}{(n-r)!}

With repeats (p alike, q alike, …)

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

TrickBoxes first. Draw a box for each position and write the number of choices in it, left to right:

\boxed{8}\,\boxed{7}\,\boxed{6}

for gold, silver and bronze among

8

runners. Multiply. Fill any restricted boxes first (see the next lesson).

Section 4 of 14

Worked examples

Example 1 · Ordered choices0 / 2 steps

How many

4

-letter codes can be made from

10

different letters if no letter is repeated?

Example 2 · Repeated letters0 / 2 steps

Find the number of arrangements of the letters of NEEDLESS.

Example 3 · Coloured beads0 / 1 steps

Seven beads,

3

red,

2

blue and

2

green, are threaded in a line. Beads of the same colour are identical. How many different patterns are there?

Section 5 of 14

Explore the graph

How fast factorials grow

Loading the interactive graph…

Try this

The steepest curve passes through $n!$ at each whole number $n$ (the point shows $k!$ ). The others are $2^n$ and $n^3$ .
At which $n$ does $n!$ overtake $n^3$ ? And $2^n$ ?
$10!$ is $3\,628\,800$ , off the top of this graph. Why do arrangement counts become huge so quickly?

Factorials eventually outgrow every power and every exponential.

In the real worldWhy long passwords win. A password of

8

characters from

62

letters and digits allows

62^8\approx2\times10^{14}

possibilities;

12

characters allow about

3\times10^{21}

, millions of times more. Every extra character multiplies the count, which is the multiplication principle protecting your accounts.

Section 6 of 14

Common mistakes

IncorrectBANANA:

6!=720

Correct

\dfrac{6!}{3!\,2!}=60

Swapping identical letters makes no new word.

IncorrectDividing by

3!+2!

CorrectDivide by

3!\times2!

Each group of repeats divides separately.

Incorrect

0!=0

Correct

0!=1

There is exactly one way to arrange nothing.

IncorrectUsing

^nP_r

when order does not matter

CorrectUse

^nC_r

That is a selection (see the combinations lesson).

how many ways can 10 people stand in a line?

Calculator

$3\,628\,800$

so the queue for the canteen is basically a miracle

Section 7 of 14

Exam tip

Exam tipShow the calculation, not just the number. Write

\dfrac{8!}{4!\,2!}

before evaluating it. Answers in permutations are large and easy to mistype; the expression earns the method mark even if the calculator slips.

Section 8 of 14

Guided practice

SA-9709-P5-PC-G011 markL1 FoundationCalculatorArrangements

In how many ways can

6

different books be arranged in a row on a shelf?

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P5-PC1-0011 markL1 FoundationCalculatorArrangements

In how many different orders can

5

people stand in a line for a photograph?

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC-0111 markL1 FoundationCalculatorArrangements

In how many ways can a gold, a silver and a bronze medal be awarded to

3

8

runners?

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC1-0022 marksL1 FoundationCalculatorArrangements

How many

4

-letter codes can be made from

10

different letters if no letter may be used more than once?

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC-0012 marksL1 FoundationCalculatorRepeats

Find the number of different arrangements of the letters of the word BANANA.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC1-0032 marksL1 FoundationCalculatorRepeats

Find the number of different arrangements of the letters of the word LETTER.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC-0022 marksL2 StandardCalculatorRepeats

Find the number of different arrangements of the letters of the word STATISTICS.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC1-0062 marksL2 StandardCalculatorRepeats

Seven beads,

3

red,

2

blue and

2

green, are threaded in a line. Beads of the same colour are identical. How many different patterns are possible?

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PC1-0042 marksL2 StandardCalculatorRepeats

Find the number of different arrangements of the letters of the word NEEDLESS.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P5-PC-0142 marksL2 StandardCalculatorRepeats

Find the number of different arrangements of the letters of MISSISSIPPI.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Multiply the choices for each position.
$n$ different objects: $n!$ ; $r$ of them in order: $^nP_r$ .
Repeats: divide by the factorial of each group of identical objects.
$0!=1$ .

Section 12 of 14

Quick quiz

Q1 · SA-9709-P5-PC-Q021 markL1 FoundationCalculatorArrangements

0!

is equal to

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P5-PC-Q031 markL1 FoundationCalculatorRepeats

The number of arrangements of the letters AABB is

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P5-PC1-0051 markL1 FoundationCalculatorArrangements

Eight runners finish a race with no ties. In how many different orders can they finish?

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 9709 AS & A Level Mathematics
Where: Paper 5, Probability & Statistics 1 · 5.3 Permutations and combinations
Expected: Understand the terms permutation and combination; solve problems about arrangements of objects in a line, including those involving repetition (for example, arranging the letters of the word NEEDLESS).
Calculator: Permitted. Give non-exact answers to 3 significant figures; probabilities may be fractions or decimals.
Typical style: A word with repeated letters: the total number of arrangements, then arrangements with a restriction.
Open in: 9709 · Paper 5

Section 14 of 14

Past paper questions

8 questions (73 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

9709/52 · Oct/Nov 2025Question 710 marks
Arrangements of letters with restrictions, and choosing two different letters.
Question paper, page 12 Mark scheme
9709/52 · May/June 2025Question 69 marks
Arrangements and selections of letters from a word with repeated letters and restrictions.
Question paper, page 8 Mark scheme
9709/52 · Feb/March 2025Question 610 marks
Arrangements and selections of books with restrictions.
Question paper, page 12 Mark scheme
9709/52 · Oct/Nov 2024Question 24 marks
Arrangements of letters with a restriction on a gap.
Question paper, page 4 Mark scheme
9709/52 · May/June 2024Question 710 marks
Arrangements with repeated letters and restrictions, and a probability.
Question paper, page 12 Mark scheme
9709/52 · May/June 2023Question 610 marks
Selections with restrictions and arrangements with grouping or separation.
Question paper, page 12 Mark scheme

Show 2 more

9709/52 · Oct/Nov 2022Question 710 marks
Arrangements of letters with grouping and spacing, a probability, and selections.
Question paper, page 12 Mark scheme
9709/52 · May/June 2022Question 610 marks
Arrangements with restrictions, selections and splitting into groups.
Question paper, page 10 Mark scheme

All 9709 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9709, 2026–2027 (official specification).