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Module 4 · L2 of 15 ~30 min ⚡ +95 XP available

Factorial Notation & Simple Arrangements

In Lesson 1 you multiplied $n \times (n-1) \times (n-2) \times \cdots$ to count arrangements without repetition. Mathematicians write this product so often that they invented a compact symbol for it: the factorial. This lesson introduces $n!$, explores its algebraic properties, and uses it to count arrangements of distinct objects in a row.

Today's hook — Five people line up for a photo. How many different orderings are possible? You could list them all — but that's exhausting. What if it's 10 people? Or 20? There's a one-symbol answer that handles any size. By the end of this lesson you'll use it fluently and simplify algebraic expressions with it.

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Recall — your gut answer first

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Five people — Alice, Ben, Chen, Dana, and Eli — line up for a photo. Without calculating — how many different orderings of the five people do you think are possible? Write your estimate and explain how you arrived at it.

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The two moves

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Factorials rest on two core moves: expand $n!$ into its product form to evaluate it, and cancel common factorial factors when simplifying fractions. Every factorial problem uses one or both of these.

The secret to factorial simplification: never multiply out both factorials in a fraction. Instead, write out only the extra factors that appear in the larger factorial until the smaller factorial cancels. For $\dfrac{7!}{5!}$, write $7 \times 6 \times \cancel{5!} / \cancel{5!} = 42$.

$n! = n \times (n-1)!$

$0! = 1$ by definition

There is exactly one way to arrange zero objects — do nothing. So $0! = 1$. This is a definition, not something to derive.

Cancel from the top

For $\dfrac{n!}{(n-r)!}$, write the $r$ extra factors from $n$ down: $n(n-1)\cdots(n-r+1)$. Never fully expand before cancelling.

Arrangements = $n!$

$n$ distinct objects in a row: $n$ choices for position 1, $n-1$ for position 2, … giving $n!$ total arrangements.

What you'll master

Know

Key facts

$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$
$0! = 1$ (by definition)
$n$ distinct objects can be arranged in $n!$ ways

Understand

Concepts

Why $n!$ counts all arrangements of $n$ distinct objects (link to Lesson 1)
How the recursive identity $n! = n \times (n-1)!$ works
Why factorials grow so rapidly (and why $20! > 2 \times 10^{18}$)

Can do

Skills

Evaluate factorials and factorial fractions by hand
Simplify expressions like $\dfrac{n!}{(n-r)!}$ and $\dfrac{(n+1)!}{n!}$
Count arrangements of distinct objects in a line

Key terms

Factorial $n!$The product $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$. Defined for all non-negative integers; $0! = 1$.

$0! = 1$By definition: there is exactly one way to arrange an empty set. Required for formulas to be consistent.

Recursive identity$n! = n \times (n-1)!$ — each factorial is the previous factorial multiplied by $n$. Useful for simplification.

Arrangement (permutation)An ordering of distinct objects in a line where the order matters. $n$ objects give $n!$ arrangements.

Factorial fractionAn expression like $\dfrac{n!}{(n-r)!}$ that simplifies by cancelling common factorial factors from numerator and denominator.

Distinct objectsAll objects are different from each other — no identical copies. If some objects are identical, the formula changes (covered in later lessons).

Factorial notation

core concept

The factorial of a positive integer $n$, written $n!$, is the product of all positive integers from 1 to $n$:

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

By definition: $0! = 1$

Some values to memorise:

$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
$6! = 720$, $\quad 7! = 5040$, $\quad 10! = 3{,}628{,}800$

The recursive identity $n! = n \times (n-1)!$ is particularly useful. For example:

$7! = 7 \times 6! = 7 \times 720 = 5040$

Rapid growth. Factorials grow faster than exponentials. $10! \approx 3.6$ million, $15! \approx 1.3$ trillion, $20! \approx 2.4 \times 10^{18}$. This is why it is impossible to systematically test all possible orderings once $n$ gets large — combinatorics gives us smarter approaches.

n! = n (n-1) 2 1 for n 1; by definition 0! = 1; Recursive form: n! = n (n-1)! — use this to build up factorial values

Pause — copy the factorial definition into your book: $n! = n \times (n-1) \times \cdots \times 2 \times 1$; by definition $0! = 1$; recursive form $n! = n \times (n-1)!$.

Quick check: What is the value of $\dfrac{6!}{4!}$?

Arrangements of distinct objects

core concept

We just saw the counting principle multiplies choices at each stage. That raises a question: how do restrictions like "no repetition" change the count? This card answers it → arrangement formulas with and without repetition.

The number of ways to arrange $n$ distinct objects in a line is $n!$.

$$\text{Arrangements of } n \text{ distinct objects} = n!$$

This follows directly from the Fundamental Counting Principle (Lesson 1):

Position 1: $n$ choices
Position 2: $n-1$ choices (one used)
Position 3: $n-2$ choices
$\vdots$
Position $n$: 1 choice

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

For a subset of $r$ objects chosen from $n$ (where $r \leq n$), arranged in a line:

$$\frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$$

This is the number of ordered selections of $r$ objects from $n$ — also written $^nP_r$ (covered in Lesson 3).

n distinct objects in a line: n! arrangements; This follows from Lesson 1: n (n-1) 1 = n!

Pause — copy the linear arrangement formula into your book: $n$ distinct objects in a line $= n!$ arrangements, from the Counting Principle: $n \times (n-1) \times \cdots \times 1 = n!$.

Did you get this? True or false: $0! = 0$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · ARRANGING ALL OBJECTS

In how many ways can 5 people be arranged in a row?

5 distinct people, all positions filled $\Rightarrow$ use $n!$ with $n = 5$.

We are arranging all 5 people (not choosing a subset), and all are distinct — so the formula is $5!$.

PROBLEM 2 · FACTORIAL FRACTION

Evaluate $\dfrac{7!}{5!}$.

Write the extra factors in the numerator: $7! = 7 \times 6 \times 5!$

Recognise that $7!$ contains $5!$ as a factor. Expand only the two extra terms rather than computing the full factorials.

PROBLEM 3 · ALGEBRAIC SIMPLIFICATION

Simplify $\dfrac{(n+1)!}{n!}$ and hence find $\dfrac{(n+1)!}{n!}$ when $n = 7$.

Use the recursive identity: $(n+1)! = (n+1) \times n!$

The factorial of $(n+1)$ is $(n+1)$ multiplied by $n!$. This is the recursive form $k! = k \times (k-1)!$ applied with $k = n+1$.

Fill the gap: The letters of the word MATHS can be arranged in different ways.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

$0! = 0$ (it equals 1)

Students often write $0! = 0$ by analogy with multiplication by zero. But $0! = 1$ by definition — there is exactly one way to arrange an empty set. Getting this wrong will break your $^nP_r$ and $^nC_r$ formulas whenever $r = n$.

Trap 02

$(a + b)! \neq a! + b!$

Factorials do not distribute over addition. $(3+2)! = 5! = 120$, but $3! + 2! = 6 + 2 = 8$. Always compute what is inside the factorial first before applying the $!$ symbol.

Trap 03

Forgetting to cancel before expanding

Computing $\dfrac{8!}{6!}$ as $\dfrac{40320}{720} = 56$ wastes time and risks arithmetic errors. Instead: $\dfrac{8!}{6!} = 8 \times 7 = 56$. Always cancel the smaller factorial before multiplying anything out.

Did you get this? True or false: $(3+2)! = 3! + 2! = 8$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Evaluate $\dfrac{8!}{6!}$.

In how many ways can the letters of the word MATHS be arranged?

Simplify $\dfrac{(n+2)!}{n!}$.

A shelf holds 7 different books. In how many orders can they be arranged on the shelf?

Explain why $\dfrac{n!}{(n-2)!} = n(n-1)$ and verify with $n = 6$.

Odd one out: Three of these expressions equal 60. Which one does NOT?

Revisit your thinking

Earlier you estimated the number of ways to arrange 5 people in a row.

By the Fundamental Counting Principle (Lesson 1), or directly using factorial notation:

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

There are exactly 120 orderings. If you under-estimated, this is a classic case of underestimating how quickly products grow. For 10 people, the answer is $10! = 3{,}628{,}800$ — more than 3.6 million orderings!

Factorial notation is compact notation for a pattern that grows extremely rapidly. This rapid growth is both useful (strong encryption) and challenging (brute-force search quickly becomes impossible).

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 31 mark

Q1. Evaluate $\dfrac{8!}{6!}$. (1 mark)

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ApplyBand 31 mark

Q2. In how many ways can the letters of the word MATHS be arranged? (1 mark)

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AnalyseBand 52 marks

Q3. Simplify $\dfrac{(n+1)!}{n!}$ and explain, using the recursive identity, why the result is $n+1$. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $\dfrac{8!}{6!} = 8 \times 7 = 56$ · 2. $5! = 120$ (all 5 letters distinct) · 3. $\dfrac{(n+2)!}{n!} = (n+2)(n+1)$ · 4. $7! = 5040$ · 5. $\dfrac{n!}{(n-2)!} = n(n-1)$ because $n! = n \times (n-1) \times (n-2)!$ so cancelling $(n-2)!$ leaves $n(n-1)$; when $n=6$: $6 \times 5 = 30$, verified by $\dfrac{6!}{4!} = \dfrac{720}{24} = 30$.

Q1 (1 mark): $\dfrac{8!}{6!} = 8 \times 7 = 56$ [1].

Q2 (1 mark): MATHS has 5 distinct letters. Arrangements $= 5! = 120$ [1].

Q3 (2 marks): By the recursive identity, $(n+1)! = (n+1) \times n!$ [1]. Therefore $\dfrac{(n+1)!}{n!} = \dfrac{(n+1) \times n!}{n!} = n+1$ (cancel $n!$) [1]. The identity $k! = k \times (k-1)!$ applied with $k = n+1$ shows that the factorial grows by exactly one multiplicative factor at each step.

Boss battle · The Factorial Forge

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering factorial and arrangement questions. A lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 1