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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 calculatinghow 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 confidencethat 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