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Module 4 · L3 of 15 ~35 min ⚡ +95 XP available

Permutations ($^nP_r$)

You have 10 athletes competing for gold, silver and bronze. That is a selection problem — but it's also an ordering problem, because first place matters. How many podium possibilities exist? The permutation formula $^nP_r = \dfrac{n!}{(n-r)!}$ answers every question like this in one step. In this lesson you'll derive it, practise it, and learn exactly when to reach for it.

Today's hook — You have 5 letters: A, B, C, D, E. Before looking up the formula, how many different three-letter arrangements (order matters, no repeats) do you think exist? Jot your estimate. You'll check it after card 05.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

You have 5 letters: A, B, C, D, E. Without using a formula — estimate how many different three-letter arrangements (order matters, no repeats) you can make. Write your reasoning below.

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The two moves
+5 XP to read

Every permutation question comes down to two decisions: recognise that order matters, then apply $^nP_r = \dfrac{n!}{(n-r)!}$ or the equivalent step-by-step count.

Think of filling $r$ labelled slots from a pool of $n$ objects. The first slot has $n$ choices, the second has $n-1$, and so on down to $n-r+1$ choices for the last slot. Multiply those together:

$n \times (n-1) \times \cdots \times (n-r+1) = \dfrac{n!}{(n-r)!}$

Slot 1 n Slot 2 n−1 Slot r n−r+1 × = n! / (n−r)! = ⁿPᵣ
$^nP_r = \dfrac{n!}{(n-r)!}$
Order matters
AB and BA are different permutations. If order doesn't matter, you want combinations ($^nC_r$) — not covered until Lesson 6.
Special cases
$^nP_n = n!$ (arrange all), $^nP_1 = n$ (pick one), $^nP_0 = 1$ (choose none — one way to do nothing).
r cannot exceed n
$^nP_r$ is undefined (or zero by convention) when $r > n$. You can't choose 6 objects from a pool of 4.
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What you'll master
Know

Key facts

  • $^nP_r = \dfrac{n!}{(n-r)!}$ counts ordered arrangements of $r$ objects from $n$
  • $^nP_n = n!$, $^nP_1 = n$, $^nP_0 = 1$
  • $^nP_r$ is zero (or undefined) when $r > n$
Understand

Concepts

  • Why the denominator $(n-r)!$ cancels the objects that were never selected
  • The connection between slot-by-slot counting and the formula
  • The distinction between permutations and combinations
Can do

Skills

  • Evaluate $^nP_r$ by formula and by step-by-step multiplication
  • Set up and solve permutation word problems
  • Prove basic identities such as $^nP_{n-1} = n!$
04
Key terms
PermutationAn ordered arrangement of objects. AB and BA are different permutations of the same two letters.
$^nP_r$The number of ways to arrange $r$ objects chosen from $n$ distinct objects, where order matters.
$n$ (pool size)The total number of distinct objects available to choose from.
$r$ (selection size)The number of objects being arranged. Must satisfy $0 \leq r \leq n$.
Factorial $n!$$n! = n \times (n-1) \times \cdots \times 2 \times 1$, with $0! = 1$ by convention.
Without replacementEach object can only be selected once. This is the default assumption in permutation problems unless stated otherwise.
05
The permutation formula
core concept

A permutation is an arrangement of objects where order matters. When we select $r$ objects from $n$ distinct objects and arrange them in a row, we use:

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

Why does this formula work? We fill $r$ positions one at a time. The first position can be filled in $n$ ways, the second in $n-1$ ways (since one object is used), and so on until the $r$-th position has $n-r+1$ choices. This gives:

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

The $(n-r)!$ in the denominator exactly cancels the $n-r$ factorial terms we did not use.

Example: Three-letter arrangements from A, B, C, D, E (no repeats):

$^5P_3 = \dfrac{5!}{(5-3)!} = \dfrac{5!}{2!} = \dfrac{120}{2} = 60$

So there are 60 arrangements. Check your estimate from card 01!

Real-world example. Ten athletes compete for gold, silver and bronze. The number of possible podium outcomes is $^{10}P_3 = \dfrac{10!}{7!} = 10 \times 9 \times 8 = 720$. Order matters because swapping gold and silver gives a different result.

Permutation: ordered arrangement of r objects from n — use when order matters; ^nP_r = n!{(n-r)!} — or equivalently n(n-1)(n-2)(n-r+1)

Pause — copy the permutation formula into your book: $^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$ — $r$ descending factors starting from $n$.

Quick check: Which expression correctly counts the number of ways to arrange 4 objects chosen from 9 distinct objects?

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Evaluating $^nP_r$ step by step
core concept

We just saw that $^nP_r = \frac{n!}{(n-r)!}$ counts ordered arrangements of $r$ objects from $n$. That raises a question: the formula gives the same answer whether you use it as a fraction or as $r$ descending factors — which form is faster to compute by hand? This card answers it → write $r$ descending factors starting at $n$: $n \times (n-1) \times \cdots \times (n-r+1)$, avoiding the need to compute large factorials.

You can evaluate $^nP_r$ in two equivalent ways:

  • Using the formula: $^nP_r = \dfrac{n!}{(n-r)!}$. Calculate numerator and denominator separately, then divide.
  • Using the product: $^nP_r = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)$, writing exactly $r$ factors starting from $n$.

The product form is often faster for calculation without a calculator. For example:

$^7P_4 = 7 \times 6 \times 5 \times 4 = 840$   (four factors, starting at 7, decreasing by 1 each time)

Check: $\dfrac{7!}{3!} = \dfrac{5040}{6} = 840$ ✓

Notation note. $^nP_r$, $P(n,r)$, and $P_r^n$ all mean the same thing. The HSC uses $^nP_r$. Some textbooks and calculators use $P(n,r)$ or the notation $nPr$ on a key.

Product form: write r descending factors starting from n:   n, n-1, n-2, , n-r+1; Formula form: ^nP_r = n!{(n-r)!}

Pause — copy both computation forms into your book: product form — write $r$ descending factors from $n$; formula form — $^nP_r = n!/(n-r)!$; use product form when $r$ is small to avoid computing large factorials.

Did you get this? True or false: $^8P_3 = 8 \times 7 \times 6 = 336$.

PROBLEM 1 · LETTER ARRANGEMENTS

How many three-letter arrangements can be made from the letters A, B, C, D, E if no letter is repeated?

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$n = 5$ (five letters), $r = 3$ (choosing three), order matters $\Rightarrow$ use $^nP_r$
Identify $n$ and $r$. Since AB and BA are different arrangements, order matters.
PROBLEM 2 · EVALUATE

Evaluate $^7P_4$.

1
$^7P_4 = \dfrac{7!}{(7-4)!} = \dfrac{7!}{3!}$
Apply the formula with $n=7$, $r=4$.
PROBLEM 3 · WORD PROBLEM

A club has 12 members. In how many ways can a president, vice-president and secretary be elected (assuming one person cannot hold two roles)?

1
Three distinct roles $\Rightarrow$ order matters (president ≠ vice-president). Use $^{12}P_3$.
The three positions are labelled and different, so swapping two people gives a different outcome.

Fill the gap: $^6P_2 = 6 \times $ $= 30$ .

Trap 01
Confusing $n$ and $r$
$n$ is the total pool of objects; $r$ is how many you select and arrange. In $^5P_3$, you have 5 objects but only arrange 3. Writing $^3P_5$ instead is undefined — always identify the pool and the selection size first.
Trap 02
Forgetting that order matters
AB and BA are different permutations. If the problem asks for "teams" or "groups" where members are interchangeable, you need combinations ($^nC_r$), not permutations. Key trigger words for permutations: arrange, order, first/second/third, ranking.
Trap 03
Using $r > n$
$^nP_r$ requires $r \leq n$ (without replacement). If you try to select 6 from 4 objects, $(n-r)! = (-2)!$ is undefined. Always check the constraint — if $r > n$, the answer is 0.

Did you get this? True or false: selecting a president, vice-president and treasurer from 10 candidates requires a permutation rather than a combination.

Work mode · how are you completing this lesson?
1

Evaluate $^6P_2$ using the formula and verify using the product form.

2

How many ways can a president and vice-president be chosen from a committee of 8 people?

3

A PIN code uses 4 different digits from 0–9. How many distinct PIN codes are possible?

4

Evaluate $^{10}P_3$ using the product form only (no calculator for factorials).

5

Show that $^nP_{n-1} = n!$. (Hint: substitute $r = n-1$ into the formula.)

Odd one out: Three of these evaluations are correct. Which one is NOT?

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Revisit your thinking

Earlier you estimated the number of three-letter arrangements from A, B, C, D, E.

The exact answer is $^5P_3 = 5 \times 4 \times 3 = \mathbf{60}$. Did your intuition come close? The key insight is that each successive slot has one fewer choice because we cannot repeat a letter — slot 1 has 5, slot 2 has 4, slot 3 has 3. Multiply those choices together and you have the answer.

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Multiple choice
+5 XP per correct · +25 XP all-correct

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

02
Short answer
ApplyBand 31 mark

Q1. Evaluate $^6P_2$. (1 mark)

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ApplyBand 42 marks

Q2. How many ways can a president and vice-president be chosen from a committee of 8 people? (2 marks)

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AnalyseBand 52 marks

Q3. Show that $^nP_{n-1} = n!$. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $^6P_2 = 6 \times 5 = 30$  ·  2. $^8P_2 = 8 \times 7 = 56$  ·  3. $^{10}P_4 = 10 \times 9 \times 8 \times 7 = 5040$  ·  4. $^{10}P_3 = 10 \times 9 \times 8 = 720$  ·  5. $^nP_{n-1} = \dfrac{n!}{(n-(n-1))!} = \dfrac{n!}{1!} = n!$

Q1 (1 mark): $^6P_2 = \dfrac{6!}{4!} = 6 \times 5 = 30$ [1].

Q2 (2 marks): $n=8$, $r=2$. The two roles are distinct so order matters [1]. $^8P_2 = 8 \times 7 = \mathbf{56}$ ways [1].

Q3 (2 marks): Substitute $r = n-1$: $^nP_{n-1} = \dfrac{n!}{(n-(n-1))!} = \dfrac{n!}{1!} = \dfrac{n!}{1} = n!$ [2]. (Award 1 mark for correct substitution, 1 mark for reaching $n!$.)

01
Boss battle · The Arrangement Master
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
02
Science Jump · platform challenge

Climb platforms by answering permutation questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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