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.
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.
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:
$n \times (n-1) \times \cdots \times (n-r+1) = \dfrac{n!}{(n-r)!}$
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 $r > n$
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
Skills
- Evaluate $^nP_r$ by formula and by step-by-step multiplication
- Set up and solve permutation word problems
- Prove basic identities such as $^nP_{n-1} = n!$
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:
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:
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!
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?
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: $\dfrac{7!}{3!} = \dfrac{5040}{6} = 840$ ✓
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: $^8P_3 = 8 \times 7 \times 6 = 336$.
Worked examples · 3 in a row, reveal as you go
How many three-letter arrangements can be made from the letters A, B, C, D, E if no letter is repeated?
Evaluate $^7P_4$.
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)?
Fill the gap: $^6P_2 = 6 \times $ $= 30$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: selecting a president, vice-president and treasurer from 10 candidates requires a permutation rather than a combination.
Activities · practice with the ideas
Evaluate $^6P_2$ using the formula and verify using the product form.
How many ways can a president and vice-president be chosen from a committee of 8 people?
A PIN code uses 4 different digits from 0–9. How many distinct PIN codes are possible?
Evaluate $^{10}P_3$ using the product form only (no calculator for factorials).
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?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $^6P_2$. (1 mark)
Q2. How many ways can a president and vice-president be chosen from a committee of 8 people? (2 marks)
Q3. Show that $^nP_{n-1} = n!$. (2 marks)
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 $n!$.)
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering permutation questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.