Permutations | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

An ordered selection uses different roles or positions

A permutation counts a selection with distinguishable positions or roles. Gold, silver and bronze assigned to three people form a different result when two medal winners exchange places, even though the same people receive medals.

With $n$ distinct objects and $r$ positions, without repetition, take $r$ descending factors. The formula requires integers $0 \leq r \leq n$ ; its denominator removes the unused tail of the factorial, not the orders of the selected objects.

n P_{r} = \frac{n!}{(n - r)!}

Build restrictions into the counting model

For two objects that must stand together, temporarily treat them as one block. Arrange this block with the remaining objects, then multiply by the number of possible orders inside it. Each allowed row has one such block representation.

Indistinguishable repeated symbols need different treatment: divide by the factorial of each repeated group because interchanging identical symbols creates no new arrangement. These counts assume a straight row or labelled positions; unlabelled circular arrangements require separate consideration.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Award different medals

Nine finalists receive gold, silver and bronze, with no ties. Count the outcomes.

Worked solution

1
Assign the gold medal.
$9 choices$
Any finalist can receive gold.
2
Assign silver and bronze without reusing a finalist.
$\begin{aligned} silver & : 8 choices \\ bronze & : 7 choices \end{aligned}$
Each medal must go to a different person, and the medal roles are distinguishable.
3
Count the ordered outcomes.
$9 P_{3} = 9 \times 8 \times 7 = 504$
Exchanging medal roles changes the outcome, so no division by $3!$ is needed.

Answer

504.

EXAMPLE 02

Adjacent books

Five different books include two named novels that must stand together. Count the rows.

Worked solution

1
Combine the two named novels into one block.
$1 + 3 = 4 units$
The block and the other three books can be arranged as four distinct units.
2
Arrange the four units.
$4! = 24$
Ordering the units determines where the adjacent novels sit among the other books.
3
Arrange the novels inside the block.
$2! = 2$
Either of the two distinct novels may stand first within their block.
4
Count all permitted rows.
$N = 4! \times 2! = 24 \times 2 = 48$
Every block order admits both internal orders, with no duplicate rows.

Answer

48.

03 / YOUR TURN

Try it yourself.

Check your understanding

Choose a president and secretary from 6 people. How many outcomes?

Show answer & reasoning

30.

There are 6 choices for president, then 5 different people for secretary. The roles are distinct, so $6 \times 5 = 30$ outcomes.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Count ordered selections of 3 different letters from A, B, C, D, E.
2 marks
Seven runners compete for first and second. Count the ordered results.
2 marks
How many rows of 6 distinct students have Ava first?
2 marks
How many rows of 6 distinct students have Ava and Ben adjacent?
3 marks
Count distinct arrangements of the letters in LEVEL.
3 marks
From 8 people choose 3 ordered speakers. One particular person must speak first. Count the schedules.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Count ordered selections of 3 different letters from A, B, C, D, E.
2 marks
$5 P_{3} = 5 \times 4 \times 3 = 60$ .
60.
Seven runners compete for first and second. Count the ordered results.
2 marks
$7 \times 6 = 42$ .
42.
How many rows of 6 distinct students have Ava first?
2 marks
Fix Ava, then arrange the other 5: $5! = 120$ .
120.
How many rows of 6 distinct students have Ava and Ben adjacent?
3 marks
Arrange the pair block and 4 others in $5!$ ways.
Multiply by 2 orders of the pair: $240$ .
240.
Count distinct arrangements of the letters in LEVEL.
3 marks
There are 5 letters with L repeated twice and E repeated twice.
$\frac{5!}{2! 2!} = 30$ .
30.
From 8 people choose 3 ordered speakers. One particular person must speak first. Count the schedules.
2 marks
Fix that person first.
Choose the next two in $7 \times 6 = 42$ ways.
42.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: counting, binomial expansion, data, samples, summary measures and introductory normal distributions. These lessons provide the Counting and Statistics strand of the approved Methods preparation route.

Official curriculum source ↗

In this lesson

Understand the idea.

An ordered selection uses different roles or positions

Build restrictions into the counting model

See the method in action.

Award different medals

Assign the gold medal.

Assign silver and bronze without reusing a finalist.

Count the ordered outcomes.

Adjacent books

Combine the two named novels into one block.

Arrange the four units.

Arrange the novels inside the block.

Count all permitted rows.

Try it yourself.

Check your understanding

Keep practising