Count ordered arrangements using permutations

Alberta Grade 12 Mathematics

Permutations, Combinations, and Binomial Theorem

Math 30-1 study topic PC2

A school club needs a president, vice-president, and secretary. Choosing three students is not enough: each student must be assigned a different role. If Maya is president and Luis is vice-president, that arrangement differs from one in which Luis is president and Maya is vice-president. This lesson shows how to count arrangements when order or assigned positions matter. It uses the multiplication principle, factorials, and permutation notation.

What you will learn

Recognize when the order of selected items matters.
Use the multiplication principle to count ordered choices.
Calculate permutations using factorial notation and the formula for selecting and arranging some of a set.
Explain why two similar counting situations may require different methods.

1. Review: choices and order

Before counting arrangements, recall the multiplication principle. If one step can be done in a certain number of ways, and a second step can then be done in another number of ways, multiply the counts to find the number of complete outcomes. For example, if there are four choices for a first position and three remaining choices for a second position, there are twelve ordered outcomes.

An arrangement is ordered when changing the positions of items creates a different outcome. The letters AB and BA are different ordered arrangements. If a task only asks which two letters were selected, without assigning positions, AB and BA describe the same selection. This lesson focuses on ordered arrangements.

4\times3=12

Ask whether switching the selected items or their positions changes the outcome.
When order matters, count the available choices for each position.

2. Permutations and factorials

A permutation is an ordered arrangement of items chosen from a set. To arrange all items in a set of

n

distinct items, there are

n

choices for the first position,

n-1

for the second, and so on, until one choice remains. Multiplying these counts gives the total.

The symbol n! is read “

n

factorial.” It means the product of the positive whole numbers from

n

down to one. By definition,

0!=1

. Factorial notation is useful because it writes a long product compactly.

Sometimes a task asks you to fill only

r

positions from

n

distinct items, with no item used more than once. The first position has

n

choices, the next has

n-1

, and the pattern continues for

r

positions. The notation

{}_nP_r

means the number of ways to arrange

r

items chosen from

n

. Its formula is

n!/(n-r)!

. Here,

n

and

r

are whole numbers with

0\leq r\leq n

.

The denominator removes the unused factors from n!. For instance, when only three positions are filled from a set of seven, the needed product is

7\times6\times5

. The factorial formula gives the same product after the common factors cancel. The formula applies because the choices are distinct and no item is repeated.

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

For an arrangement of all $n$ distinct items, use n!.
For $r$ ordered positions filled from $n$ distinct items without repetition, use ${}_nP_r$ .
Check that $r$ does not exceed $n$ .

3. A table for ordered choices

A slots table makes the choice at each position visible. A slot is one position to be filled. The number above each slot is the number of choices available at that step. Each choice reduces the next count when repetition is not allowed.

For example, suppose five distinct books are available and three different books will be placed in first, second, and third positions. The table shows five choices for the first position, four for the second, and three for the third. Multiplying gives the total number of arrangements. The slot counts are not all five, because a book already placed cannot be selected again.

5\times4\times3={}_5P_3=60

A slot count must reflect what has already been chosen.
Multiply the counts for all positions to get the total.

4. Choosing a method and checking the result

Start by identifying what counts as a different outcome. If the task assigns distinct positions, such as first, second, and third, order matters. If it only asks for a group and does not distinguish positions, the permutation method is not the right count. The wording and purpose of the task determine whether an arrangement is ordered.

Next, identify the number of available distinct items and the number of positions. If every item is used, count down to one and use a factorial. If only some items are used, count down for just the required number of positions or apply the permutation formula. State whether repetition is allowed; the formula in this lesson assumes that an item cannot be used twice.

Check the size of the answer. For

r

positions chosen without repetition, the first slot has

n

choices and later slots have fewer. The total should be a whole number and should not exceed the count you would get if every slot had

n

choices. A calculator can help with large factorials, but write the setup first so the meaning of the calculation is clear.

{}_nP_r=n(n-1)\cdots(n-r+1)

Decide whether order matters before calculating.
Use the stated number of distinct items and positions.
The permutation formula counts without repetition.

Choices for three positions from five books

Position	Choices remaining
First	5
Second	4
Third	3
Total arrangements	$5\times4\times3=60$

Worked example

Three roles from six students

Six students are available to fill the roles of chair, recorder, and presenter. No student can fill more than one role. How many assignments are possible?

Three distinct roles

The slot counts decrease because each student can be assigned only once.

Check whether order matters
The roles are different, so assigning one student to chair and another to recorder is different from switching those assignments. Order matters because the positions have distinct names.
Count choices for each role
There are six choices for chair. After that choice, five students remain for recorder, then four for presenter. Multiplying counts the complete assignments.
$6\times5\times4$
Use permutation notation
There are three ordered positions filled from six distinct students without repetition, so the permutation notation represents the same product.
${}_6P_3=\frac{6!}{3!}=120$

Answer: There are 120 possible assignments.

Check: The first choice has six options, and each later position has fewer options. The count is therefore less than

6^3=216

, as expected when students cannot be reused.

Worked example

Arranging a full set

A display has four different paintings. In how many orders can all four paintings be placed in a row?

Four painting positions

All four paintings are arranged, so the choices decrease to one.

Count the choices by position
Any of the four paintings can go first. Once it is placed, three remain for the second position, then two for the third, and one for the last.
$4\times3\times2\times1$
Write the product as a factorial
All four distinct paintings are being arranged, so the descending product is four factorial.
$4!=24$

Answer: There are 24 possible orders.

Check: For the first position there are four choices; for the last position there is only one remaining painting. The product includes each position exactly once.

Worked example

A code with distinct digits

A four-digit code is made by arranging four different digits chosen from the digits 1 through 8. No digit can repeat. How many codes are possible?

Four code positions

Each used digit reduces the choices for the next position.

Identify the positions and choices
A code has four ordered positions. There are eight choices for its first digit, then seven, six, and five choices as digits are used.
$8\times7\times6\times5$
Apply the permutation formula
Four positions are filled from eight distinct digits without repetition. Substitute eight for $n$ and four for $r$ .
${}_8P_4=\frac{8!}{4!}=1680$

Answer: There are 1,680 possible codes.

Check: The order of digits changes the code, so permutations apply. The count is less than the number of four-position strings that would allow all eight digits in every position.

Common mistakes and how to avoid them

Using the total number of items for every position when items cannot be repeated.

Correction: Reduce the available choices after each position is filled. For example, three positions from five distinct items use counts of five, four, and three.

Using a permutation when switching positions does not create a different outcome.

Correction: First decide whether the task distinguishes the positions. Permutations count ordered arrangements.

Using n! when only some of the items are arranged.

Correction: For

r

positions from

n

items, use

{}_nP_r

, or multiply the

r

descending choice counts.

Forgetting that the permutation formula assumes no repetition.

Correction: Check the wording. If an item may be reused, the decreasing-choice formula in this lesson does not describe the task.

Lesson summary

An ordered arrangement is one in which changing positions creates a different outcome.
Use the multiplication principle to count choices position by position.
Arrange all $n$ distinct items in n! ways.
Arrange $r$ items chosen from $n$ distinct items without repetition in ${}_nP_r=n!/(n-r)!$ ways.

Check your understanding

Question 1

Five runners are assigned first, second, and third place. No ties are possible. How many ordered results are possible?

15
60
125
120

Show answer and explanation

60

The places are distinct, so order matters. The counts are five choices for first, four for second, and three for third:

{}_5P_3=5\times4\times3=60

.

Question 2

How many ways can all six different photographs be arranged in a row?

36
120
720
6

Show answer and explanation

720

All six photographs are arranged, so use a factorial:

6!=720

.

Key terms

Ordered arrangement: A placement of items in positions where switching positions creates a different result.
Permutation: An ordered arrangement of selected items from a set.
Factorial: The product of the positive whole numbers from a given whole number down to one; by definition, $0!=1$ .
Multiplication principle: A counting rule that multiplies the number of choices at each step to count complete outcomes.
Repetition: Using the same item more than once in an arrangement.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation PC2. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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PC2 · Count ordered arrangements using permutations

Alberta Grade 12 Mathematics

Math 30-1 study topic PC2

What you will learn

1. Review: choices and order

2. Permutations and factorials

3. A table for ordered choices

4. Choosing a method and checking the result

Choices for three positions from five books

Three roles from six students

Arranging a full set

A code with distinct digits

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through Math 30-1

About this lesson and its review