Factorials | Year 11 | Ascent Library

01 / LEARN

The idea, step by step

Factorials count arrangements of distinct objects

For a positive integer n, n factorial is the product of all whole numbers from n down to 1. For n different books in a row, there are n choices for the first place, n−1 for the next and so on, giving n! arrangements. The definition applies to non-negative integers in this course; do not interpret 4.5! using this definition.

n! = n (n - 1) \dots 2 \times 1, 0! = 1

Cancel shared factors before evaluating

Write a larger factorial as a product ending in the smaller one. For example, 8! = 8×7×6!, so 8!/6! = 8×7. This keeps arithmetic small and exposes the counting structure. The convention 0! = 1 agrees with there being exactly one way to arrange an empty collection, and makes selection formulas work at their endpoints.

A common trap

Treating $6!$ as $6 \times 5$ or as $6^{6}$ .

Instead: The factors decrease all the way to 1. Write the product if the meaning is unclear.

02 / WORKED EXAMPLES

See the thinking

Example 1

Arrange a display

Six different postcards are placed in a row. Count the arrangements.

There are 6, 5, 4, 3, 2 and 1 choices for the successive positions.
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ .

720.

Example 2

Simplify first

Evaluate $\frac{9!}{7! 2!}$ .

$9! = 9 \times 8 \times 7!$ .
$\frac{9!}{7! 2!} = \frac{9 \times 8}{2} = 36$ .

36.

Check your understanding

What is $5! / 3!$ ?

Show answer

20.

$5! = 5 \times 4 \times 3!$ , so the ratio is $5 \times 4$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Evaluate $4!$ .
2 marks
Simplify $8! / 6!$ .
2 marks
Evaluate $7! / (5! 2!)$ .
2 marks
How many orders can 5 different runners finish in, assuming no ties?
2 marks
Solve $n! = 120$ for a non-negative integer n.
2 marks
Explain why $0!$ is 1 rather than 0 in the expression $5! / (5! 0!)$ .
3 marks

Read the worked solutions

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

Evaluate $4!$ .
2 marks
$4! = 4 \times 3 \times 2 \times 1 = 24$ .
24.
Simplify $8! / 6!$ .
2 marks
$8! = 8 \times 7 \times 6!$ .
56.
Evaluate $7! / (5! 2!)$ .
2 marks
$7 \times 6 / 2 = 21$ .
21.
How many orders can 5 different runners finish in, assuming no ties?
2 marks
$5! = 120$ .
120.
Solve $n! = 120$ for a non-negative integer n.
2 marks
$4! = 24$ , $5! = 120$ , and factorials increase for integers at least 2.
n = 5.
Explain why $0!$ is 1 rather than 0 in the expression $5! / (5! 0!)$ .
3 marks
With $0! = 1$ , $5! / (5! 0!) = 1$ .
Using zero would cause division by zero and break the counting interpretation.
Choosing all 5 objects should have one possible selection, so the expression must be 1.

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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 ↗

Back to Counting and introductory statistics