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YEAR 11 · COUNTING AND INTRODUCTORY STATISTICS · 20 MIN

Factorials

Use factorial notation to describe descending products and simplify expressions before calculating.

In this lesson

Before you start
  • Multiplication principle
01 / LEARN

Understand the idea.

Factorials count arrangements of distinct objects

For a positive integer n, factorial notation abbreviates the descending product from n to 1. Arranging distinct objects in a row produces this product because every filled position leaves one fewer object available.

This course uses factorials of non-negative integers. The convention 0!=1 counts one empty arrangement: there is exactly one way to put no objects into no positions. It also keeps counting formulas valid when every object is selected.

n!=n⁢(n−1)⁢⋯2×1,0!=1

Cancel shared factors before evaluating

To simplify a factorial fraction, expand the larger factorial only as far as the smaller one. For instance, 8!=8×7×6! exposes the common factor 6! without evaluating either large factorial.

Cancel common factors by division, then calculate the remaining product. A factorial sign applies to the whole number immediately before it; subtracting factorial arguments or cancelling individual digits is not a valid shortcut.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Arrange a display

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

Worked solution

  1. 1

    Count the choices by position.

    6, 5, 4, 3, 2, 1

    The postcards are distinct and no postcard can occupy two positions.

  2. 2

    Express the descending product.

    N=6!=6×5×4×3×2×1

    Every row is determined by filling all six positions in order.

  3. 3

    Evaluate the product.

    N=720

    The multiplication principle counts each distinct postcard row once.

Answer

720.

EXAMPLE 02

Simplify first

Evaluate 9!7!2!.

Worked solution

  1. 1

    Expose the common factorial.

    9!=9×8×7!

    The descending product from 7 to 1 is exactly the denominator factor 7!.

  2. 2

    Cancel the common factor.

    9!7!2!=9×82!

    Dividing numerator and denominator by the nonzero factor 7! leaves the ratio unchanged.

  3. 3

    Evaluate the remaining factorial and fraction.

    9×82×1=36

    The remaining denominator is 2!, which equals 2.

Answer

36.

03 / YOUR TURN

Try it yourself.

Check your understanding

What is 5!/3!?

Show answer & reasoning

20.

Write 5!=5×4×3!. Dividing by 3! leaves 5×4=20.

Keep practising

6 worksheet questions · 13 marks

Read the questions online
  1. Evaluate 4!.
    2 marks
  2. Simplify 8!6!.
    2 marks
  3. Evaluate 7!5!2!.
    2 marks
  4. How many orders can 5 different runners finish in, assuming no ties?
    2 marks
  5. Solve n!=120 for a non-negative integer n.
    2 marks
  6. 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.

  1. Evaluate 4!.
    2 marks

    4!=4×3×2×1=24.

    24.
  2. Simplify 8!6!.
    2 marks

    8!=8×7×6!.

    56.
  3. Evaluate 7!5!2!.
    2 marks

    7×62=21.

    21.
  4. How many orders can 5 different runners finish in, assuming no ties?
    2 marks

    5!=120.

    120.
  5. 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.
  6. 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 ↗

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