Combinations | Year 11 | Ascent Library

01 / LEARN

The idea, step by step

Remove the ordering that does not matter

A combination is a selection in which order does not matter. A committee containing Ali, Bo and Chen is unchanged when their names are listed in a different order. The permutation formula counts each r-person group r! times, so dividing by r! counts each group once.

(\begin{matrix} n \\ r \end{matrix}) = \frac{n!}{r! (n - r)!}

Use cases or complements for restricted groups

If a group needs specified numbers from separate categories, choose within each category and multiply. If a condition says at least one, it may be easier to count all groups then subtract groups with none. Cases must not overlap. The identity n choose r = n choose n−r follows because choosing who is included also determines who is excluded.

A common trap

Using combinations for captain and vice-captain.

Instead: Roles make order matter. Decide what counts as the same outcome before choosing a formula.

02 / WORKED EXAMPLES

See the thinking

Example 1

A committee

A club chooses 3 committee members from 8 people. Count the groups.

$(\begin{matrix} 8 \\ 3 \end{matrix}) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ .
No roles were specified, so order does not matter.

56.

Example 2

At least one specialist

Choose 3 people from 5 designers and 4 programmers. How many groups contain at least one programmer?

All groups: $(\begin{matrix} 9 \\ 3 \end{matrix}) = 84$ .
No programmers: $(\begin{matrix} 5 \\ 3 \end{matrix}) = 10$ .
Subtract: $84 - 10 = 74$ .

74.

Check your understanding

How many pairs can be chosen from 7 people?

Show answer

21.

$(\begin{matrix} 7 \\ 2 \end{matrix}) = 7 \times 6 / 2 = 21$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Choose 4 students from a group of 9. Count the groups.
2 marks
Choose 2 red balls from 6 distinct red balls and 1 blue ball from 4 distinct blue balls. Count the selections.
2 marks
Choose a team of 3 from 7 people, including one named person. Count the teams.
2 marks
Choose 3 from 5 adults and 3 children. How many groups have exactly 2 adults?
2 marks
How many 3-person groups from 8 include at least one of two named people?
3 marks
Explain why $(\begin{matrix} 12 \\ 10 \end{matrix}) = (\begin{matrix} 12 \\ 2 \end{matrix})$ . Evaluate it.
3 marks

Read the worked solutions

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

Choose 4 students from a group of 9. Count the groups.
2 marks
$(\begin{matrix} 9 \\ 4 \end{matrix}) = 126$ .
126.
Choose 2 red balls from 6 distinct red balls and 1 blue ball from 4 distinct blue balls. Count the selections.
2 marks
$(\begin{matrix} 6 \\ 2 \end{matrix}) (\begin{matrix} 4 \\ 1 \end{matrix}) = 15 \times 4 = 60$ .
60.
Choose a team of 3 from 7 people, including one named person. Count the teams.
2 marks
Fix the named person and choose 2 from the remaining 6.
$(\begin{matrix} 6 \\ 2 \end{matrix}) = 15$ .
15.
Choose 3 from 5 adults and 3 children. How many groups have exactly 2 adults?
2 marks
$(\begin{matrix} 5 \\ 2 \end{matrix}) (\begin{matrix} 3 \\ 1 \end{matrix}) = 10 \times 3 = 30$ .
30.
How many 3-person groups from 8 include at least one of two named people?
3 marks
$(\begin{matrix} 8 \\ 3 \end{matrix}) - (\begin{matrix} 6 \\ 3 \end{matrix}) = 56 - 20 = 36$ .
36.
Explain why $(\begin{matrix} 12 \\ 10 \end{matrix}) = (\begin{matrix} 12 \\ 2 \end{matrix})$ . Evaluate it.
3 marks
An included group of 10 determines one excluded pair.
$(\begin{matrix} 12 \\ 2 \end{matrix}) = 12 \times 11 / 2 = 66$ .
Both count choosing the 2 people left out; the value is 66.

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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.

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