Understand the idea.
Remove the ordering that does not matter
A combination is an unordered group. Listing the same committee members in another order does not create a different committee. First count ordered selections, then divide by , the number of listings of each group.
For distinct objects selected without replacement, applies to integers . Choosing the included objects also determines the excluded , so .
Use cases or complements for restricted groups
For a fixed number from each of several separate categories, choose within each category and multiply. For several permitted category counts, add their totals only when the cases are disjoint and together cover every allowed selection.
An at least one condition often has a simple complement. Count every group of the required size, then subtract groups containing none of the required type. This removes only forbidden groups and avoids counting multi-specialist groups several times.
See the method in action.
A committee
A club chooses 3 committee members from 8 people. Count the groups.
Worked solution
- 1
Count ordered lists of three members.
There are three successive selections without replacement.
- 2
Remove the order of the listing.
Each committee appears in six orders, but none of its members has a distinct role.
- 3
Count the committees.
Dividing by six counts each three-person group exactly once.
56.
At least one specialist
Choose 3 people from 5 designers and 4 programmers. How many groups contain at least one programmer?
Worked solution
- 1
Count all three-person groups.
The five designers and four programmers give nine distinct people.
- 2
Count groups with no programmer.
Every member of a forbidden group must be a designer.
- 3
Subtract the forbidden groups.
Every remaining group contains at least one programmer, including groups with two or three.
74.
Try it yourself.
Check your understanding
How many pairs can be chosen from 7 people?
Show answer & reasoning
21.
Ordered pairs number , but each unordered pair is listed twice. Thus .
Keep practising
6 worksheet questions · 14 marks
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 . 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
.
126. - Choose 2 red balls from 6 distinct red balls and 1 blue ball from 4 distinct blue balls. Count the selections.2 marks
.
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.
.
15. - Choose 3 from 5 adults and 3 children. How many groups have exactly 2 adults?2 marks
.
30. - How many 3-person groups from 8 include at least one of two named people?3 marks
.
36. - Explain why . Evaluate it.3 marks
An included group of 10 determines one excluded pair.
.
Both count choosing the 2 people left out; the value is 66.
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 ↗