Count unordered selections with combinations
Remove repeated orderings when only membership matters, and restore role assignments when the problem requires them.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.2. State whether order matters, objects are distinguishable, repetition is permitted and positions or rows are labelled. Count mutually exclusive cases before adding them; account for internal arrangements when using blocks. These lessons include seating in two or more rows. Arrangements in a circle are excluded from this syllabus section. Unless a question says otherwise, people and named roles are distinguishable, and selection is without replacement. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate factorials and ordered selections.
- Identify which positions or roles are labelled.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which outcomes are identical for an ordinary two-person committee?
What you will learn
- Derive nCr by dividing ordered selections by r!.
- Separate selecting a group from allocating its roles.
Membership is the complete outcome
For the names A, B and C, selecting two gives AB, AC and BC: three sets. Ordered pairs give AB, BA, AC, CA, BC and CB: six. Each set occurs 2! times in the ordered list. When order is irrelevant, divide the ordered count by r! to remove this repeated counting.
The formula is nCr = n!/[r!(n − r)!] for integer 0 ≤ r ≤ n. Seven candidates choosing a committee of three give 7C3 = 35. The symmetry nCr = nC(n − r) follows because choosing the included people determines the excluded people. Choosing no one or everyone each gives one set.
Selection and roles can be successive stages
Choosing a three-person committee from six and then choosing its chair gives 6C3 × 3 = 60. A chair is singled out, but the other two members have no different roles. Equivalently choose the chair first in six ways and two other members from the remaining five: 6 × 5C2 = 60.
Do not use 6P3 = 120 for this committee-with-chair model: that incorrectly distinguishes the two ordinary members. Conversely, using only 6C3 = 20 ignores the chair. A formula is justified by the definition of the complete outcome, not a keyword such as ‘choose’.
Pause and explain
Select two members from A, B and C with no roles. How many sets?
Worked example
Select two names from A, B and C, first without roles and then for two labelled roles.
Show the worked solution
- Without roles, list AB, AC and BC: three sets.
- For two labelled roles, each set has two orders: AB/BA, AC/CA, BC/CB.
- Thus 3C2 = 3 and 3P2 = 6.
Answer 3 unordered sets; 6 ordered assignments.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Choose a three-person committee with no roles from seven candidates. Count the committees.
Give me a hint
Divide ordered three-person selections by the number of internal orders.
Compare my reasoning
- Ordered selections number 7 × 6 × 5 = 210.
- Each committee appears 3! = 6 times in that list.
- The count is 210/6 = 7C3 = 35.
35 committees.
Look for these in your work
- Define a committee as a set of members.
- Explain the division by six.
Select two different books from eight for an unordered reading shortlist. Count shortlists.
Give me a hint
Swapping the order in which the books are named does not change the shortlist.
Compare my reasoning
- Ordered choices give 8 × 7 = 56.
- Each pair is named twice, once in each order.
- Divide by 2! to get 8C2 = 28.
28 shortlists.
Look for these in your work
- Use no replacement.
- Remove only the duplicate ordering.
Select a three-person committee from six candidates and appoint one member as chair; the other two have equal roles. Count outcomes in two ways.
Give me a hint
Either choose the members first or choose the chair first.
Compare my reasoning
- Choose the group then chair: 6C3 × 3 = 20 × 3 = 60.
- Choose chair then the remaining pair: 6 × 5C2 = 6 × 10 = 60.
- Do not assign an order to the two ordinary members.
60 outcomes by either method.
Look for these in your work
- Include the distinguished chair in each outcome.
- Keep ordinary members unordered.
Common mistakes
- A combination records membership rather than sequence.
- Divide by r! only when all those orders represent the same outcome.
- Treat named roles as an additional stage when appropriate.
What can you explain now?
Why does nCr equal nC(n − r)?
Compare with the explanation
Each selected set determines exactly one complementary excluded set.
This is a one-to-one correspondence, not an assumption that order matters.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Ask whether swapping selected names changes the outcome; combine membership choices with any distinct roles.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.