Count unordered selections with combinations
Use nCr to count groups of r distinct items chosen from n when rearranging the same group does not create a new outcome.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which formula counts an unordered selection of r items from n?
What you will learn
- Use nCr = n!/[r!(n − r)!] to count selections
- Explain why each group is counted r! times by ordered arrangements
- Use symmetry nCr = nC(n − r) to simplify a calculation
A group has no internal order
For a selection of r items from n distinct items, the formula is nCr = n!/[r!(n − r)!]. The factor r! removes the different orders in which the same selected group could be listed.
For example, selecting A, B and C gives the same group as C, B and A. Count each group once, even though its members can be listed in 3! ways.
Choose the smaller side when it helps
Since selecting r items to include is equivalent to selecting the n − r items to leave out, nCr = nC(n − r). Thus 12C10 = 12C2, which is quicker to calculate.
Combinations apply to unranked teams, committees or sets. No selected item repeats, and there is no later arrangement stage in the selection question.
Pause and explain
Why is 10C8 equal to 10C2?
Worked example
How many different committees of 3 people can be chosen from 8 volunteers?
Show the worked solution
- Committee members have no ranks or positions, so order does not matter.
- Use 8C3 = 8!/[3!5!].
- Cancel 5!: (8 × 7 × 6)/(3 × 2 × 1) = 56.
Answer 56 committees.
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.
How many unranked teams of 3 can be formed from 8 players?
Give me a hint
A team is the same whichever order its members are listed.
Compare my reasoning
- Use 8C3 = 8!/(3!5!).
- Cancel 5! to get (8 × 7 × 6)/(3 × 2 × 1).
- Evaluate to 56.
56 teams.
Look for these in your work
- Unordered group identified
- Combination formula used
- Count simplified
How many groups of 4 can be selected from 12 distinct candidates?
Give me a hint
Use the complement symmetry to replace 12C4 only if it makes the arithmetic clearer.
Compare my reasoning
- The group is unordered, so calculate 12C4.
- Use 12C4 = (12 × 11 × 10 × 9)/(4 × 3 × 2 × 1).
- The quotient is 495.
495 groups.
Look for these in your work
- Selection model chosen
- Factorials simplified correctly
- Final integer checked
A greenhouse has 7 distinct seed varieties. How many ways can a grower select 2 varieties for a trial bed?
Give me a hint
The pair has no order and no variety is selected twice.
Compare my reasoning
- Use 7C2.
- Calculate 7 × 6/(2 × 1).
- Simplify to 21 pairs.
21 selections.
Look for these in your work
- Pair treated as unordered
- No repeated variety
- Combination count correct
Common mistakes
- Counting every listing order as a different group
- Dividing by the wrong number of arrangements
- Using nCr when positions are labelled
- Forgetting to simplify factorials before multiplying
What can you explain now?
How many different ways can a group of 2 be selected from 5 people?
Compare with the explanation
5C2 = 10
The two selected people can be listed in 2! orders, but each pair is one group.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why does the combination formula divide by r!, and when can symmetry make nCr easier to calculate?
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.