Arrange repeated objects without counting identical copies twice
Derive the factorial correction for indistinguishable copies and retain every required repetition.
Section 5.2 · Permutations and combinations
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.
- Distinguish physical individuals from indistinguishable symbols.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many A copies occur in BANANA?
What you will learn
- Count distinct linear arrangements of repeated symbols.
- Explain why factorial division removes duplicate labelled descriptions.
Identical copies do not create new visible outcomes
LEVEL has five letters: two L, two E and one V. If temporary subscripts distinguish L₁, L₂ and E₁, E₂, there are 5! labelled arrangements. Each visible string occurs 2! × 2! times under these swaps. Hence there are 5!/(2!2!) = 30 distinct strings.
For n objects with indistinguishable groups of sizes r₁, r₂, …, the count is n!/(r₁!r₂!…). The sizes include every repeated group and sum to n when singleton groups are included. This formula counts arrangements using every specified copy; it is not a method for selecting a subset with replacement.
The outcome determines whether copies are distinguishable
BANANA has three A, two N and one B, so its distinct strings number 6!/(3!2!) = 60. AABBCC gives 6!/(2!2!2!) = 90. Removing repeated letters before counting would change the required objects rather than remove duplicates.
Six people wearing badges A, A, B, B, C, C are still six distinguishable people if their identities matter. Their seating orders number 6!, not 90. If the recorded outcome is only the visible badge string, then 90 is appropriate. State precisely what observers can distinguish and what the question counts.
Pause and explain
How many distinct strings use all five letters of LEVEL?
Worked example
Count distinct strings using every letter of LEVEL exactly once.
Show the worked solution
- There are five positions, giving 5! arrangements if every copy is labelled.
- Swaps of the two L copies and the two E copies leave the visible string unchanged.
- Divide by 2!2!: 120/4 = 30.
Answer 30 distinct strings.
Try a topic test
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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.
Count distinct strings made from all letters of LEVEL.
Give me a hint
Identify each repeated group before forming the denominator.
Compare my reasoning
- There are five letters: L twice, E twice and V once.
- Temporary labelled copies would give 5! = 120 orders.
- Divide by 2!2! to obtain 30 visible strings.
30 strings.
Look for these in your work
- Keep all five copies.
- Remove both sources of identical-copy duplication.
Count distinct strings made from all letters of BANANA.
Give me a hint
The multiplicities are three A, two N and one B.
Compare my reasoning
- Six positions give 6! = 720 labelled orders.
- Each visible string appears 3!2! = 12 times under identical-copy swaps.
- Divide 720/12 = 60.
60 strings.
Look for these in your work
- Identify both repeated groups correctly.
- Explain the denominator as duplicate descriptions.
Count distinct strings using AABBCC, then compare with seating six different people who wear those badge letters when identities matter.
Give me a hint
A visible letter string and an identified seating arrangement are different outcomes.
Compare my reasoning
- Strings have multiplicities 2, 2 and 2, so count 6!/(2!2!2!) = 90.
- Six individual people remain distinct even with repeated badges.
- Their seating orders number 6! = 720 when identities are recorded.
90 badge strings; 720 identified seating orders.
Look for these in your work
- State what is distinguishable.
- Do not transfer identical-symbol division to different people.
Common mistakes
- Repeated symbols are retained, not discarded.
- Divide once for each indistinguishable group.
- People do not become identical just because they share a label.
What can you explain now?
Do six different people with repeated badge letters have only 90 seating orders?
Compare with the explanation
No; if individual identities count, they have 6! = 720.
The count 90 records only the badge strings AABBCC.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Temporarily label identical copies, count their orders, then divide out copy swaps that change no visible outcome.
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.