Seat people in two or more labelled rows
Count row membership and seat positions separately, respecting named rows and restrictions within each row.
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
- Use combinations and permutations.
- Apply fixed-position and adjacency restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In two labelled rows of three seats, how many distinguishable seats are there?
What you will learn
- Count seating in labelled rows with distinguishable seats.
- Apply same-row or opposite-row restrictions without dividing away row labels.
Labelled seats distinguish every placement
Six different people fill a front row of three numbered seats and a back row of three numbered seats. All six seats are distinguishable, so there are 6! = 720 seatings. Equivalently choose the front-row members in 6C3 ways, arrange them in 3! ways and arrange the rest in the back row in 3! ways: 20 × 6 × 6 = 720.
Do not divide by 2! merely because there are two equal-sized rows. Interchanging front and back changes occupied labelled seats. For three labelled rows of two seats each, successively choose memberships in 6C2 × 4C2 ways and arrange each pair in (2!)³ ways, also giving 720. A circular seating convention is outside this syllabus section.
Row membership and within-row order are separate conditions
Require A and B to be in the same row of the two-row layout, not necessarily adjacent. Choose their row in two ways, their third row member from the other four in four ways, and arrange each row in 3! ways. The count is 2 × 4 × 6 × 6 = 288. Adjacency within that row would be a stronger requirement with a different count.
Require A and B to be in opposite rows instead. Choose which row contains A in two ways and choose A’s two row companions from the four other people in 4C2 ways. Arrange both rows in (3!)² ways: 2 × 6 × 36 = 432. The counts 288 and 432 add to all 720 seatings. Adjacent seats mean neighbours within one row here; the end of one row is not adjacent to the start of another.
Pause and explain
A and B must share one of two labelled three-seat rows. What count is correct?
Worked example
Seat A–F in front and back rows of three numbered seats, with A and B in the same row.
Show the worked solution
- Choose their labelled row in two ways.
- Choose its third person from C–F in four ways.
- Arrange each row in 3! ways: 2 × 4 × (3!)² = 288.
Answer 288 seatings.
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.
Seat six different people in front and back rows of three numbered seats. Count unrestricted seatings in two ways.
Give me a hint
Either fill six labelled seats directly or choose the front-row members first.
Compare my reasoning
- Six distinct seats give 6! = 720 seatings.
- Alternatively choose front membership in 6C3 = 20 ways.
- Arrange both rows in 3! ways each: 20 × 6 × 6 = 720.
720 seatings by either method.
Look for these in your work
- Preserve row and seat labels.
- Account for within-row order after membership selection.
In that two-row layout, A and B must be in opposite rows. Count seatings.
Give me a hint
Choose which row contains A, then choose A’s companions from the other four people.
Compare my reasoning
- Choose A’s row in two ways.
- Choose two companions from C–F in 4C2 = 6 ways; B is in the other row.
- Arrange both rows in (3!)² = 36 ways: 2 × 6 × 36 = 432.
432 seatings.
Look for these in your work
- Exclude B from A’s row companions.
- Retain both row orders and the row choice.
A–F sit in three labelled rows—front, middle and back—with two numbered seats in each. A is fixed at front seat 1 and B must sit in the back row. Count seatings.
Give me a hint
Choose B’s back seat before placing the remaining four people.
Compare my reasoning
- A’s seat is fixed in one way.
- Choose B’s back seat in two ways.
- Arrange C–F in the four unused distinguishable seats: 2 × 4! = 48.
48 seatings.
Look for these in your work
- Keep all three row labels and the remaining seat positions distinguishable.
- Apply B’s row condition after fixing A.
Common mistakes
- Front and back remain labelled even when row sizes match.
- Same row does not necessarily mean adjacent.
- State which seat relationships count as adjacency.
What can you explain now?
Why is there no division by 2! for front and back rows?
Compare with the explanation
Swapping the rows changes the labelled seats occupied by each person.
The swapped placement is a different outcome under this seat model.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Choose row memberships, arrange each row, and keep every named row and numbered seat distinct.
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.