Use blocks for objects that must stay together
Treat adjacent objects as one unit, then count their permitted internal orders and any repeated units.
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
- Count permutations and repeated objects.
- Interpret adjacent as consecutive positions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many outer units remain when A and B form one block among five people?
What you will learn
- Use a block to enforce adjacency in a linear arrangement.
- Account for internal order and indistinguishable objects outside a block.
A block reduces the number of outer units
Five different people A, B, C, D and E stand in a row with A and B adjacent. Replace them by one block, leaving four units: [AB], C, D and E. Arrange these units in 4! ways and arrange the two block members in 2! ways. The count is 4! × 2! = 48.
If A must be immediately to the left of B, the block has only one allowed internal order, AB, and the count becomes 4! = 24. For six different people with A, B and C occupying three consecutive positions in any order, there are four outer units and 3! internal orders, giving 4! × 3! = 144.
A repeated block needs its own outcome check
For strings using AABB with the A copies adjacent, regard AA as one fixed block. The outer units are [AA], B and B, so their arrangements number 3!/2! = 3. They are AABB, BAAB and BBAA. The identical A copies do not give an extra 2! internal factor.
A block method works when every valid arrangement is represented exactly once by a block location and permitted internal order. Overlapping conditions such as ‘A next to B or C’ require disjoint cases or an overlap correction; separate block counts can include the same arrangement twice. Confirm exactly which adjacency is required.
Pause and explain
A and B must be adjacent among five different people. How many rows?
Worked example
Arrange A, B, C, D, E in a row with A next to B.
Show the worked solution
- Replace A and B by a single unit, giving four outer units.
- Their outer orders number 4! = 24.
- The block may be AB or BA, so multiply by two to get 48.
Answer 48 arrangements.
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.
Arrange A, B, C, D, E in a row with A and B adjacent, in either order.
Give me a hint
Count the block and the other three people as four units.
Compare my reasoning
- There are 4! = 24 outer orders.
- The block has two orders, AB and BA.
- Multiply 24 × 2 = 48.
48 arrangements.
Look for these in your work
- Enforce adjacency through one block.
- Include both allowed internal orders.
Six different people A–F form a row. A, B and C must occupy three consecutive positions in any order. Count arrangements.
Give me a hint
Three members form one block alongside the other three people.
Compare my reasoning
- The outer units are [ABC], D, E and F, giving 4! = 24 orders.
- The block’s members have 3! = 6 internal orders.
- The count is 24 × 6 = 144.
144 arrangements.
Look for these in your work
- Use four outer units rather than three or six.
- Preserve every permitted internal order.
Count distinct strings using AABB in which the A copies are adjacent. List the outcomes to check your count.
Give me a hint
Use a fixed AA block and two identical B units.
Compare my reasoning
- Arrange [AA], B, B in 3!/2! = 3 ways.
- The AA block has only one visible internal order.
- List AABB, BAAB and BBAA to confirm all three.
3 strings: AABB, BAAB, BBAA.
Look for these in your work
- Correct for identical B copies outside the block.
- Do not multiply for swapping identical A copies.
Common mistakes
- Count the new outer units carefully.
- Include internal orders only when they are permitted and distinct.
- Overlapping block cases may double-count arrangements.
What can you explain now?
What changes if A must be immediately before B?
Compare with the explanation
Only the internal order AB is allowed, so the count is 24.
Adjacency alone allows both AB and BA; directed adjacency does not.
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Make it stick
Form the required block, arrange outer units, then count distinct permitted internal orders.
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.