Check mixed restrictions with disjoint cases and small lists
Combine selection and arrangement carefully, correct overlapping exclusions and verify a count through a second method.
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, repeated-object counts, blocks and gaps.
- Distinguish disjoint cases from overlapping descriptions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Are AB-adjacent and CD-adjacent rows necessarily disjoint?
What you will learn
- Handle more than one condition without double-counting.
- Check a counting model by listing a small case or using an independent method.
An overlap correction protects the complement
Arrange A, B, C, D in a row with neither A next to B nor C next to D. All rows number 4! = 24. The forbidden AB adjacency has 2!3! = 12 rows, and CD adjacency also has 12. Their overlap has two blocks in 2! outer orders and 2! internal orders for each block, giving 8.
Subtracting both forbidden counts removes overlap twice, so add it back once: 24 − 12 − 12 + 8 = 8 valid rows. This is the two-set inclusion–exclusion principle. Check that the blocks are disjoint; for overlapping named pairs such as AB and AC, the intersection has a different structure and cannot use two independent blocks.
Select first, then arrange under the remaining conditions
Choose three people from A, B, C, D, E, F and place them in a row with A selected but not at an end. A must occupy the middle. Choose two of the other five in 5C2 = 10 ways and arrange them at the ends in 2! ways, giving 20. Alternatively choose the left end in five ways and the right in four, also giving 20.
For AABBCC strings with different endpoint letters, first count equal endpoints. There are three disjoint choices AA, BB or CC for the end pair, each leaving four letters with two identical pairs: 4!/(2!2!) = 6 strings. From the total 90 subtract 18 equal-end strings to get 72. A direct count chooses ordered different endpoint types in 3 × 2 ways and arranges remaining multiplicities 1, 1, 2 in 4!/2! ways, also giving 72. A small exhaustive list checks the model; it does not replace an explained general method.
Pause and explain
A–D rows forbid both AB and CD adjacency. What is the count?
Worked example
Arrange A–D with neither AB nor CD adjacent.
Show the worked solution
- There are 24 unrestricted rows; each forbidden adjacency has 12 rows.
- Both forbidden adjacencies occur together in 2! × 2! × 2! = 8 rows.
- Use 24 − 12 − 12 + 8 = 8 to avoid subtracting their overlap twice.
Answer 8 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 with neither A next to B nor C next to D. Use an overlap correction.
Give me a hint
The intersection consists of two disjoint adjacent blocks.
Compare my reasoning
- All rows: 24; each forbidden block count: 2!3! = 12.
- Both blocks together: 2! outer orders × 2! × 2! internal orders = 8.
- Valid rows: 24 − 12 − 12 + 8 = 8.
8 arrangements.
Look for these in your work
- Describe the intersection explicitly.
- Exclude each forbidden arrangement once.
Select three people from A–F and arrange them in a row, with A included but not at either end. Count outcomes in two ways.
Give me a hint
A must be the middle person; only the two end people are to be chosen and ordered.
Compare my reasoning
- Choose two of the other five in 5C2 = 10 ways.
- Arrange the selected pair at the ends in 2! ways, giving 20.
- Alternatively choose left end then right end in 5 × 4 = 20 ways.
20 arrangements.
Look for these in your work
- Use the middle-position implication.
- Make membership and position methods agree.
Count distinct AABBCC strings with different letters at the two ends. Check using both a complement and direct endpoint choices.
Give me a hint
Equal endpoints leave two identical pairs; different endpoints leave multiplicities one, one and two.
Compare my reasoning
- All strings number 90; equal endpoints number 3 × 4!/(2!2!) = 18.
- The complement gives 90 − 18 = 72.
- Directly choose ordered different endpoint types in 3 × 2 ways, then arrange the middle in 4!/2! = 12 ways: 72.
72 strings by either method.
Look for these in your work
- Treat equal endpoint cases as disjoint.
- Recalculate remaining repetitions after fixing endpoints.
Common mistakes
- Check whether forbidden cases overlap before subtracting.
- Selection counts and row orders describe different stages.
- Verify repeated-letter restrictions using the remaining multiplicities.
What can you explain now?
Why add the intersection back when subtracting two forbidden events?
Compare with the explanation
Arrangements in both forbidden events were subtracted twice.
They should be excluded once, so the second subtraction must be corrected.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Write the outcome and restrictions; correct overlaps, then verify with a second count or a small exhaustive list.
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.