Keep objects apart with complements and gaps
Subtract forbidden adjacency for a pair, or place separated objects into gaps around an existing 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 adjacency blocks and combinations.
- Count orders of distinct and repeated objects.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many gaps surround a row of three objects, including both ends?
What you will learn
- Count non-adjacent pairs using a complement.
- Use gaps for several separated objects and distinguish identical from distinct copies.
A complement handles one forbidden pair
Among five different people A–E, there are 5! = 120 unrestricted rows. A and B together give 2!4! = 48 forbidden rows. Therefore A and B not adjacent gives 120 − 48 = 72. These two cases partition all rows, so subtraction is exact.
For several objects that must be mutually separated, separate pairwise forbidden counts can overlap. A gap construction is often easier. Arrange three different girls first in 3! ways. There are four gaps: before the first girl, between successive girls and after the last. Put two different boys into two different gaps to ensure no boy adjacency.
Choose gaps and then allocate distinct objects
For three different girls and two different boys, choose two of the four gaps in 4C2 ways and assign boys in 2! ways. The total is 3! × 4C2 × 2! = 72. Both end gaps count. At most one boy per gap enforces the restriction; placing both in one gap would allow adjacency.
For four identical A and three identical B with no adjacent B, the A row has five gaps. Choose three gaps in 5C3 = 10 ways. There is no 4! factor for identical A and no 3! factor for identical B. If the separated objects outnumber available gaps, there are zero arrangements, rather than a factorial expression with a negative argument.
Pause and explain
Three distinct girls and two distinct boys, no adjacent boys: what count is correct?
Worked example
Seat three distinct girls and two distinct boys in one row with no adjacent boys.
Show the worked solution
- Arrange the girls in 3! = 6 ways, opening four gaps.
- Choose two different gaps in 4C2 = 6 ways.
- Assign the boys in 2! = 2 ways, giving 6 × 6 × 2 = 72.
Answer 72 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 not adjacent. Use a complement.
Give me a hint
Subtract the adjacent-block count from all five-person rows.
Compare my reasoning
- All rows number 5! = 120.
- Adjacent A and B give 2! × 4! = 48.
- The valid complement has 120 − 48 = 72 rows.
72 arrangements.
Look for these in your work
- Identify the forbidden condition exactly.
- Subtract from the correct unrestricted universe.
Arrange four different girls and three different boys in a row, with no adjacent boys.
Give me a hint
The girl row creates five gaps and the boys need three distinct gaps.
Compare my reasoning
- Arrange girls in 4! = 24 ways.
- Choose three of five gaps in 5C3 = 10 ways and order boys in 3! = 6 ways.
- Multiply 24 × 10 × 6 = 1440.
1440 arrangements.
Look for these in your work
- Include both end gaps.
- Distinguish choosing gaps from assigning boys.
Count distinct strings using four identical A and three identical B with no adjacent B.
Give me a hint
There is one visible A row, surrounded by five gaps.
Compare my reasoning
- Write AAAA; it creates five possible B gaps.
- Choose three different gaps in 5C3 = 10 ways.
- Identical copies contribute no additional internal ordering factors.
10 strings.
Look for these in your work
- Use at most one B per gap.
- Avoid factorials for indistinguishable copies.
Common mistakes
- Include the gaps at both ends.
- Assign distinct objects after choosing gaps.
- Do not multiply for identical copies or use overlapping complements uncorrected.
What can you explain now?
Can three mutually non-adjacent B copies be placed around just one A?
Compare with the explanation
No; a single A creates only two gaps.
Each occupied gap can contain at most one B under the non-adjacency restriction.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
For one pair, subtract adjacency; for many separated objects, build the base row and fill distinct gaps.
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.