Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 5: Probability & Statistics 1 · official syllabus
PERMUTATIONS AND COMBINATIONS LAB
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Four short missions on selections, repeated letters and adjacency restrictions in a line. Decide whether order matters, predict the rule, then calculate and test your count.
1234
Mission 1 of 4
Choose a committee, not roles
Choose a committee of two people from six distinct people A to F. Neither member has a special role. Set the number of possible committees. The grid shows each unordered pair once above the diagonal; the reversed pair below the diagonal is not a new committee.
If the two committee members exchange places in the list, the committee...
Row and column labels name the two people. Only the upper triangle is counted; the diagonal would choose one person twice and the lower triangle repeats reversed pairs.
Your count is 30. Avoid counting each pair in both orders.
Study notes
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Decide whether order matters 5.2
A selection chooses which objects are included. An arrangement also specifies their order or positions. Choosing Alice and Ben for one committee is the same selection as choosing Ben and Alice. Giving Alice the chair role and Ben the secretary role is a different arrangement from reversing the roles.
For n distinct objects, n!=n(n-1)...(1), with 0!=1. Ordered choices of r distinct objects without replacement have nPr=n!/(n-r)! possibilities. Unordered selections have nCr=n!/(r!(n-r)!). Use 0<=r<=n; a factorial formula does not make an impossible choice possible.
Worked example: choosing two people from six gives 6C2=15 committees. Choosing a chair and a secretary from the same six, with different people in the two roles, gives 6P2=6(5)=30. The extra factor 2 counts the two assignments of roles to each selected pair.
Multiply stages, add disjoint cases 5.2
Multiply the number of choices at successive stages when every valid partial choice has the counted number of continuations. If that number varies, split into cases rather than multiplying unrelated averages. State whether reuse or replacement is allowed.
Add counts for mutually exclusive cases that together cover the required outcomes. Avoid double-counting an arrangement that satisfies more than one verbal description. A complement can be simpler: required count = total count - count of excluded outcomes.
Worked example: a three-person committee is chosen from four boys and three girls. There are 7C3=35 total committees. Those with no girl number 4C3=4, so 31 contain at least one girl. 'At least one' includes one, two or three girls, not only exactly one.
Repeated objects reduce the number of distinct arrangements 5.2
When n objects include indistinguishable groups of sizes a, b, and so on, the number of distinct linear arrangements is n!/(a!b!...); the group sizes, including any singletons, add to n. Swapping equal letters does not create a new visible word.
Worked example: BALLOON has seven letters, with two Ls and two Os. Its distinct arrangements number 7!/(2!2!)=1260. Dividing by 2 only once would remove only one duplicated pair. No condition here requires the arrangement to be an English word.
If a restriction fixes some repeated objects or combines them into a block, recount the remaining multiplicities before applying a factorial formula. Do not blindly use the unrestricted denominator after the objects being arranged have changed.
Use a block for required adjacency 5.2
When two particular distinct people must stand together, treat them as one block, arrange that block with the other people, then count the internal orders of the block. A block represents a constraint, not an extra person.
Worked example: six distinct people A, B, C, D, E and F stand in a line. If A and B must be adjacent, arrange five units in 5! ways and choose AB or BA in 2 ways. The count is 5!(2)=240.
The internal factor depends on the condition. If A must stand immediately to the left of B, only AB is allowed and the count is 5!=120. If three particular distinct people form one consecutive block with no internal restriction, its internal factor is 3!.
Use a complement for forbidden adjacency 5.2
For the same six distinct people, the unrestricted count is 6!=720. Since 240 have A and B adjacent, 720-240=480 have them non-adjacent. Every arrangement is in exactly one of these two cases, so subtraction is valid.
Do not subtract two overlapping sets of forbidden arrangements without handling their overlap. For more complicated restrictions, fix positions, use gaps, or split into disjoint cases and check that every valid arrangement is counted once.
Worked example with fixed ends: if A and B must occupy the two ends of a six-person line, choose their order in 2! ways and arrange the remaining four people in 4! ways. The result is 48. This is a different condition from requiring A and B to be adjacent.
Rows and positions must be distinguishable 5.2
For people sitting in two labelled rows, decide whether the seats and rows are distinct and whether any seats are empty. Six distinct people filling six distinct seats in two labelled rows still have 6!=720 arrangements when there are no restrictions.
For three specified seats in the front row and three in the back, another count is 6C3 x 3! x 3!=720. Select the front-row group, then arrange each row. Dividing by two because there are two rows would incorrectly identify front and back when they are labelled.
These examples concern linear arrangements and selections, matching section 5.2. The official syllabus excludes questions about objects arranged in a circle; do not substitute a circular-arrangement rule for a line or a set of labelled seats.
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