Count ordered selections with permutations
Assign distinct objects to distinct roles or positions using decreasing choice counts and factorial notation.
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 the product rule.
- Recognise when two outcomes differ only in order.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In a no-tie podium, does swapping first and second place change the outcome?
What you will learn
- Calculate ordered selections without replacement using nPr.
- Explain factorials and distinguish ordered roles from a committee.
Roles make order matter
Choosing a captain, deputy and secretary from five students gives 5 × 4 × 3 = 60 assignments. The same student cannot hold two of these roles. Assigning A as captain and B as deputy is different from assigning B as captain and A as deputy, even if the third student is unchanged.
For r ordered positions filled by distinct objects selected from n, without replacement, nPr = n!/(n − r)!. Here n and r are non-negative integers with 0 ≤ r ≤ n. A factorial multiplies integers down to 1; 0! = 1. Selecting and arranging all n objects gives n!, while r = 0 gives one empty assignment.
State the assumptions before using a formula
Seven different books arranged in four numbered display positions have 7 × 6 × 5 × 4 = 840 outcomes. We are selecting four books and placing them in order; dividing by 4! would lose distinct displays. A repeated choice is not allowed in this model because each physical book is used at most once.
If each position can independently use any of n symbols with repetition allowed, the count is n^r instead. Neither nPr nor nCr describes every problem. For a podium with three distinct places and no ties, six competitors give 6P3 = 120; if ties are permitted, the outcome model needs further rules.
Pause and explain
How many assignments fill three different roles from five students, without replacement?
Worked example
Assign captain, deputy and secretary to three different students from five.
Show the worked solution
- Choose captain in five ways.
- Choose deputy in four remaining ways and secretary in three.
- Multiply 5 × 4 × 3 = 5P3 = 60; do not divide by 3! because roles differ.
Answer 60 assignments.
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.
Assign three different roles to three different people from five candidates. How many assignments?
Give me a hint
Choose the first role, then remove its person before choosing the next.
Compare my reasoning
- First role: five choices; second role: four remaining choices.
- Third role: three remaining choices.
- The product is 5 × 4 × 3 = 60.
60 assignments.
Look for these in your work
- Use the no-replacement condition.
- Retain the distinction between roles.
Seven different books are available for four numbered display positions. Use each book at most once. Count displays.
Give me a hint
A different order in numbered positions gives a different display.
Compare my reasoning
- The position counts are 7, 6, 5 and 4.
- Multiply them: 7 × 6 × 5 × 4 = 840.
- Equivalently, 7P4 = 7!/3!; no division by 4! is appropriate.
840 displays.
Look for these in your work
- Select and arrange in one count.
- Explain why positions preserve order.
Six runners compete for first, second and third place, with no ties. Count podiums and explain why 6C3 would be insufficient.
Give me a hint
A set of three names can occupy the three places in several orders.
Compare my reasoning
- Choose the places successively in 6, 5 and 4 ways.
- The count is 6P3 = 120.
- 6C3 = 20 selects only the names; each set has 3! = 6 podium orders.
120 podiums; 6C3 omits the six orders for each selected set.
Look for these in your work
- Use the explicit no-tie condition.
- Distinguish selection from place assignment.
Common mistakes
- Distinct roles are ordered even without a physical row.
- Use decreasing choices only when replacement is forbidden.
- Do not cancel the order by dividing by r! for labelled positions.
What can you explain now?
Why is 5P3 different from 5C3?
Compare with the explanation
The three selected students receive different roles in a permutation.
Each three-person set has 3! distinct role assignments.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
When roles or positions distinguish outcomes, use ordered choices; nPr assumes no replacement.
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.