Grade 11 · Statistics 1 · Cambridge A-Level 9709 · Age 16–17
Counting is the foundation of probability. Before you can calculate how likely something is, you need to count how many ways it can happen. In Cambridge A-Level Statistics 1 (9709), permutations and combinations give you the systematic tools to count arrangements and selections, whether order matters or not. These techniques are essential for every probability question involving choosing, arranging, or grouping.
n! counts all orderings of n distinct items
ⁿPᵣ = n!/(n−r)! — ordered r from n
ⁿCᵣ = n!/(r!(n−r)!) — unordered r from n
Items together: treat as one unit; items apart: total − together
n!/(p!q!…) divides out repeated orderings
(n−1)! fixes one item to remove rotational duplicates
(n−1)!/2 also removes reflection duplicates
ⁿCᵣ = ⁿCₙ₋ᵣ — choosing r or leaving n−r
The factorial of a positive integer n, written n!, is the product of all positive integers from 1 up to n:
By convention, 0! = 1. This is not arbitrary — it makes combinatorial formulas consistent when r = n or r = 0.
Factorials often appear as fractions. Cancel common factors rather than expanding fully — this avoids enormous numbers.
The number of ways to arrange n distinct objects in a row is n!. This is because there are n choices for the first position, (n−1) for the second, and so on.
For n ≥ 13, factorials exceed a billion. Always use your calculator. On a Casio fx-991, press the number then the x! button (or MATH → PRB → !). Cambridge exams expect exact integer answers for counting questions — never leave factorial answers as decimals unless explicitly asked.
If Task A can be done in m ways and Task B in n ways (independently), then A and B together can be done in m × n ways. This multiplicative principle extends to any number of independent tasks.
A permutation is an ordered selection. We choose r items from n distinct items, and the order in which we pick them matters. For example, in a race, finishing 1st, 2nd, 3rd from a field of 10 is a permutation — swapping the 1st and 2nd place finishers gives a different result.
This counts the number of ways to fill r ordered positions from n distinct objects.
When all n items are selected (r = n), we get ⁿPₙ = n!/(n−n)! = n!/0! = n!/1 = n!. This confirms that the number of arrangements of all n items is n!.
The key test: does swapping the selected items give a different outcome? If yes, order matters — use ⁿPᵣ. If no, order doesn't matter — use ⁿCᵣ (see Learn 3).
On a Casio fx-991: enter n, press SHIFT, then nPr (above ÷), enter r, press =. Alternatively, use the formula directly. Always verify with a small example you can count manually.
A combination is an unordered selection. We choose r items from n distinct items, and the order does not matter. Choosing Alice, Bob, and Carol for a committee is the same as choosing Carol, Bob, and Alice — it's the same committee.
ⁿCᵣ is also often written as nCr or as the binomial coefficient notation with brackets.
Each unordered group of r items can be arranged in r! ways. So the number of ordered selections (ⁿPᵣ) equals the number of unordered selections (ⁿCᵣ) times the orderings within each selection (r!):
Choosing r items to include is the same as choosing (n−r) items to exclude. This symmetry can save calculation time.
Choosing r items from n is equivalent to either including a specific item (then choosing r−1 from the remaining n−1) or excluding it (then choosing r from the remaining n−1):
This identity generates Pascal's Triangle and is used in the binomial expansion.
When a restriction specifies that certain items must or must not be selected, handle them first, then count the remaining choices.
If k specific items must always be adjacent (together), treat them as a single "super-item". This reduces n items to (n−k+1) items to arrange, and the k items within the super-item can be arranged in k! ways among themselves.
Count the total arrangements with no restriction, subtract those where the unwanted items ARE together.
To place items so that no two of a group are adjacent: first arrange the other items, then insert the restricted items into the gaps between them.
When some items are identical, swapping them gives the same arrangement — so n! overcounts. We divide by the factorials of the counts of each repeated item.
In a circular arrangement, rotations of the same arrangement are considered identical. Fix one person to remove duplicates — effectively we arrange the remaining (n−1) people in (n−1)! ways.
For a necklace or bracelet, flipping it over gives the same arrangement (unlike a circular table where you can't flip people). So we also divide by 2.
Fix one object to remove rotational duplicates, then apply the identical-objects formula to the rest. These problems can be complex — work systematically.
8 fully worked examples covering the key question types. Study the method, not just the answer.
Find the number of distinct arrangements of all the letters in the word MATHS.
A committee of 4 is chosen from 10 people. How many ways can this be done?
A 4-digit password is formed from the digits 1–9, with no digit repeated. How many passwords are possible?
8 people stand in a row. Person A and Person B must always be next to each other. How many arrangements?
How many distinct arrangements are there of the letters in the word ARRANGE?
5 men and 3 women sit around a circular table. In how many ways can they sit if all women must sit together?
From 6 boys and 5 girls, a committee of 5 is chosen with exactly 2 boys and 3 girls. How many ways?
A committee of 4 is chosen from 7 men and 4 women. Find the number of ways the committee includes at least one woman.
These mistakes appear repeatedly in Cambridge A-Level examinations. Learn to recognise and avoid them.
All the formulas you need for Permutations & Combinations in Cambridge A-Level Statistics 1.
| Formula | Expression | When to Use |
|---|---|---|
| Factorial | n! = n × (n−1) × … × 2 × 1 0! = 1 | Count all orderings of n distinct items |
| Row Arrangements | n! | Arrange all n distinct items in a line |
| Permutation ⁿPᵣ | n! / (n−r)! | Ordered selection of r from n; order matters |
| Combination ⁿCᵣ | n! / [r!(n−r)!] | Unordered selection of r from n; order doesn't matter |
| Relationship | ⁿCᵣ = ⁿPᵣ / r! | ⁿCᵣ removes internal orderings from ⁿPᵣ |
| Symmetry | ⁿCᵣ = ⁿCₙ₋ᵣ | Use when n−r is smaller than r (easier computation) |
| Pascal's Identity | ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁ | Builds Pascal's Triangle; used in proof and binomial theorem |
| Items Together | (n−k+1)! × k! | k items must be adjacent in a row of n |
| Items Never Together | Total − Together | Complement: subtract cases where items are adjacent |
| Identical Objects | n! / (p! × q! × r! × …) | n objects with p of type 1, q of type 2, etc. |
| Circular Arrangements | (n−1)! | n distinct objects around a circle (rotations identical) |
| Necklace / Bracelet | (n−1)! / 2 | Circular + reflections are also identical |
| Fundamental Principle | m × n × … | Independent choices; one task AND another |
| 0! Convention | 0! = 1 | Required for formulas when r = 0 or r = n |
Three key proofs for Cambridge A-Level. Understanding these deepens your grasp of why the formulas work, not just how to apply them.
Claim: ⁿCᵣ = ⁿPᵣ / r!
Argument: Count the number of ordered arrangements of r items chosen from n: that's ⁿPᵣ = n!/(n−r)!.
Now, every unordered selection of r items corresponds to exactly r! ordered arrangements (the r! permutations of those same r items within their group).
So the number of unordered selections is:
ⁿCᵣ = ⁿPᵣ / r! = [n!/(n−r)!] / r! = n! / [r!(n−r)!]
This confirms the combination formula is simply the permutation formula with the internal orderings removed. ∎
Claim: The number of ways to choose r items from n equals the number of ways to choose n−r items from n.
Algebraic proof:
ⁿCₙ₋ᵣ = n! / [(n−r)! × (n−(n−r))!] = n! / [(n−r)! × r!] = n! / [r! × (n−r)!] = ⁿCᵣ
Combinatorial argument: Every selection of r items to include corresponds uniquely to a selection of n−r items to exclude (the leftovers). So both counts are equal. ∎
Application: ²⁰C₁₇ = ²⁰C₃, which is far easier to calculate.
Claim: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁
Algebraic proof: Starting from the left-hand side and finding a common denominator:
ⁿCᵣ + ⁿCᵣ₊₁ = n!/[r!(n−r)!] + n!/[(r+1)!(n−r−1)!]
Factor out n! / [(r+1)!(n−r)!] from both terms:
= [n!/(r+1)!(n−r)!] × [(r+1) + (n−r)]
= [n!/(r+1)!(n−r)!] × (n+1)
= (n+1)! / [(r+1)!(n−r)!] = ⁿ⁺¹Cᵣ₊₁ ∎
Combinatorial argument: Fix one item X from n+1 items. Selections of r+1 from n+1 either exclude X (choose r+1 from remaining n → ⁿCᵣ₊₁) or include X (choose r more from remaining n → ⁿCᵣ). These two cases are mutually exclusive and exhaustive, so ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁. ∎
Enter n and r to calculate ⁿPᵣ and ⁿCᵣ with full working shown. Use this to check your manual calculations.
Enter the word or sequence to count distinct arrangements, accounting for repeated letters.
Calculate each value exactly. Enter integers only (no decimals for factorial results).
All answers are integers. Use the formula ⁿPᵣ = n!/(n−r)! or build the product directly.
All answers are integers. Use ⁿCᵣ = n!/[r!(n−r)!].
Apply grouping, complement, or fixed-position methods. All answers are integers.
Use n!/(p!q!…) for repeated items; (n−1)! for circular arrangements. All answers are integers.
Mixed permutations, combinations, identical objects, restrictions, and circular arrangements. 100% triggers confetti!
Multi-step problems requiring careful analysis. These mirror the difficulty of harder Cambridge exam questions.
8 Cambridge S1-style questions with mark schemes. Attempt each before revealing the solution. Work under timed conditions: allow about 8 minutes per question.
Find the number of different ways that the letters of the word STATISTICS can be arranged.
A committee of 5 people is to be chosen from 8 men and 5 women. Find the number of ways in which the committee can be formed if it must contain at least 3 women.
6 boys and 2 girls are to stand in a row. Find the number of arrangements in which the 2 girls are not next to each other.
From a group of 12 people (7 men and 5 women), a team of 4 is chosen. Find the probability that the team contains exactly 2 men, given that it contains at least 1 woman.
Five people — Alice, Bob, Carol, Dan, and Eve — sit in a row. Find the number of arrangements in which Alice sits to the left of Bob (not necessarily adjacent).
8 people sit around a circular table. Find the number of arrangements if two specific people, X and Y, must sit directly opposite each other. (Assume the table has 8 equally spaced seats.)
A bag contains 10 balls: 4 red, 3 blue, and 3 green. 3 balls are selected at random. Find the number of selections that include at least one ball of each colour.
The letters of the word ARRANGE are written on separate cards. (i) Find the number of distinct arrangements of all 7 letters. (ii) Find the number of arrangements where the two R's are separated (not next to each other).
5 questions drawn from Cambridge A-Level 9709 Statistics 1 past papers on Permutations & Combinations. Attempt under exam conditions before revealing the mark scheme.
Find the number of ways in which the 9 letters of the word SEVENTEEN can be arranged if (i) there are no restrictions, (ii) the arrangement starts and ends with the letter E.
A team of 4 is chosen from 5 teachers and 10 students. Find the number of ways the team can be chosen if it must include at least 2 teachers.
How many different arrangements are there of the 8 letters in the word FOOTBALL? How many of these arrangements have the two letters L not next to each other?
6 people are to sit in a row. Two of the people, Alan and Beth, refuse to sit next to each other. Find the number of possible arrangements.
A committee of 4 people is to be chosen from 4 women and 5 men. (i) Find the number of possible committees. (ii) Find the number of possible committees that include more women than men. (iii) One of the men is the boss. Find the number of possible committees that include the boss.