What is the difference between permutations and combinations in Cambridge International A Levels Mathematics?

In Cambridge International A Levels Mathematics, permutations refer to the arrangement of objects in a specific order, where the sequence matters. Combinations, on the other hand, refer to the selection of objects where the order does not matter. For example, the permutation of three letters A, B, C can be ABC, ACB, BAC, BCA, CAB, CBA, while the combination of the same letters would simply be ABC.

How do you calculate permutations and combinations in Probability and Statistics 1?

To calculate permutations, use the formula nPr = n! / (n-r)!, where n is the total number of items, and r is the number of items to arrange. For combinations, use the formula nCr = n! / [r!(n-r)!]. These formulas are essential in Probability and Statistics 1 for solving problems related to arranging or selecting items.

When should I use permutations instead of combinations in mathematical problems?

Use permutations when the order of arrangement is important, such as in seating arrangements or race results. Use combinations when the order does not matter, such as selecting a committee or choosing lottery numbers. Understanding the context of the problem is crucial in deciding whether to use permutations or combinations.

Can you provide an example of a real-world application of permutations and combinations?

A real-world application of permutations is in scheduling, where the order of tasks is crucial. For combinations, consider a scenario in a lottery draw, where the order of the numbers drawn does not matter. Both concepts are widely used in fields like cryptography, logistics, and game theory, which are relevant to the Cambridge International A Levels curriculum.

What are some common mistakes to avoid when solving permutations and combinations problems?

Common mistakes include confusing when to use permutations versus combinations, miscalculating factorials, and not considering restrictions in a problem. It's important to carefully read the problem to determine if order matters and to apply the correct formula. Practice and familiarity with different types of problems can help avoid these errors.

Why is "Using $\binom{n}{r}$ when order matters" a common mistake in Permutations and Combinations ?

Why it happens: Not reading question carefully. How to avoid it: Permutation = arrangement (ORDER MATTERS). Combination = selection (ORDER DOES NOT). Read 'arranged' vs 'chosen / committee / selected'. Source: 9709 Examiner Reports 2022-2024

Why is "Forgetting to multiply when items must be together" a common mistake in Permutations and Combinations ?

Why it happens: Treating identical items as not-internally-arrangeable. How to avoid it: If two DIFFERENT items must be together, treat as ONE unit BUT multiply by 2! for internal arrangement. For two IDENTICAL items, no internal multiplier. Source: 9709 Examiner Reports 2022-2024

Probability and Statistics 1Cambridge International A Levels Mathematics (9709)

Permutations and Combinations

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Permutations and Combinations Study Notes — Cambridge International A Level Mathematics 9709 S1 (2024-2026 syllabus)

Counting arrangements and selections. Factorials. Constraints like 'together' or 'separated'. Foundation for probability.

What you’ll learn

Mapped to the Cambridge IGCSE 9709 syllabus (2024-2026).

S1.3.1 — Compute permutations.
S1.3.2 — Compute combinations.
S1.3.3 — Handle constraints (together, separated, distinct groups).

Factorials and permutations

Order matters.

Factorial. $n!$ = number of arrangements of $n$ DISTINCT items.

$^nP_r$ = arrangements of $r$ items chosen from $n$ : $\frac{n!}{(n-r)!}$ .

Repeated items. Arrangements of $n$ items with $n_1$ identical of type 1, $n_2$ of type 2, etc.: $\frac{n!}{n_1! n_2! \cdots}$

Example. Arrangements of MISSISSIPPI (11 letters: 1 M, 4 I, 4 S, 2 P). $\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = 34650$ .

Cambridge tip. Count repeated letters carefully.

$n!$ counts arrangements.
Divide by repeats' factorials.
$^nP_r$ for $r$ from $n$ .

Combinations

Order doesn't matter.

Combination. Selection of $r$ from $n$ , order doesn't matter: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Also written $^nC_r$ .

Compare to permutation. $^nP_r = r! \cdot \binom{n}{r}$ — permutation = combination × arrangements.

Cambridge tip. Read carefully: "arranged" → permutation; "chosen / selected / committee" → combination.

$\binom{n}{r}$ when order doesn't matter.
$^nP_r = r! \binom{n}{r}$ .
Look for keyword.

See the full worked example for permutations and combinations →

Constraints

Together, separated, groups.

'Together'. Treat group as ONE unit.

Example. Arrangements of LETTER with both Es together:

Treat (EE) as one unit.
Now arranging 5 units: (EE), L, T, T, R.
$\frac{5!}{2!} = 60$ (divide by 2! for repeated T).

'Not together / separated'. Total arrangements − arrangements with them together.

'Distinct groups'. Multiply independent choices.

Example. Committee of 3 men from 6 AND 2 women from 5: $\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200$ .

'At least'. Add cases or subtract complement.

Cambridge tip. Draw a clear plan of CASES before computing.

Together: treat as one.
Not together: subtract from total.
Distinct groups: multiply.

See the full worked example for permutations and combinations →

How it’s examined

Permutations and combinations appear every S1. Most-tested: arrangement with constraint (8 marks), committee selection (5-8 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/52 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Permutations and Combinations

Step-by-step solutions to past-paper-style questions on permutations and combinations , written exactly the way a tutor would explain them at the board.

1Arrangements with constraint (6 marks)
Extended• Adapted from 9709/52 May/Jun 2024• permutations
Question
How many 6-letter arrangements of LETTER include both Es together? (6 marks)
Step-by-step solution
1. Step 1
  Treat both Es as ONE unit. Now arranging 5 units: (EE), L, T, T, R.
2. Step 2
  Arrangements with repeated T. $\frac{5!}{2!} = 60$ .
3. Step 3
  EEs treated as one unit; no further internal arrangement (both same letter).
Answer
$60$ arrangements.
2Combinations (5 marks)
Extended• combinations
Question
A committee of 4 is chosen from 10 people. How many ways? (5 marks)
Step-by-step solution
1. Step 1
  Order doesn't matter → combinations.
  $\binom{10}{4} = \dfrac{10!}{4!6!} = \dfrac{10 \cdot 9 \cdot 8 \cdot 7}{24} = 210$
Answer
$210$ .
3Mixed arrangement (8 marks)
Extended• mixed
Question
A committee of 3 men and 2 women is to be selected from 6 men and 5 women. How many possible committees? (8 marks)
Step-by-step solution
1. Step 1
  Choose men. $\binom{6}{3} = 20$ .
2. Step 2
  Choose women. $\binom{5}{2} = 10$ .
3. Step 3
  Multiply (independent choices).
  $20 \times 10 = 200$
Answer
$200$ committees.

Key Formulae — Permutations and Combinations

The formulae you need to memorise for permutations and combinations on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

Factorial
$n! = n(n-1)(n-2) \cdots 2 \cdot 1, \quad 0! = 1$
$n$
non-negative integer
When to use
Counting arrangements (orderings).
Permutations $^nP_r$
$^nP_r = \dfrac{n!}{(n-r)!}$
When to use
Arrangements of $r$ from $n$ where ORDER matters.
Combinations $^nC_r$ or $\binom{n}{r}$
$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$
When to use
Choosing $r$ from $n$ where ORDER does NOT matter.
Arrangements with repeats
$\dfrac{n!}{n_1! n_2! \cdots n_k!}$
$n_i$
number of identical items of type $i$
When to use
Arrangements of $n$ items where some are identical (e.g. letters of MISSISSIPPI).

Key Definitions and Keywords — Permutations and Combinations

Definitions to memorise and the exact keywords mark schemes credit for permutations and combinations answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9709 sitting.

Permutation
Examiner keyword
Arrangement where ORDER matters. ABC and BAC are different.
Combination
Examiner keyword
Selection where ORDER does NOT matter. {A, B, C} and {C, B, A} are the same.
Factorial
$n! = n \times (n-1) \times \ldots \times 2 \times 1$ . $0! = 1$ by convention.

Common Mistakes and Misconceptions — Permutations and Combinations

The traps other students keep falling into on permutations and combinations questions — taken from recent Cambridge International A Level 9709 examiner reports and mark schemes — and how to avoid them.

✕Using $\binom{n}{r}$ when order matters
9709 Examiner Reports 2022-2024
Why it happens
Not reading question carefully.
How to avoid it
Permutation = arrangement (ORDER MATTERS). Combination = selection (ORDER DOES NOT). Read 'arranged' vs 'chosen / committee / selected'.
✕Forgetting to multiply when items must be together
9709 Examiner Reports 2022-2024
Why it happens
Treating identical items as not-internally-arrangeable.
How to avoid it
If two DIFFERENT items must be together, treat as ONE unit BUT multiply by 2! for internal arrangement. For two IDENTICAL items, no internal multiplier.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Permutations and Combinations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Permutations and Combinations

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Arrangements with constraint (6 marks)

2Combinations (5 marks)

3Mixed arrangement (8 marks)

Factorial

Permutations nPr^nP_rnPr​

Combinations nCr^nC_rnCr​ or (nr)\binom{n}{r}(rn​)

Arrangements with repeats

Permutation

Combination

Factorial

✕Using (nr)\binom{n}{r}(rn​) when order matters

✕Forgetting to multiply when items must be together

Practice questions

Past paper style quiz

Assess your understanding

Permutations and Combinations — frequently asked questions

Permutations $^nP_r$

Combinations $^nC_r$ or $\binom{n}{r}$

✕Using $\binom{n}{r}$ when order matters