# Permutations and Combinations

> Additional Mathematics · CIE IGCSE 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u11-permutations-and-combinations/

This guide covers all permutation and combination content required for CIE IGCSE Additional Mathematics 0606, including factorial calculations, distinguishing counting methods, and solving constrained arrangement and selection problems.

**Prerequisites:** Understanding of basic counting multiplication principle; Familiarity with algebraic fraction simplification

## Learning objectives

- Distinguish permutations (order matters) from combinations (order does not matter)
- Calculate factorials, $^nP_r$ and $^nC_r$ values using the rule $0! = 1$
- Solve permutation arrangement problems with simple must include/exclude constraints
- Solve combination selection problems with simple must include/exclude constraints

## 1. Factorial Notation and Key Rules

**Factorial** — For a positive integer n, n! is the product of all positive integers from 1 to n. The special case $0! = 1$ is a standard rule to avoid undefined values in permutation calculations.

*Notation:* n! = n \times (n-1) \times (n-2) \times \dots \times 1

*Example:* 5! = 5 × 4 × 3 × 2 × 1 = 120

Factorials are the foundation of permutation and combination calculations. For non-calculator papers, always cancel common factorial terms before multiplying to avoid large, error-prone calculations.

**Worked example:** Simplify $\frac{7!}{4!}$ without using a calculator.

1. Step 1: Expand the larger factorial to include the smaller factorial as a term:

   $$7! = 7 \times 6 \times 5 \times 4!$$
2. Step 2: Cancel the common 4! term in numerator and denominator:

   $$\frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5$$
3. Step 3: Calculate the final value:

   $$7 \times 6 \times 5 = 210$$

> **tip**
>
> You do not need to calculate full factorial values for most non-calculator questions: cancelling first will always simplify your working.

## 2. Permutations (Order Matters)

**Permutation** — An ordered arrangement of r distinct objects selected from n objects, with no repetition allowed. Use permutations when swapping two selected items creates a different outcome.

*Notation:* ^nP_r = \frac{n!}{(n-r)!}

*Example:* Number of ways to arrange 2 letters from {A,B,C}: $^3P_2 = 6$ (AB, BA, AC, CA, BC, CB)

Common permutation use cases include arranging people in a line, awarding ranked prizes, or arranging distinct letters in a word.

**Worked example:** How many ways can 4 distinct books be arranged on a shelf?

1. Step 1: Identify n=4 and r=4 (we are arranging all 4 books):

   $$^4P_4 = \frac{4!}{(4-4)!} = \frac{4!}{0!}$$
2. Step 2: Substitute $0! = 1$ and calculate:

   $$\frac{4!}{1} = 4 \times 3 \times 2 \times 1 = 24$$

> **Exam tip**
>
> If the question uses words like 'arrange', 'order', or 'rank', it is almost always a permutation problem.

## 3. Combinations (Order Does Not Matter)

**Combination** — An unordered selection of r distinct objects selected from n objects, with no repetition allowed. Use combinations when swapping two selected items creates the same outcome.

*Notation:* ^nC_r = \frac{n!}{r!(n-r)!}

*Example:* Number of ways to select 2 letters from {A,B,C}: $^3C_2 = 3$ (AB, AC, BC)

Common combination use cases include selecting teams, choosing items for a trip, or picking questions to answer on an exam paper.

**Worked example:** How many ways can a 3-person committee be selected from 8 eligible students?

1. Step 1: Identify n=8 and r=3 (committee positions have no hierarchy, so order does not matter):

   $$^8C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$$
2. Step 2: Cancel common 5! terms:

   $$\frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{6}$$
3. Step 3: Simplify to get the final answer:

   $$8 \times 7 = 56$$

> **info**
>
> You may also see $^nC_r$ written as $\binom{n}{r}$ in some resources: both notations are acceptable for your 0606 exam.

## 4. Solving Constrained Counting Problems

Most exam problems include simple constraints: either specific objects must be included in the selection/arrangement, or specific objects must be excluded. Always handle the fixed constraint first, then calculate the remaining free choices.

**Worked example:** A group of 10 students includes 2 siblings. How many 4-person teams can be selected that include both siblings?

1. Step 1: Account for the 'must include' constraint: both siblings are already selected, so we only need to choose 2 more members from the remaining 8 students.
2. Step 2: Calculate the combination for the remaining positions:

   $$^8C_2 = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28$$

**Check your understanding**

1. Is the number of ways to award 1st, 2nd and 3rd place prizes to 10 runners a permutation or combination problem?

   - Permutation
   - Combination

   *Why:* Order matters for ranked prizes: 1st place is different from 2nd place, so this is a permutation problem.

> **Exam tip**
>
> For 'must exclude' constraints, simply reduce the n value by the number of excluded objects, then calculate the permutation or combination as normal.

## 5. Algebraic Problems with $^nP_r$ and $^nC_r$

A very common exam question gives you an equation involving $^nP_r$ or $^nC_r$ and asks you to find the value of n. The method is always the same: replace each term with its factorial definition, cancel the common factorial factors, and you are left with a simple polynomial equation (usually a quadratic) to solve. Remember to reject any solution that is negative or too small for the expression to be defined (you need $n \geq r$).

**Worked example:** Given that $^nC_4 = 6\,{}^nC_2$, find the value of n.

1. Step 1: Write each combination using $^nC_r = \frac{n!}{r!(n-r)!}$:

   $$\frac{n!}{4!(n-4)!} = 6 \times \frac{n!}{2!(n-2)!}$$
2. Step 2: Divide both sides by $n!$ (which is non-zero), then cross-multiply:

   $$2!\,(n-2)! = 6 \times 4!\,(n-4)!$$
3. Step 3: Substitute $2! = 2$ and $4! = 24$, expand $(n-2)! = (n-2)(n-3)(n-4)!$, then cancel $(n-4)!$:

   $$2(n-2)(n-3) = 144$$
4. Step 4: Divide by 2 and expand into a quadratic:

   $$(n-2)(n-3) = 72 \;\Rightarrow\; n^2 - 5n - 66 = 0$$
5. Step 5: Factorise and solve, rejecting the negative root:

   $$(n-11)(n+6) = 0 \;\Rightarrow\; n = 11$$

**Worked example:** Show that $^nC_3 + {}^nC_2 = \frac{1}{6}(n^3 - n)$, and hence solve $^nC_3 + {}^nC_2 = 35$.

1. Step 1: Write each term as a product using the factorial definition:

   $$^nC_3 = \frac{n(n-1)(n-2)}{6}, \quad {}^nC_2 = \frac{n(n-1)}{2}$$
2. Step 2: Put both terms over a common denominator of 6:

   $$^nC_3 + {}^nC_2 = \frac{n(n-1)(n-2)}{6} + \frac{3n(n-1)}{6}$$
3. Step 3: Factor out $n(n-1)$ from the numerator:

   $$= \frac{n(n-1)[(n-2)+3]}{6} = \frac{n(n-1)(n+1)}{6}$$
4. Step 4: Since $n(n-1)(n+1) = n(n^2-1) = n^3 - n$, the identity is shown:

   $$^nC_3 + {}^nC_2 = \frac{1}{6}(n^3 - n)$$
5. Step 5: Hence set the expression equal to 35 and solve the resulting cubic by inspection (n must be a positive integer with $n \geq 3$):

   $$\frac{1}{6}(n^3 - n) = 35 \;\Rightarrow\; n^3 - n - 210 = 0 \;\Rightarrow\; n = 6$$

> **Exam tip**
>
> After expanding, always cancel the common $n!$ (or the largest shared factorial) before multiplying out. This turns a messy factorial equation into a simple quadratic or cubic you can factorise.

## 6. Together, Not Together, and At Least / At Most

Two constraint types appear again and again in 0606 counting questions. The first is when certain items must stay together (adjacent) or must be kept apart (not together) in a row. The second is when a selection must contain at least, or at most, a given number of a certain kind of item. Each has a reliable standard method.

**Block (bundling) method for 'together'** — When k specific items must be adjacent in a row, tie them together into a single block. Arrange that block alongside the other separate items as units, giving (number of units)! arrangements, then multiply by k! for the internal order of the items inside the block. For 'not together', use the complement: (total arrangements) minus (arrangements with the items together).

*Notation:* \text{arrangements} = (\text{units})! \times k!

*Example:* 2 people must be adjacent among 5 in a row: tie them into 1 block, giving 4 units, so $4! \times 2! = 48$.

**Worked example:** Five different people stand in a row for a photograph. (a) In how many arrangements do two particular people, A and B, stand next to each other? (b) In how many arrangements are A and B NOT next to each other?

1. Part (a), Step 1: Tie A and B together into a single block. This block, together with the other 3 people, gives 4 units to arrange in a row:

   $$\text{units} = 4 \;\Rightarrow\; 4! = 24$$
2. Part (a), Step 2: Inside the block, A and B can be ordered as AB or BA, giving $2!$ internal arrangements. Multiply the two stages:

   $$4! \times 2! = 24 \times 2 = 48$$
3. Part (b), Step 1: Find the total number of unrestricted arrangements of all 5 people:

   $$5! = 120$$
4. Part (b), Step 2: 'Not together' is the complement of 'together', so subtract the part (a) result:

   $$120 - 48 = 72$$

**Worked example:** A team of 8 players is chosen from a squad of 15. Four of the players are close friends who will only play if all four are chosen together, or none of them is chosen. In how many ways can the team be selected?

1. Step 1: Selection is unordered, so use combinations. Unlike the row block method there is no internal $\times k!$ here, because the team has no order. The condition 'all four together or none' splits into two separate cases: all four friends in, or all four out.
2. Step 2: Case 1 — all four friends included. They fill 4 of the 8 places, so choose the remaining 4 players from the other 11:

   $${}^{11}C_{4} = 330$$
3. Step 3: Case 2 — all four friends excluded. Choose all 8 players from the other 11:

   $${}^{11}C_{8} = {}^{11}C_{3} = 165$$
4. Step 4: The two cases are mutually exclusive, so add them:

   $$330 + 165 = 495$$

**Worked example:** A team of 4 people is chosen from 6 men and 4 women. In how many ways can the team be chosen if it must contain at least 2 women? Selection is unordered.

1. Step 1: 'At least 2 women' means exactly 2, exactly 3, or exactly 4 women. Split into cases and count each with combinations, since order does not matter.
2. Step 2: Exactly 2 women (and so 2 men): choose 2 of the 4 women and 2 of the 6 men:

   $${}^{4}C_{2} \times {}^{6}C_{2} = 6 \times 15 = 90$$
3. Step 3: Exactly 3 women (and 1 man):

   $${}^{4}C_{3} \times {}^{6}C_{1} = 4 \times 6 = 24$$
4. Step 4: Exactly 4 women (and 0 men):

   $${}^{4}C_{4} \times {}^{6}C_{0} = 1 \times 1 = 1$$
5. Step 5: Add the cases for the final total:

   $$90 + 24 + 1 = 115$$
6. Check with the complement method (total teams minus those with 0 or 1 woman); both methods agree:

   $${}^{10}C_{4} - ({}^{6}C_{4} + {}^{4}C_{1}\,{}^{6}C_{3}) = 210 - (15 + 80) = 115$$

> **Exam tip**
>
> Reach for the complement (total minus unwanted) whenever the wanted cases outnumber the unwanted ones: 'at least 1' is fastest as total minus (none), and 'not together' is fastest as total minus (together).

## 7. Forming Numbers and Passwords

A very common 0606 question gives you a set of digits and asks how many k-digit numbers you can make from them, usually with an extra condition: the number must be even, odd, divisible by 5, or greater than some value. Treat each digit position as a slot to be filled and use the multiplication principle, but always deal with the restricted positions first.

**Position (slot) method for forming numbers** — Fill the most restricted position first, then work through the rest. The last (units) digit is fixed by any even/odd/divisibility condition: even numbers end in 0, 2, 4, 6 or 8; odd numbers end in 1, 3, 5, 7 or 9; multiples of 5 end in 0 or 5. The first digit can never be 0, or the number would have fewer digits. When 0 is an allowed last digit AND is banned from the first slot, the two restrictions clash over the same digit, so split the count into cases on the last digit.

*Notation:* \text{total} = (\text{choices for restricted slots}) \times (\text{choices for the remaining slots})

*Example:* Three-digit even numbers from {1,2,3,4,5}, no repeats: the units digit must be 2 or 4 (2 ways), then the first two slots take 2 of the other 4 digits, $^4P_2 = 12$, giving $2 \times 12 = 24$.

**Worked example:** Using the digits 1, 2, 3, 4, 5 with no digit repeated, how many three-digit even numbers can be formed?

1. Step 1: A number is even exactly when its units (last) digit is even. Among {1,2,3,4,5} the even digits are 2 and 4, so fill this restricted position first:

   $$\text{units digit} \in \{2,4\} \Rightarrow 2 \text{ choices}$$
2. Step 2: One digit is now used, leaving 4 digits for the hundreds and tens positions. Arrange 2 of these 4 in order (there is no 0 to worry about here):

   $$^4P_2 = 4 \times 3 = 12$$
3. Step 3: Multiply the stages using the multiplication principle:

   $$2 \times 12 = 24$$
4. Step 4 (check): There are $^5P_3 = 60$ three-digit numbers in total. The odd ones have units digit in {1,3,5} (3 choices) with the other two slots giving $^4P_2 = 12$, i.e. $3 \times 12 = 36$. So the even count is $60 - 36 = 24$, which agrees.

   $$60 - 36 = 24$$

**Worked example:** Using the digits 0, 1, 2, 3, 4 with no digit repeated: (a) how many three-digit numbers can be formed, and (b) how many of them are even?

1. Part (a), Step 1: The hundreds (first) digit cannot be 0, otherwise the result is not a three-digit number. Choose it from {1,2,3,4}:

   $$\text{hundreds digit} \in \{1,2,3,4\} \Rightarrow 4 \text{ choices}$$
2. Part (a), Step 2: The tens and units positions are filled from the 4 digits that remain (this pool now includes 0). Arrange 2 of these 4:

   $$^4P_2 = 4 \times 3 = 12$$
3. Part (a), Step 3: Multiply the stages:

   $$4 \times 12 = 48$$
4. Part (b), Step 4: An even number ends in an even digit; among {0,1,2,3,4} those are 0, 2 and 4. Because 0 is also banned from the hundreds slot, split into cases on the units digit so the two rules do not clash.
5. Part (b), Step 5: Case 1 — units digit = 0. The hundreds digit is then any of {1,2,3,4} (4 choices) and the tens digit is any of the 3 that remain:

   $$1 \times 4 \times 3 = 12$$
6. Part (b), Step 6: Case 2 — units digit = 2 or 4 (2 choices). Now 0 is still unused but banned from the hundreds slot, so the hundreds digit has $4 - 1 = 3$ choices, and the tens digit is any of the 3 that remain:

   $$2 \times 3 \times 3 = 18$$
7. Part (b), Step 7: Add the cases for the total number of even three-digit numbers:

   $$12 + 18 = 30$$
8. Part (b), Step 8 (check): From part (a) there are 48 numbers in total. The odd ones have units digit in {1,3} (2 choices), a non-zero hundreds digit (3 choices) and any tens digit (3 choices), giving $2 \times 3 \times 3 = 18$ odd. So the even count is $48 - 18 = 30$, which agrees.

   $$48 - 18 = 30$$

> **Exam tip**
>
> Whenever 0 is one of your digits, ask two questions before counting: can 0 be the first digit (never), and can 0 be the last digit (only if the number may be even or a multiple of 5). If 0 is caught by both rules, split into cases on the last digit so you never miscount the first-digit choices.

The same slot method works for codes and passwords built from a mixed set of characters — letters, digits and symbols together. Fill any restricted positions first, then fill the rest, and use the complement (total minus unwanted) for an 'at least one' condition.

**Worked example:** A password is 4 characters long and is formed from 5 different letters, 3 different digits and 2 different symbols (10 distinct characters in total). No character may be repeated. (a) How many passwords are possible? (b) How many begin with a letter and end with a digit? (c) How many contain at least one letter?

1. Part (a): There are 10 distinct characters filling 4 ordered positions with no repeats, so this is a straightforward permutation:

   $$^{10}P_4 = 10 \times 9 \times 8 \times 7 = 5040$$
2. Part (b), Step 1: Fill the two restricted positions first. The first character is one of the 5 letters and the last is one of the 3 digits:

   $$5 \times 3 = 15$$
3. Part (b), Step 2: The two middle positions are filled from the 8 characters that remain, arranged in order:

   $$^8P_2 = 8 \times 7 = 56$$
4. Part (b), Step 3: Multiply the stages:

   $$15 \times 56 = 840$$
5. Part (c): 'At least one letter' is quickest by the complement. A password with no letter at all uses only the 5 non-letter characters (3 digits + 2 symbols):

   $$^5P_4 = 5 \times 4 \times 3 \times 2 = 120$$
6. Subtract these from the total in part (a):

   $$5040 - 120 = 4920$$

## Common pitfalls

- **Wrong:** Using permutations for unordered selections (e.g. $^5P_2$ for 2-person teams from 5 people)
  - Why it fails: Permutations count each unordered pair twice (e.g. Alice+Bob and Bob+Alice as separate) when they are identical for unordered selections.
  - Correct: Use combinations for unordered selections, permutations for ordered arrangements.
- **Wrong:** Forgetting $0! = 1$, leading to division by zero when calculating $^nP_n$
  - Why it fails: The definition of factorial explicitly sets $0! = 1$ to avoid undefined values when arranging all n objects.
  - Correct: Memorize $0! = 1$ as a non-negotiable rule for all factorial calculations.
- **Wrong:** Attempting to solve circular arrangement problems with standard permutation formulas
  - Why it fails: Circular arrangements have different counting rules and are explicitly excluded from the 0606 syllabus.
  - Correct: All 0606 arrangement problems are linear, so use standard $^nP_r$ formulas only.
- **Wrong:** Mixing permutations and combinations in the same problem
  - Why it fails: 0606 problems are designed to use either permutations OR combinations, not both; mixing methods leads to overcounting.
  - Correct: First confirm if order matters, then use only the relevant formula for the entire problem.
- **Wrong:** Calculating full factorial values instead of cancelling first on non-calculator papers
  - Why it fails: Large factorial values lead to error-prone multiplications and waste time.
  - Correct: Cancel common factorial terms in numerator and denominator before multiplying.

## Cheatsheet

| Concept | Formula | Use Case |
| --- | --- | --- |
| Factorial | $n! = n \times (n-1) \times ... \times 1$, $0! = 1$ | Foundational calculation for all counting problems |
| Permutation | $^nP_r = \frac{n!}{(n-r)!}$ | Ordered arrangements, no repetition of objects |
| Combination | $^nC_r = \frac{n!}{r!(n-r)!}$ | Unordered selections, no repetition of objects |
| Must include (combination / selection) | Fix the k required items, choose the rest: $^{n-k}C_{r-k}$ | Selecting r from n where k specific items must be included (order does not matter) |
| Must include (permutation / arrangement) | Place the k required items in positions ($^{r}P_{k}$), then arrange the other $r-k$ from the remaining ($^{n-k}P_{r-k}$) | Arranging r from n where k specific items must be included (order matters) |
| Must exclude constraint | Reduce n by number of excluded items, keep r the same | Problems forbidding specific objects from selection |

## What's next

Now that you have mastered permutations and combinations for CIE IGCSE Additional Mathematics 0606, you can apply these counting skills to other syllabus areas, most directly the binomial theorem, where the same $^nC_r$ coefficients reappear in the expansion of $(a+b)^n$. Permutation and combination questions make up an average of 5-8% of your total exam marks, so practice past paper questions to build speed and accuracy, especially for non-calculator paper problems where you will need to simplify factorial fractions manually. Remember that circular arrangements, repeated objects, and mixed permutation/combination problems are out of scope for 0606, so you do not need to practice these for your exam. Next, move on to the Series unit (binomial theorem and progressions), where these counting skills are reused to write out binomial expansions.

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