Counting Principles

IB Mathematics: Analysis and Approaches HL· 45 min read

1. Fundamental Counting Rules★☆☆☆☆⏱ 10 min

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All counting problems are built from two fundamental rules that apply to different logical scenarios: the product rule for independent sequential choices, and the sum rule for mutually exclusive options.

📘 Definition

Product Rule

n \times m

If there are $n$ ways to complete one independent task, and $m$ ways to complete a second independent task, there are $n \times m$ ways to complete both tasks in sequence.

Example:

Choosing 1 shirt from 5 options and 1 tie from 3 options gives $5 \times 3 = 15$ total outfit combinations.

The sum rule applies when you select exactly one option from mutually exclusive groups. If group A has $n$ options and group B has $m$ options, there are $n + m$ total possible selections.

📐 Worked Example

A café offers 6 appetizers, 12 main courses, and 4 desserts. How many total 3-course meals are possible? How many total single-item orders (appetizer OR main OR dessert)?

1
3-course meals require choosing one of each item (independent sequential choices), so apply the product rule:
2
$6 \times 12 \times 4 = 288$
3
Single-item orders are mutually exclusive (you choose one item only), so apply the sum rule:
4
$6 + 12 + 4 = 22$
5
Final results: 288 3-course meals, 22 single-item orders.

2. Permutations of Distinct Elements★★☆☆☆⏱ 15 min

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Permutations are used when we count ordered arrangements of distinct elements. Order matters here: switching two elements creates a new outcome (e.g. arranging people in a line, creating a password).

📘 Definition

Permutation

P(n,k) = \frac{n!}{(n-k)!}

The number of ordered selections of $k$ distinct elements from a set of $n$ total distinct elements. For arrangements of all $n$ elements, $P (n, n) = n!$ .

📐 Worked Example

How many 3-digit passwords can be created from digits 1-9 with no repeated digits? How many ways can 4 people be arranged in a row for a photo?

1
For passwords, we need ordered selections of 3 distinct digits from 9, so calculate $P (9, 3)$ :
2
$P(9,3) = \frac{9!}{(9-3)!} = 9 \times 8 \times 7 = 504$
3
For arranging 4 people, we arrange all 4 distinct people, so calculate $P (4, 4) = 4!$ :
4
$4! = 4 \times 3 \times 2 \times 1 = 24$
5
Final results: 504 passwords, 24 arrangements.

Exam tip:

Always confirm if repetition is allowed before choosing the permutation formula.

3. Combinations (Unordered Selections)★★★☆☆⏱ 15 min

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Combinations are used when the order of selection does not matter. Switching two elements does not create a new outcome (e.g. choosing a committee of students, selecting marbles from a bag).

📘 Definition

Combination

\binom{n}{k} = \frac{n!}{k!(n-k)!}

The number of unordered selections of $k$ distinct elements from a set of $n$ total distinct elements, where $0 \leq k \leq n$ .

📐 Worked Example

How many ways can a 3-person committee be selected from a group of 8 students? How many ways to select 2 apples and 3 oranges from 5 apples and 7 oranges?

1
Committee membership is unordered, so calculate $(3 8)$ :
2
$\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
3
For fruit selection, calculate combinations separately for apples and oranges, then multiply by the product rule:
4
$\binom{5}{2} \times \binom{7}{3} = 10 \times 35 = 350$
5
Final results: 56 committees, 350 fruit selections.

4. Permutations with Repeats & Restrictions★★★★☆⏱ 20 min

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IB exam problems often include permutations with repeated identical elements or restricted positions (e.g. vowels must go at the start of a word). These require adjusted counting formulas.

For permutations of $n$ items with repeated identical groups of size $n_{1}, n_{2}, ..., n_{k}$ , the number of distinct arrangements is given by $\frac{n !}{n _{1} ! n _{2} ! ... n _{k} !}$ .

📐 Worked Example

How many distinct arrangements are there of the letters in the word "STATISTICS"?

1
Count total letters and repeated groups: 10 total letters, 3 S, 3 T, 2 I, 1 A, 1 C.
2
Substitute into the repeated permutation formula:
3
$\frac{10!}{3! \cdot 3! \cdot 2! \cdot 1! \cdot 1!} = 50400$

For problems with restricted positions, always count the restricted positions first, then count permutations for the remaining positions.

Exam tip:

For restricted arrangements, fix the restricted elements first to avoid overcounting.

5. Common Pitfalls

Wrong move:

Using permutations for unordered selections

Why:

Permutations count order, so they overcount unordered outcomes by a factor of $k!$

Correct move:

Confirm if order matters: use combinations for unordered selections

Wrong move:

Using the permutation formula for selections with repetition

Why:

Permutations of distinct elements assume no repeats, leading to undercounting when repeats are allowed

Correct move:

Use the product rule $n^{k}$ for ordered selections where repetition is allowed

Wrong move:

Adding instead of multiplying (or vice versa) for AND/OR scenarios

Why:

AND (both events) requires multiplication, OR (one event) requires addition

Correct move:

Identify the logical connective first: multiply for AND, add for OR

Wrong move:

Not dividing by factorials for repeated identical items

Why:

Swapping identical items does not create a new arrangement, leading to overcounting

Correct move:

Divide by the factorial of the count of each identical group when counting arrangements

6. Quick Reference Cheatsheet

Rule	When to use	Formula
Product Rule	Independent sequential choices (A AND B)	$n \times m$
Sum Rule	Mutually exclusive options (A OR B)	$n + m$
Permutation (no repeat)	Order matters, distinct elements	$\frac{n !}{( n - k )!}$
Combination	Order does not matter, no repeat	$\frac{n !}{k ! ( n - k )!}$
Permutation with repeats	Distinct arrangements of identical items	$\frac{n !}{n _{1} ! n _{2} ! ... n _{k} !}$
Ordered with repetition	Repeats allowed, order matters	$n^{k}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · 1
6-mark permutation counting problem
2023 · 2
4-mark combination problem

What's Next

Counting principles are the foundational building block for almost all combinatorics and probability topics in IB AA HL. You will immediately use these rules to expand binomials in the binomial theorem, calculate probabilities of compound events, and find probabilities for discrete distributions like the binomial distribution. Advanced counting problems with restrictions also often appear as short questions in Paper 1 and longer problems in Paper 3, so mastering these rules will prevent simple errors that cost you marks in higher-weight questions.