Counting principles, permutations and combinations

IB Mathematics: Applications and Interpretation HL· Unit 1: Number and Algebra, Topic 7· 15 min read

1. The Fundamental Counting Principle★☆☆☆☆⏱ 4 min

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📘 Definition

Fundamental Counting Principle

If there are $n_{1}$ ways to complete the first task, $n_{2}$ ways to complete the second, and so on for $k$ independent tasks, the total number of ways to complete all tasks is the product $n_{1} \times n_{2} \times ... \times n_{k}$ .

Example:

A 3-digit PIN with independent digits has $10 \times 10 \times 10 = 1000$ total possible outcomes.

📐 Worked Example

A coffee shop offers 4 drink sizes, 8 coffee bean types, and 3 sweetener options. How many unique combinations of one size, one bean, and one sweetener are possible?

1
Identify the number of independent options for each choice:
2
$4 \text{ (sizes)} \times 8 \text{ (beans)} \times 3 \text{ (sweeteners)}$
3
Calculate the product to get the total number of combinations:
4
$4 \times 8 \times 3 = 96$

Exam tip:

Always confirm choices are independent before multiplying. For dependent selection (without replacement), adjust the number of options for each subsequent step.

2. Permutations of Distinct Items★★☆☆☆⏱ 5 min

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📘 Definition

Permutation (distinct items)

$P (n, r)$ or $^{n} P_{r}$

An ordered arrangement of $r$ items selected from $n$ distinct items, where no repetition is allowed. Order matters, so swapping two items creates a new permutation.

Example:

3-letter codes from 5 distinct letters are permutations, since ABC is different from CBA.

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

📐 Worked Example

How many ways can 5 different paintings be arranged in a row on a wall?

1
We arrange all 5 distinct items, so $n = 5$ , $r = 5$ :
2
$P(5,5) = \frac{5!}{(5-5)!} = \frac{120}{0!}$
3
Recall that $0! = 1$ by definition, so:
4
$P(5,5) = 120$

✓ Quick check

Test your understanding: How many 3-letter code words can be made from A,B,C,D,E with no repetition?

How many total codes are possible?
- 20
- 60
- 10
- 120
Reveal answer
60 —
$P (5, 3) = \frac{5 !}{( 5 - 3 )!} = 5 \times 4 \times 3 = 60$ . Correct, since order matters for code words.

3. Permutations with Repeated Elements★★★☆☆⏱ 6 min

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📘 Definition

Permutations of a multiset

When arranging $n$ total items with repeated identical elements, we divide by the factorial of the count of each repeated element to avoid counting identical arrangements multiple times.

Example:

Arranging the letters of "BOOK" requires dividing by $2!$ for the two identical O's.

\text{Total distinct permutations} = \frac{n!}{n_1! \times n_2! \times ... \times n_k!}

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📐 Worked Example

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

1
Count total letters: $n = 6$ . Count repeated letters: A = 3, N = 2, B = 1:
2
$\text{Total arrangements} = \frac{6!}{3! \times 2! \times 1!}$
3
Simplify the expression:
4
$\frac{720}{6 \times 2 \times 1} = 60$

Exam tip:

Do not forget to divide by the factorial of every repeated element, not just one repeated group.

4. Combinations for Unordered Selections★★★☆☆⏱ 7 min

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📘 Definition

Combination

$(r n)$ or $C (n, r)$

An unordered selection of $r$ items from $n$ distinct items. Order does not matter, so swapping two items does not create a new combination.

Example:

Selecting a 3-person committee from 12 people is a combination, since all roles are identical.

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

📐 Worked Example

A class of 12 students selects a 3-person committee. How many different possible committees are there?

1
Order does not matter (no distinct roles), so $n = 12$ , $r = 3$ :
2
$\binom{12}{3} = \frac{12!}{3! \times 9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1}$
3
Calculate the result:
4
$2 \times 11 \times 10 = 220$

5. Common Pitfalls

Wrong move:

Treating permutations as combinations (or vice versa)

Why:

Forgetting that order changes the count for arrangements, leading to wrong final values

Correct move:

Ask: does swapping two items create a new outcome? If yes, use permutations; if no, use combinations

Wrong move:

Forgetting that $0! = 1$

Why:

Leads to undefined or incorrect results when permuting all $n$ items ( $r = n$ )

Correct move:

Always remember $0!$ is defined as 1, not 0, by convention

Wrong move:

Not dividing by factorials for repeated elements in permutations

Why:

Counts identical arrangements as distinct, leading to an overestimate of the total number of outcomes

Correct move:

For every group of identical repeated items, divide by the factorial of the size of the group

Wrong move:

Adding instead of multiplying for the fundamental counting principle

Why:

Confuses "and" (complete all tasks) with "or" (choose one option from mutually exclusive groups)

Correct move:

Multiply for "and" scenarios, add for mutually exclusive "or" scenarios

Wrong move:

Using fundamental counting principle without adjusting for dependent selection

Why:

Keeps the original number of options for every step even when items are removed after selection

Correct move:

Reduce the number of available options by 1 for each subsequent step when selection is without replacement

6. Quick Reference Cheatsheet

Rule	Formula	When to use
Fundamental Counting Principle	$n_{1} \times n_{2} \times ... \times n_{k}$	Independent sequential choices
Permutation (distinct, no repetition)	$P (n, r) = \frac{n !}{( n - r )!}$	Ordered selection from $n$ distinct items
Permutation (repeated items)	$\frac{n !}{n _{1} ! n _{2} ! ... n _{k} !}$	Arrangements with identical repeated items
Combination (unordered)	$(r n) = \frac{n !}{r ! ( n - r )!}$	Unordered selection from $n$ distinct items

7. Frequently Asked

Do I need to memorize formulas for permutations and combinations?

While your GDC can calculate values, you must recognize when to use which formula, so memorizing the difference between the two is critical for exam success.

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.

2021 · 1
Counting seating arrangements
2022 · 1
Combinations for team selection
2023 · 2
Permutations with repeated elements

Going deeper

SyllabusIB Mathematics AI HL Official SyllabusOfficial reference for Unit 1 content

What's Next

Counting principles are the foundational building block for calculating probabilities of discrete events, which you will use extensively across all statistics topics in IB AI HL. You will apply these rules to calculate probabilities for binomial distributions, expected value, and combinatorial hypothesis testing. Permutations and combinations problems appear regularly in both Paper 1 and Paper 2, often worth 5-8 marks per question. Mastering the distinction between permutations and combinations is the most critical skill from this topic to carry forward.