Permutations and Combinations

Introduction

students, in mathematics, sometimes the order of choices matters, and sometimes it does not. That single difference is the heart of permutations and combinations 🔢. In this lesson, you will learn how to count arrangements and selections accurately, use factorial notation, and choose the correct method for a problem. These ideas are important in probability, algebraic reasoning, and many real-world situations such as arranging books on a shelf, choosing a team, or creating passwords.

Learning objectives

Explain the main ideas and terminology behind permutations and combinations.
Apply IB Mathematics: Analysis and Approaches HL reasoning or procedures related to permutations and combinations.
Connect permutations and combinations to the broader topic of Number and Algebra.
Summarize how permutations and combinations fit within Number and Algebra.
Use evidence or examples related to permutations and combinations in IB Mathematics: Analysis and Approaches HL.

A key skill is deciding whether a situation is about arranging objects in order or selecting objects without regard to order. Once that is clear, the counting process becomes much easier.

Core ideas: what is being counted?

A permutation is an arrangement where order matters. If three students, A, B, and C, line up, then $ABC$, $ACB$, $BAC$, $BCA$, $CAB$, and $CBA$ are all different permutations. The same three people arranged in a different order give a different result.

A combination is a selection where order does not matter. If students chooses 2 students from A, B, and C to form a pair, then the pairs $AB$, $AC$, and $BC$ are the only combinations. The pair $AB$ is the same selection as $BA$ because the order is irrelevant.

This difference is crucial. Many counting mistakes happen because students count ordered arrangements when the problem only asks for selections.

Factorials

The factorial notation is central to both topics. For a positive integer $n$,

$$n! = n(n-1)(n-2)\cdots 3\cdot 2\cdot 1$$

Also, $0! = 1$. This value is not random; it is chosen so that many counting formulas work correctly.

For example:

$5! = 5\cdot 4\cdot 3\cdot 2\cdot 1 = 120$
$3! = 6$
$0! = 1$

Factorials grow very quickly, which is why they often appear in problems involving large numbers of arrangements.

Permutations: order matters

Suppose students wants to arrange $r$ objects chosen from $n$ distinct objects. The number of permutations is

$$^nP_r = \frac{n!}{(n-r)!}$$

This formula counts how many ordered arrangements are possible when no object is repeated.

Why this formula works

Imagine choosing the first position in an arrangement. There are $n$ choices. For the second position, there are $n-1$ choices. For the third, there are $n-2$ choices, and so on, until $r$ positions are filled. So the total is

$$n(n-1)(n-2)\cdots (n-r+1)$$

This product is exactly the same as

$$\frac{n!}{(n-r)!}$$

because the terms from $(n-r)!$ cancel out.

Example 1: arranging books

If 4 different books are arranged on a shelf, the number of possible orders is

$$^4P_4 = \frac{4!}{(4-4)!} = \frac{4!}{0!} = 24$$

So there are 24 ways to arrange the books. This is a permutation because the order of the books on the shelf matters.

Example 2: choosing leaders

Suppose 8 students are available, and students wants to choose a president, vice-president, and secretary. Since these jobs are different, order matters. The number of ways is

$$^8P_3 = \frac{8!}{5!} = 8\cdot 7\cdot 6 = 336$$

A person chosen as president is not the same as that person chosen as secretary, so this is a permutation.

Special note: repeated objects

If some objects are identical, the number of different arrangements is smaller. For example, the word $LEVEL$ has 5 letters, but some repeat. The number of distinct arrangements is

$$\frac{5!}{2!\,2!}$$

because $L$ appears twice and $E$ appears twice. This idea is useful in symbolic manipulation because repeated factors in factorial expressions often simplify neatly.

Combinations: order does not matter

When students selects $r$ objects from $n$ distinct objects and order does not matter, the number of combinations is

$$^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

This formula is read as “$n$ choose $r$.”

Why this formula works

Each combination can be arranged in $r!$ different ways. For example, if a group of 3 students is selected, those 3 students can be ordered in $3! ways. A permutation counts every order separately, but a combination counts the group only once. Therefore,

$$^nC_r = \frac{^nP_r}{r!}$$

Substituting the permutation formula gives

$$^nC_r = \frac{n!}{r!(n-r)!}$$

Example 3: choosing a committee

If 10 students are available and students chooses 4 for a committee, the number of combinations is

$$^{10}C_4 = \frac{10!}{4!6!} = 210$$

This is a combination because the committee is the same regardless of the order in which the 4 students are selected.

Example 4: selecting a pizza

If a pizza shop offers 6 toppings and students chooses 2 toppings, then the number of possible topping pairs is

$$^6C_2 = \frac{6!}{2!4!} = 15$$

The order of toppings does not matter, so this is a combination.

How to decide: permutation or combination?

A quick rule is this:

If the problem involves arranging, ranking, ordering, or assigning distinct positions, use a permutation.
If the problem involves choosing, selecting, or forming a group, use a combination.

A useful test is to ask: “Would switching the order create a new result?” If yes, use a permutation. If no, use a combination.

Example 5: mixed reasoning

A class has 12 students. students chooses 3 students to receive a prize, and then ranks them gold, silver, and bronze.

Choosing 3 students without ranking is a combination:

$$^{12}C_3$$

Ranking those 3 students is a permutation:

$$^{12}P_3$$

Because gold, silver, and bronze are different positions, the order matters.

Links to Number and Algebra

Permutations and combinations belong to Number and Algebra because they use integer counting, factorial notation, and algebraic formulas. They also support symbolic manipulation, especially when simplifying expressions like

$$\frac{n!}{(n-r)!}$$

$$\frac{n!}{r!(n-r)!}$$

These formulas require careful cancellation and reasoning with variables. In IB Mathematics: Analysis and Approaches HL, this connects with algebraic structure because students must recognize patterns, simplify expressions, and use notation accurately.

Permutations and combinations also support work in sequences and series. Factorials are examples of products with regular structure, and counting formulas often appear inside algebraic proofs or probability expressions. They are also used in binomial expansions, where coefficients are combinations:

$$\binom{n}{r}$$

These coefficients help build terms in expressions like

$$\left(a+b\right)^n$$

So counting methods are not isolated topics; they connect to many other parts of the syllabus.

Common mistakes and how to avoid them

A frequent error is using a combination when the question needs a permutation. For example, if students is assigning roles, the positions are different, so order matters.

Another mistake is forgetting that $0! = 1$. This becomes important when simplifying expressions such as

$$^nP_n = \frac{n!}{0!} = n!$$

A third mistake is mixing up the meaning of the notation. Remember:

$^nP_r$ counts ordered arrangements.
$^nC_r$ counts unordered selections.

Careful reading is essential. In exam questions, keywords such as “arrange,” “order,” “rank,” and “select” give clues about which formula to use.

Conclusion

Permutations and combinations are fundamental counting tools in IB Mathematics: Analysis and Approaches HL. students should now be able to explain the difference between order matters and order does not matter, use the formulas

$$^nP_r = \frac{n!}{(n-r)!}$$

and

$$^nC_r = \frac{n!}{r!(n-r)!}$$

to solve problems, and connect these methods to algebraic reasoning in Number and Algebra. These ideas are especially important because they support probability, binomial expansion, and other higher-level mathematical topics. Understanding when to count arrangements and when to count selections is a powerful skill ✅

Study Notes

A permutation counts arrangements where order matters.
A combination counts selections where order does not matter.
Factorial notation is $n! = n(n-1)(n-2)\cdots 1$, and $0! = 1$.
The permutation formula is $^nP_r = \frac{n!}{(n-r)!}$.
The combination formula is $^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.
If swapping two selected objects creates a new outcome, use a permutation.
If swapping two selected objects does not create a new outcome, use a combination.
Repeated objects reduce the number of distinct arrangements.
These ideas connect to algebra through factorial manipulation and to probability through counting outcomes.
In IB Mathematics: Analysis and Approaches HL, permutations and combinations support binomial expansion, proof, and structured reasoning.