Unit 8 · Statistics & ProbabilityS6 · Counting & EV · Telemetry

Counting: Permutations & Combinations

Use the multiplication principle and decide whether order matters.

28 min
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Start with the warm-up.

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Warm-up

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Key terms

factorial: The product n(n-1)...1.
permutation: A selection where order matters.
combination: A selection where order does not matter.

Start by naming the quantity each problem asks for. That habit blocks the most common ACT trap.

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Take a calm first shot. Label what you know, write one relationship, and make the best choice you can.

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The big idea

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Build the picture before the arithmetic

Slots suggest multiplication or permutations. Groups, committees, and subsets suggest combinations. Restrictions are easiest when you handle the special slot first.

Key idea

The decision that matters

Ask whether swapping two selected people creates a new outcome. If yes, order matters; if no, divide away the repeated orders.

Combinations

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

Write this relationship before inserting numbers. It keeps labels and units attached to the work.

A compact example

Choosing 3 of 8 students gives $\binom83=56$ committees.

Cue	Best move
Committee or subset	Order does not matter: use a combination
Ranks or positions	Order matters: use a permutation

Trap

Using a permutation for an unordered team

This error produces a believable answer. Pause before computing and say what each number means.

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Skip check · 2 questions

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Worked example

Example 1: Choose a committee

Problem

How many 3-person committees can be chosen from 8 students?

A committee is unchanged when its members are listed in a different order, so use a combination.

$\binom83=\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=56$ .

Which counting method gives the number of committees?

Desmos supports nCr(8,3) and returns 56; typing the short product is just as fast here.

Why does choosing a committee require a combination?

Worked example

Example 2: Handle a restriction first

Problem

A 4-digit code uses distinct digits from 0 through 9 and cannot begin with 0. How many codes are possible?

The first digit has 9 choices: 1 through 9.

After that, 9 digits remain for the second slot, then 8, then 7.

How should the no-leading-zero restriction be counted?

The count is $9\cdot9\cdot8\cdot7=4,536$ .

Why are there 9 choices in each of the first two positions?

Faded example · you finish it

Example 3: You finish it

Problem

Six finalists compete for gold, silver, and bronze. Complete the count.

Decide whether swapping two selected people creates a new outcome before choosing nCr or nPr.

Which completion is correct?

Which reason validates the result?

Desmos way

Check a large count

For an unordered selection, type nCr(8,3) and read 56.

For ordered medals, type nPr(6,3) and read 120. Choose the model before touching the calculator.

Check a large count

Degree mode

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Spot the mistake

Where did it go wrong?

One line has an error. Click the line where it first goes wrong.

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Optional · 4 questions · Hardest levelChallenge set: the kind of question that decides a 36

These are last-ten-questions hard. Take your time, use Desmos when it helps, and expect to miss some: that's where the learning is.