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Module 4 · L6 of 15 ~35 min ⚡ +95 XP available

Combinations, $^nC_r$

You're picking 5 cards from a deck of 52. The order you pick them doesn't matter, {A, K, Q, J, 10} is the same hand no matter which card you reached for first. That's the essence of a combination: we're choosing a group, not forming a sequence. Combinations are the engine behind committee selection, lottery mathematics, and every "choose $r$ from $n$" problem in the HSC.

Today's hook, A class of 10 students needs to elect a committee of 3. Does the order the three people are chosen matter? No, the committee is the same group regardless. By the end of this lesson you'll know exactly when to use combinations instead of permutations, and you'll be able to evaluate $^{10}C_3$ in under 10 seconds.

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Recall, your gut answer first

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You want to choose 3 friends from a group of 10 to invite to your birthday party. Without any formulado you think the number of possible groups is closer to 100, 500, or more than 1000? Write your estimate and reason.

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The core idea

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There is one key question to ask before every combinatorics problem: does order matter? If yes, use permutations. If no, use combinations.

A permutation is an arrangement the order in which objects are placed matters. A combination is a selection only which objects are chosen matters, not how they are ordered. The combination formula divides permutations by $r!$ to cancel out all the internal orderings of the chosen $r$ objects.

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

Order test

Ask: "Does swapping two chosen objects give a different result?" Committee/team/group = no $\Rightarrow$ combination. Captain/president/ordered list = yes $\Rightarrow$ permutation.

Symmetry shortcut

$^nC_r = \,^nC_{n-r}$. Choosing 3 from 10 is equivalent to choosing who doesn't come (7 from 10). Use whichever is easier to compute.

Link to permutations

$^nP_r = \,^nC_r \times r!$: first choose $r$ objects (combinations), then arrange them ($r!$ ways). Permutation = combination $\times$ arrangement.

What you'll master

Know

Key facts

$^nC_r = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}$
$^nC_0 = \,^nC_n = 1$ and $^nC_1 = n$
$^nC_r = \,^nC_{n-r}$ (symmetry property)
$^nP_r = \,^nC_r \times r!$

Understand

Concepts

Why combinations divide permutations by $r!$
The difference between selecting a group (combination) and filling ranked roles (permutation)
Why $^nC_r = \,^nC_{n-r}$ makes intuitive sense

Can do

Skills

Evaluate $^nC_r$ for any given $n$ and $r$
Decide whether a problem requires combinations or permutations
Apply $^nC_r$ to committee, team, and selection problems

Key terms

CombinationA selection of $r$ objects from $n$ distinct objects where order does not matter. Denoted $^nC_r$ or $\binom{n}{r}$.

$^nC_r$Read "n choose r". Equals $\dfrac{n!}{r!(n-r)!}$. Also written $\binom{n}{r}$ (binomial coefficient notation).

Order irrelevantThe defining property of combinations: the set $\{A,B,C\}$ is the same selection as $\{C,A,B\}$, both count as one combination.

Symmetry property$^nC_r = \,^nC_{n-r}$. Choosing $r$ objects is equivalent to choosing the $n-r$ objects not selected.

Permutation–combination link$^nP_r = \,^nC_r \times r!$: choose which $r$ objects, then arrange them internally.

Binomial coefficientThe notation $\binom{n}{r}$ is identical to $^nC_r$. It appears as the coefficients in the expansion of $(a+b)^n$, more on this in Lesson 9.

What is a combination, and where does the formula come from?

core concept

A combination is a selection where order does not matter. The number of ways to choose $r$ objects from $n$ distinct objects (without regard to order) is written $^nC_r$ or $\binom{n}{r}$.

Derivation: Start with permutations. There are $^nP_r = \dfrac{n!}{(n-r)!}$ ordered arrangements of $r$ objects from $n$. Each unordered selection (combination) corresponds to $r!$ ordered arrangements (one for each way to rearrange the chosen $r$ objects). So:

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

Dividing by $r!$ "undoes" the internal ordering, it removes all the rearrangements of the chosen group that permutations would count as distinct.

Real-world connection. Lotto: to win Oz Lotto you pick 7 numbers from 1–47. The number of possible tickets is $^{47}C_7 = \dfrac{47!}{7! \cdot 40!} = 62{,}891{,}499$. Your chance of winning is $\dfrac{1}{62{,}891{,}499} \approx 1.6 \times 10^{-8}$. Order definitely doesn't matter, the winning numbers are drawn as a set.

Combination: selection where order does not matter; ^nC_r = n{r} = n!{r!(n-r)!}

Pause, copy the combination formula into your book: $^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$, number of ways to choose $r$ objects from $n$ when order does not matter.

Quick check: How many ways can a committee of 3 be chosen from 8 people?

Permutation vs combination, the order test

core concept

We just saw that $^nP_r = \frac{n!}{(n-r)!}$ counts ordered selections. That raises a question: when order doesn't matter (choosing a committee, not assigning roles), dividing by $r!$ removes the overcounting, but why exactly $r!$? This card answers it → each set of $r$ chosen objects can be arranged in $r!$ orders, so $^nC_r = ^nP_r / r! = \frac{n!}{r!(n-r)!}$.

The single most important skill in this topic is identifying which formula to use. Apply the order test:

Does the arrangement matter? (e.g., 1st place, 2nd place, 3rd place; president, vice-president, treasurer) → Permutation $^nP_r$
Is only the group important? (e.g., a committee, a team, a hand of cards, a selection of books) → Combination $^nC_r$

The relationship between them: $^nP_r = \,^nC_r \times r!$. You can think of this as: to make an ordered arrangement, first choose which $r$ objects ($^nC_r$ ways), then order them ($r!$ ways).

$$^nP_r = \,^nC_r \times r! \qquad \Longrightarrow \qquad ^nC_r = \frac{^nP_r}{r!}$$

Symmetry property. $^nC_r = \,^nC_{n-r}$. Proof: $\dfrac{n!}{r!(n-r)!} = \dfrac{n!}{(n-r)!r!}$, the denominators are the same. Intuitively: choosing 3 people from 10 to include is the same count as choosing 7 people to exclude. Use this to simplify calculations: $^{20}C_{17} = \,^{20}C_3 = \dfrac{20 \times 19 \times 18}{3!} = 1140$.

Order matters permutation ^nP_r; order irrelevant combination ^nC_r; Keywords for combinations: choose, select, committee, team, group, hand, subset

Pause, copy the order-test decision rule into your book: order matters → use $^nP_r$; order irrelevant → use $^nC_r$; keywords for combinations: choose, select, committee, team, group, hand, subset.

Did you get this? True or false: selecting a president, vice-president, and treasurer from 10 candidates requires a combination, not a permutation.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC COMBINATION

How many ways can a committee of 3 be chosen from 8 people?

Order doesn't matter for a committee, use $^nC_r$ with $n=8$, $r=3$.

A committee is an unordered group; swapping who was chosen first doesn't create a new committee.

PROBLEM 2 · EVALUATE nCr

Evaluate $^{10}C_4$.

$^{10}C_4 = \dfrac{10!}{4! \cdot 6!} = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$

Write out only the top $r$ factors of $n!$ (cancel the $6!$) and divide by $r! = 4!$.

PROBLEM 3 · SYMMETRY PROPERTY

Show that $^nC_r = \,^nC_{n-r}$, and use it to evaluate $^{20}C_{17}$ efficiently.

$^nC_{n-r} = \dfrac{n!}{(n-r)!\cdot(n-(n-r))!} = \dfrac{n!}{(n-r)!\cdot r!} = \,^nC_r$ ✓

Substitute $r \to n-r$ in the formula. The denominator is just reordered, it equals $r!(n-r)!$ in both cases.

Fill the gap: $^7C_3 = \dfrac{7 \times 6 \times 5}{3!} = \dfrac{210}{6} = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using permutations for selections

Committees, teams, and groups are combinations, order is irrelevant. Using $^nP_r$ counts each group $r!$ times too many, inflating the answer. Ask: "does the order of selection matter?" For groups, the answer is always no.

Trap 02

Forgetting to divide by $r!$

The most common arithmetic error: writing $^nC_r = \dfrac{n!}{(n-r)!}$ (which is $^nP_r$) and forgetting the $r!$ in the denominator. Double-check: $^nC_r$ must be less than or equal to $^nP_r$.

Trap 03

Not using the symmetry shortcut

Evaluating $^{20}C_{17}$ by computing $20!/17!3!$ directly is error-prone. Always reduce to the smaller $r$ first: $^{20}C_{17} = \,^{20}C_3$. This saves time and reduces the chance of arithmetic errors.

Did you get this? True or false: $^{15}C_{12} = \,^{15}C_3$.

Key formulas summary

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

$$^nC_0 = \,^nC_n = 1 \qquad ^nC_1 = n$$

$$^nC_r = \,^nC_{n-r} \quad \text{(symmetry)}$$

$$^nP_r = \,^nC_r \times r!$$

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Odd one out: Which of the following requires a permutation rather than a combination?

Evaluate $^7C_3$.

A team of 4 is to be chosen from 10 players. How many different teams are possible?

Show that $^nC_r = \,^nC_{n-r}$. Provide both an algebraic proof and an intuitive explanation.

A hand of 5 cards is dealt from a standard 52-card deck. How many possible hands are there?

In your own words, explain why we divide by $r!$ when calculating $^nC_r$ from $^nP_r$.

Revisit your thinking

Earlier you estimated the number of ways to choose 3 friends from 10. The exact answer is $^{10}C_3 = \dfrac{10 \times 9 \times 8}{3!} = \dfrac{720}{6} = 120$.

Compare with the number of ways to rank 3 friends from 10 (i.e., who is your closest, second, and third friend): $^{10}P_3 = 720$. The ratio is exactly $3! = 6$, because each unordered group of 3 corresponds to $3! = 6$ ordered arrangements.

If combinations are still unclear, compare permutation and combination problems side by side: "elect a president, VP, and treasurer from 10" (permutation = 720) versus "select a 3-person committee from 10" (combination = 120). The only difference is whether the roles are labelled.

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Multiple choice

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Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 41 mark

Q1. Evaluate $^7C_3$. (1 mark)

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ApplyBand 41 mark

Q2. A team of 4 is to be chosen from 10 players. How many different teams are possible? (1 mark)

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AnalyseBand 52 marks

Q3. Show that $^nC_r = \,^nC_{n-r}$. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $^7C_3 = \frac{7 \times 6 \times 5}{6} = 35$ · 2. $^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{24} = 210$ · 3. Algebraic: $^nC_{n-r} = \frac{n!}{(n-r)!r!} = ^nC_r$. Intuitive: choosing $r$ objects to include is equivalent to choosing $n-r$ objects to exclude. · 4. $^{52}C_5 = \frac{52!}{5! \cdot 47!} = 2\,598\,960$ · 5. Each unordered selection (combination) corresponds to $r!$ ordered arrangements (permutations). Dividing by $r!$ removes these duplicate orderings so each group is counted once.

Q1 (1 mark): $^7C_3 = \dfrac{7!}{3! \cdot 4!} = \dfrac{7 \times 6 \times 5}{6} = 35$ [1].

Q2 (1 mark): $^{10}C_4 = \dfrac{10!}{4! \cdot 6!} = \dfrac{10 \times 9 \times 8 \times 7}{24} = 210$ [1].

Q3 (2 marks): $^nC_{n-r} = \dfrac{n!}{(n-r)!\cdot(n-(n-r))!} = \dfrac{n!}{(n-r)!\cdot r!}$ [1]. This equals $\dfrac{n!}{r!(n-r)!} = \,^nC_r$ [1]. Therefore $^nC_r = \,^nC_{n-r}$. $\square$

Boss battle · The Selector

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Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering combinatorics questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 5 · Circular Permutations Lesson 7 · Properties of Combinations →

Module overview · Extension 1 · Checkpoint 2