Your weak spots

Insights load after your first practice round.

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.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

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

auto-saved

The core idea

+5 XP to read

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.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that 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)

auto-saved

ApplyBand 41 mark

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

auto-saved

AnalyseBand 52 marks

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

auto-saved

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

earn bronze · silver · gold

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

Tick when you've finished the practice and review.

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

Module overview · Extension 1 · Checkpoint 2