What are splitting into groups?

To split n distinct objects into groups of sizes r_1, r_2, , r_k (with r_1 + + r_k = n):

What is symmetry the wrong way?

2017 = 203 because 20 - 17 = 3. Subtract from the top number; do not just swap digits.

What is pascal's rule sign?

nr = n - 1r - 1 + n - 1r has a plus. There is no subtraction.

§-Syllabus dot point

NSWMaths Extension 1Syllabus dot point

How do we count the number of unordered selections (combinations) of objects from a larger set?

Use the combination formula $\binom{n}{r}$ to count unordered selections, including with restrictions and complementary counting

A focused answer to the HSC Maths Extension 1 dot point on combinations. The combination formula, why you divide by $r!$ , key identities, applications including complementary counting, splitting into groups, and at-least/at-most counts, with a stepped diagram and worked examples.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to count unordered selections (combinations) of $r$ objects from $n$ distinct objects, apply the formula $\binom{n}{r}$ , use the key identities (symmetry, Pascal's rule, the sum to $2^n$ ), and handle the standard restriction types: "at least", "at most", "must include / must exclude a particular object", and splitting a set into groups. The single idea that unlocks the whole topic is that a combination is a permutation with the ordering thrown away, which is why the formula has an extra division by $r!$ .

The answer

The combination formula

The number of ways to choose $r$ unordered objects from $n$ distinct objects is

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

The notation $\binom{n}{r}$ is read " $n$ choose $r$ " and is also written ${}^n C_r$ . It is defined for integers $0 \le r \le n$ .

Why you divide by $r!$ (the link to permutations)

A combination is what you get from a permutation when order stops mattering. To choose and arrange $r$ from $n$ there are ${}^{n}P_r = \frac{n!}{(n-r)!}$ ordered lists. But every unordered selection of $r$ objects can be written as $r!$ different ordered lists (the $r!$ ways of arranging those same $r$ objects). So the ordered count over-counts the unordered count by exactly $r!$ , and

\binom{n}{r} = \frac{{}^{n}P_r}{r!} = \frac{n!}{r!\,(n-r)!}.

The picture below makes the "divide by $r!$ " concrete for $r = 3$ : a single chosen set $\{A, B, C\}$ corresponds to $3! = 6$ ordered arrangements.

Key identities

Symmetry.

\binom{n}{r} = \binom{n}{n - r}.

Choosing $r$ objects to include is the same act as choosing $n - r$ objects to exclude. This is also a labour-saver: compute whichever of the two has the smaller bottom number.

Pascal's rule.

\binom{n}{r} = \binom{n - 1}{r - 1} + \binom{n - 1}{r}.

Focus on one particular object. Either it is in the chosen subset (then choose the remaining $r - 1$ from the other $n - 1$ ) or it is not (then choose all $r$ from the other $n - 1$ ). These two cases are exclusive and exhaustive, so they add. This identity is exactly what builds Pascal's triangle row by row.

Sum identity.

\sum_{r = 0}^{n} \binom{n}{r} = 2^n.

The left side counts subsets of every size; the right side counts subsets by deciding "in or out" independently for each of the $n$ elements. Both count all subsets of an $n$ -element set.

Boundary values.

$\binom{n}{0} = \binom{n}{n} = 1$ (one way to choose nothing, one way to choose everything), and $\binom{n}{1} = n$ .

Complementary counting

For "at least $k$ " or "at most $k$ " problems it is usually far quicker to count the complement than to add up every allowed case:

(\text{at least 1}) = \text{total} - (\text{none}),

(\text{at least 2}) = \text{total} - (\text{none}) - (\text{exactly 1}).

The signal to reach for the complement is the phrase "at least one", because its opposite ("none") is a single, easy combination, whereas the direct count is a sum of several cases.

Splitting into groups

To split $n$ distinct objects into groups of sizes $r_1, r_2, \dots, r_k$ (with $r_1 + \dots + r_k = n$ ):

\binom{n}{r_1,\, r_2,\, \dots,\, r_k} = \frac{n!}{r_1! \, r_2! \cdots r_k!}.

This is the multinomial coefficient, and equivalently it is the chain $\binom{n}{r_1} \cdot \binom{n - r_1}{r_2} \cdots \binom{r_k}{r_k}$ (choose the first group, then the second from what is left, and so on). A crucial wrinkle: this counts labelled groups (team A, team B, team C). If the groups are the same size and unlabelled, divide by the number of equal-sized groups factorial, because any relabelling of them gives the same division.

Counting with constraints

"Must include object $A$ ": $A$ is already in, so choose the remaining $r - 1$ from the other $n - 1$ : $\binom{n - 1}{r - 1}$ .

"Must exclude object $A$ ": $A$ is out, so choose all $r$ from the other $n - 1$ : $\binom{n - 1}{r}$ .

"Includes $A$ or $B$ but not both": add (includes $A$ , excludes $B$ ) and (includes $B$ , excludes $A$ ).

"At least one of $A, B, C$ ": use complementary counting (total minus the selections containing none of them).

How exam questions ask about combinations

"How many [committees / teams / hands / subsets] of size $r$ ...?" with no mention of order: a plain $\binom{n}{r}$ .
"... containing at least one [type]?": complementary counting, total minus none.
"... containing exactly $k$ [type]?": choose the $k$ from that type and the rest from the others, then multiply: $\binom{a}{k}\binom{b}{r-k}$ .
"... that must include [a named person/object]?": fix it in, then $\binom{n-1}{r-1}$ .
"... that must not include [a named person/object]?": $\binom{n-1}{r}$ .
"Split / divide [people] into groups of ...": multinomial $\dfrac{n!}{r_1!\cdots r_k!}$ ; divide by the symmetry factor if equal-sized groups are unlabelled.
"In how many ways can a [defence] and a [forward line] be chosen?" (two pools): choose from each pool and multiply.
"Show that $\binom{n}{r} = \binom{n}{n-r}$ " or a Pascal's-rule identity: argue by the in/out or include/exclude meaning, not just algebra.
Watch for the order cue. "Arranged", "in a row", "ranked", "first, second, third" means a permutation, not a combination.

Exam technique

First decide whether order matters: if the wording is "select", "choose", "committee", "subset", "hand", order does not matter, so use $\binom{n}{r}$ . Use the symmetry identity to keep the arithmetic small ( $\binom{20}{18} = \binom{20}{2}$ , not the other way). For "at least one", write the complement line explicitly: "(at least one) = total $-$ (none)", then evaluate the two combinations. When two independent selections are made (defenders and forwards, men and women), multiply the two $\binom{}{}$ counts; when you are counting mutually exclusive cases (exactly 0, exactly 1, ...), add them. For a group-splitting question, state whether the groups are labelled and divide by the symmetry factor if they are not. Leave the answer as an exact integer.

Worked example

Apply the combination formula

To count the number of ways to choose $3$ books from $10$ , order does not matter. The combination formula gives:

\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = \frac{720}{6} = 120.

Final answer: $120$ ways.

Use the symmetry identity

Computing $\binom{20}{18}$ directly requires large factorials. The symmetry identity $\binom{n}{r} = \binom{n}{n-r}$ reflects that choosing $18$ items to include is the same as choosing $2$ items to exclude:

\binom{20}{18} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190.

Final answer: $190$ .

Multiply two independent selections

The defenders and forwards are chosen from separate pools, so the selections are independent. Choose $4$ defenders from $9$ and $3$ forwards from $6$ separately, then multiply by the multiplication principle:

\binom{9}{4} \cdot \binom{6}{3} = 126 \cdot 20 = 2520.

Final answer: $2520$ teams.

Count "at least one" by complementary counting

Direct counting of "at least one vowel" requires separate cases (exactly 1 vowel, exactly 2 vowels). The complement is simpler: count subsets with no vowels at all. The vowels in $\{$ A, B, C, D, E, F $\}$ are A and E; the consonants are B, C, D, F.

Total subsets of size $3$ : $\binom{6}{3} = 20$ .
No vowels (all three from B, C, D, F): $\binom{4}{3} = 4$ .

Subtract to find the answer:

\text{At least one vowel} = 20 - 4 = 16.

Final answer: $16$ subsets.

Count "exactly $k$ of a type"

A committee of $4$ is chosen from $7$ women and $5$ men. How many have exactly $2$ women? Choose the $2$ women from $7$ and the other $2$ members (men) from $5$ , then multiply:

\binom{7}{2} \cdot \binom{5}{2} = 21 \cdot 10 = 210.

Final answer: $210$ committees.

Count selections that must include a given element

"War and Peace" must be included, so it is already chosen. The remaining $3$ books are selected from the other $14$ , with no further restriction:

\binom{14}{3} = \frac{14 \cdot 13 \cdot 12}{3!} = 364.

Final answer: $364$ selections.

Split into labelled groups (multinomial coefficient)

If the three teams are labelled (team A, team B, team C), the number of ways to split $9$ people into groups of $3$ is:

\frac{9!}{3!\, 3!\, 3!} = 1680.

Adjust for indistinguishable teams:

If the three teams have no labels, then each division has been counted $3! = 6$ times (once for each permutation of which group is called A, B, or C). Divide by $6$ :

\frac{1680}{6} = 280.

Final answer: $1680$ for labelled teams, $280$ for unlabelled groups.

Verify Pascal's rule by direct calculation

Pascal's rule states $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ . To verify $\binom{6}{3} = \binom{5}{2} + \binom{5}{3}$ , compute each term. The rule says: either the sixth person is in the subset (choose the other $2$ from $5$ ) or not (choose all $3$ from $5$ ):

\binom{5}{2} + \binom{5}{3} = 10 + 10 = 20 = \binom{6}{3}. \quad \checkmark

Final answer: confirmed, $\binom{6}{3} = 20$ .

Common traps

Confusing combination with permutation: Combinations are unordered; permutations are ordered. The formulas differ by a factor of $r!$ (the combination is the permutation divided by $r!$ ). A word like "arranged" or "ranked" means permutation.
Symmetry the wrong way: $\binom{20}{17} = \binom{20}{3}$ because $20 - 17 = 3$ . Subtract from the top number; do not just swap digits.
Splitting groups: labelled vs unlabelled: Three labelled teams of $3$ ( $\frac{9!}{3!\,3!\,3!}$ ) differ from three unlabelled groups of $3$ (divide by another $3!$ ). Decide which the question means.
Pascal's rule sign: $\binom{n}{r} = \binom{n - 1}{r - 1} + \binom{n - 1}{r}$ has a plus. There is no subtraction.
Adding instead of using the complement: Counting "at least 1 woman" by adding women-only and men-only mixes cases incorrectly and double-counts. Use total minus (no women).
Multiplying mutually exclusive cases: "Exactly 1 vowel OR exactly 2 vowels" are separate cases and must be added, not multiplied. Multiply only independent stages of a single selection.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC Q72 marksHow many five-card hands are there from a standard deck of

52

cards?

Show worked answer →

Unordered selection of $5$ from $52$ .

$\binom{52}{5} = \frac{52!}{5! \, 47!} = \frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 2 \, 598 \, 960$ .

Markers reward identifying that order does not matter, the combination formula, and the arithmetic.

2022 HSC Q243 marksA committee of

4

is to be formed from

7

women and

5

men. How many committees include at least

1

woman?

Show worked answer →

Use complementary counting: $(\text{at least 1 woman}) = (\text{total}) - (\text{no women})$ .

Total committees of $4$ from $12$ : $\binom{12}{4} = 495$ .

Committees with no women (all $4$ men): $\binom{5}{4} = 5$ .

At least one woman: $495 - 5 = 490$ .

Markers reward setting up the complement, computing both combinations, and the subtraction.