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, key identities, applications including complementary counting, splitting into groups, and at-least/at-most counts, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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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 key identities, and handle restrictions like "at least", "at most", and splitting into groups.

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 ${}^n C_r$ .

The denominator $r!$ divides out the over-count: each unordered subset of size $r$ corresponds to $r!$ different ordered arrangements.

Key identities

Symmetry:

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

Choosing $r$ to include is equivalent to choosing $n - r$ to exclude.

Pascal's rule:

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

Either the $n$ -th object is in the subset (so choose $r - 1$ from the remaining $n - 1$ ) or it is not (so choose $r$ from $n - 1$ ).

Sum identity:

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

This counts all subsets of an $n$ -element set.

Boundary:

$\binom{n}{0} = \binom{n}{n} = 1$ , and $\binom{n}{1} = n$ .

Complementary counting

For "at least $k$ " or "at most $k$ " problems, it is often easier to count the complement.

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

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

This avoids tedious case work.

Splitting into groups

To split $n$ 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. Equivalently, $\binom{n}{r_1} \cdot \binom{n - r_1}{r_2} \cdot \dots \cdot \binom{r_k}{r_k}$ .

Counting with constraints

"Includes object A": A is in the subset. Choose the remaining $r - 1$ from the other $n - 1$ : $\binom{n - 1}{r - 1}$ .

"Excludes object A": A is not in the subset. Choose all $r$ from the other $n - 1$ : $\binom{n - 1}{r}$ .

"Includes A or B but not both": count includes-A-excludes-B plus includes-B-excludes-A.

"At least one of A, B, C": use complementary counting or inclusion-exclusion.

Worked example

Direct combination

How many ways to choose $3$ books from $10$ ? $\binom{10}{3} = \frac{720}{6} = 120$ .

Symmetry

$\binom{20}{18} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$ . Faster than computing $\binom{20}{18}$ directly.

Two groups

A team of $7$ has $4$ defenders and $3$ attackers. Select from $9$ defenders and $6$ attackers.

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

At least

How many three-letter subsets of $\{$ A, B, C, D, E, F $\}$ include at least one vowel?

Total: $\binom{6}{3} = 20$ .

No vowels: vowels are A, E (two of them); other letters B, C, D, F. Subsets of three from B, C, D, F: $\binom{4}{3} = 4$ .

At least one vowel: $20 - 4 = 16$ .

Includes a specific element

A book club selects $4$ books from $15$ . How many selections include "War and Peace"?

Include it: choose $3$ from the other $14$ . $\binom{14}{3} = 364$ .

Splitting into three groups

Split $9$ people into three teams of $3$ .

If teams are distinguishable (team A, B, C): $\binom{9}{3, 3, 3} = \frac{9!}{3! \, 3! \, 3!} = 1680$ .

If teams are indistinguishable (just three groups, not labelled): divide by $3!$ for the swap symmetry: $\frac{1680}{6} = 280$ .

Pascal in action

Verify $\binom{6}{3} = \binom{5}{2} + \binom{5}{3}$ .

$\binom{5}{2} = 10$ , $\binom{5}{3} = 10$ , sum $= 20 = \binom{6}{3}$ . Confirmed.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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.