# Counting Principles, Arrangements and Selections

> The multiplication principle, factorials, arrangements where order matters, selections where it does not, and counting in probability.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/counting-principles/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The school uniform
Zoe's school uniform has 3 shirts, 2 skirts and 4 pairs of socks to choose from. She wonders how many different outfits she can make. She could list them all, but there is a faster way: $3 \times 2 \times 4 = 24$.

When the choices get bigger, like the number of ways to pick a captain and vice-captain from a team, or 6 lotto numbers out of 45, listing is impossible. Counting principles give the answer in one line, and they make many probability questions easy.

**In short:** Counting principles find how many outcomes there are without listing them all. If one choice has $m$ ways and the next $n$ ways, together they have $m \times n$ ways. Use $^nP_r$ when order matters and $^nC_r$ when it does not.

## Starting point: Years 9 and 10

### The multiplication principle

If one choice can be made in $m$ ways and a second in $n$ ways, together they can be made in $m \times n$ ways. This works for any number of choices.

### Worked example: Number plates
A plate has 3 letters then 3 digits. There are $26 \times 26 \times 26 \times 10 \times 10 \times 10 = 17\,576\,000$ plates.

### Factorials

$$n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$$

So $4! = 24$ and $5! = 120$. By definition, $0! = 1$.

$n!$ counts the ways to arrange $n$ different objects in a row. Five friends can stand in a line in $5! = 120$ ways.

### Calculator tip: Factorials, nPr and nCr
The factorial key `x!` is usually `SHIFT` and another key. `nPr` and `nCr` are usually `SHIFT` on the `×` and `÷` keys. For $\binom{10}{3}$ type `10 nCr 3` to get 120. For arrangements with repeated letters, put the bottom in brackets: `10! ÷ ( 3! × 3! × 2! )`.

### Arrangements: order matters

Choosing $r$ objects from $n$ **in order** (like first, second, third):

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

### Worked example: Captain and vice-captain
From 12 players: $12 \times 11 = 132$ ways. Order matters, because being captain is different from being vice-captain. Check: $^{12}P_2 = \dfrac{12!}{10!} = 132$.

**Restrictions.** Fill the restricted places first. How many 4-digit numbers can be made from 1, 2, 3, 4, 5 with no repeats, if the number must be even? The last digit is 2 or 4 (2 ways), then the rest: $4 \times 3 \times 2$. Total $2 \times 24 = 48$.

### Selections: order does not matter

Choosing $r$ objects from $n$ when order **does not** matter (a group, a committee, a hand of cards):

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

### Worked example: A committee
Choose 3 students from 8 for a committee: $\binom{8}{3} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

### Key idea: Order or not?
- Does swapping two chosen items make a different result? Then order matters: **arrangement**, $^nP_r$.
- If not, it is a **selection**, $^nC_r$.

### Watch out: Don't add, multiply
For "this AND then that", multiply the counts. Add only for "this OR that" when the cases cannot overlap.

**Mind map**

```text
Counting
  Multiplication principle
    m ways then n ways: m × n
  Factorials
    n! arrangements in a row
    0! = 1
  Arrangements
    Order matters
    nPr = n!/(n - r)!
    Restricted places first
  Selections
    Order does not matter
    nCr = n!/(r!(n - r)!)
  Probability
    Favourable count over total count
```

### Year 11: Counting in probability
When all outcomes are equally likely,

$$P = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$

and both counts can come from counting rules.

**Example.** Five people line up at random. What is the chance that Ana is first? Total $5! = 120$. Ana first: $4! = 24$. So $P = \dfrac{24}{120} = \dfrac{1}{5}$.

**Example.** A bag has 5 red and 4 blue balls. Two are taken. $P(\text{both red}) = \dfrac{\binom{5}{2}}{\binom{9}{2}} = \dfrac{10}{36} = \dfrac{5}{18}$, the same as the tree diagram answer $\dfrac{5}{9} \times \dfrac{4}{8}$.

**Identical objects.** The letters of BOOK can be arranged in $\dfrac{4!}{2!} = 12$ ways, because swapping the two O's changes nothing.

**Keeping two people together.** Glue them into one block. Five people in a row with two friends together: arrange 4 items (the block and 3 others) in $4!$ ways, times $2!$ for the order inside the block. So $P(\text{together}) = \dfrac{4! \times 2!}{5!} = \dfrac{48}{120} = \dfrac{2}{5}$.

**Cards.** A standard pack has 52 cards: 4 suits of 13, including 4 aces. The chance three cards dealt are all hearts is $\dfrac{\binom{13}{3}}{\binom{52}{3}} = \dfrac{286}{22\,100} \approx 0.013$.

**At least one, with counting.** The chance a 5-card hand has at least one ace is $1 - \dfrac{\binom{48}{5}}{\binom{52}{5}}$, because the hands with no aces choose all 5 from the 48 other cards.

**Handshakes and pairs.** Each handshake is a pair of people with no order, so 10 people make $\binom{10}{2} = 45$ handshakes. A lottery choosing 6 numbers from 45 has $\binom{45}{6} = 8\,145\,060$ possible tickets.

### Year 12: Counting and random variables
Counting rules build probability distributions. Toss 4 coins and let $X$ be the number of heads. There are $2^4 = 16$ outcomes, and $\binom{4}{k}$ of them have $k$ heads:

| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $P(X = x)$ | $\dfrac{1}{16}$ | $\dfrac{4}{16}$ | $\dfrac{6}{16}$ | $\dfrac{4}{16}$ | $\dfrac{1}{16}$ |

The mean is $E(X) = 2$, as you would expect. Mathematics Extension 1 takes this further with the binomial distribution.

## Quick summary

- Multiplication principle: $m$ ways then $n$ ways gives $m \times n$.
- $n! = n(n - 1) \cdots 1$ arranges $n$ objects in a row. $0! = 1$.
- Order matters: $^nP_r = \dfrac{n!}{(n - r)!}$. Fill restricted places first.
- Order does not matter: $^nC_r = \dfrac{n!}{r!(n - r)!}$.
- Probability = favourable count over total count.

## Quiz

Try these questions from the chapter quiz.

1. A café offers 4 mains and 3 desserts. How many main and dessert meals are possible?
   - A. 7
   - B. 12
   - C. 24
   - D. 64
   - Answer: B. 12. $4 \times 3$.

2. What is $5!$?
   - A. 15
   - B. 25
   - C. 120
   - D. 60
   - Answer: C. 120. $5 \times 4 \times 3 \times 2 \times 1$.

3. What is $0!$?
   - A. 0
   - B. 1
   - C. undefined
   - D. 10
   - Answer: B. 1. By definition $0! = 1$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/counting-principles/#quiz