# Counting and Permutations

> The multiplication principle, factorials, arrangements with and without restrictions, repeated letters and circles.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/permutations/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The bike lock
Jay forgets the code for his bike lock. It has 4 dials, each showing 0 to 9. How long could it take to try them all?

He does not need to write out every code. The first dial has 10 choices, and for each of those the second has 10, and so on. So there are $10 \times 10 \times 10 \times 10 = 10\,000$ codes. Counting by multiplying, instead of listing, is the heart of **combinatorics**.

**In short:** A permutation is an arrangement where order matters. Multiply the number of choices at each step, so $n$ different objects can be arranged in $n!$ ways and choosing $r$ of them in order gives $^nP_r = \dfrac{n!}{(n - r)!}$. Deal with any restrictions first.

## Starting point: what you already know

- Listing outcomes in a table or tree diagram, from probability in Years 9 and 10.
- Probability $= \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ when outcomes are equally likely.
- Index notation, such as $10^4 = 10 \times 10 \times 10 \times 10$.

## The multiplication principle

### Key idea: Multiply the choices
If one choice can be made in $m$ ways and then a second in $n$ ways, the two together can be made in $m \times n$ ways. This extends to any number of steps.

### Worked example: Outfits
3 shirts, 4 pairs of pants and 2 pairs of shoes give $3 \times 4 \times 2 = 24$ outfits.

**Addition principle:** if you choose **either** one thing (in $m$ ways) **or** another (in $n$ ways), and they cannot both happen, there are $m + n$ ways. "And" means multiply, "or" means add.

## Factorials

How many ways can 5 people stand in a line? The first spot has 5 choices, then 4, then 3, 2, 1:

$$5 \times 4 \times 3 \times 2 \times 1 = 120$$

This product is written $5!$ and read "5 factorial".

$$n! = n \times (n - 1) \times \dots \times 2 \times 1, \qquad 0! = 1$$

Factorials grow very fast: $10! = 3\,628\,800$.

## Permutations

A **permutation** is an arrangement where **order matters**. The number of ways to arrange $r$ objects chosen from $n$ different objects is

$$^nP_r = \frac{n!}{(n - r)!} = n(n - 1)(n - 2)\cdots(n - r + 1)$$

### Worked example: Race medals
8 runners. How many ways can gold, silver and bronze be awarded?
$$^8P_3 = 8 \times 7 \times 6 = 336$$

## Arrangements with restrictions

**Deal with the restriction first.**

### Worked example: Must start with a vowel
How many arrangements of the letters of TRIANGLE start with a vowel?

TRIANGLE has 8 different letters with 3 vowels (I, A, E). First spot: 3 choices. The other 7 letters fill the rest in $7!$ ways.
$$3 \times 7! = 15\,120$$

### Worked example: Must be together
In how many ways can 6 people sit in a row if Ana and Ben must sit together?

Tie Ana and Ben into one block. Now arrange 5 items: $5!$ ways. Inside the block they can swap: $2!$ ways.
$$5! \times 2! = 240$$

### Exam tip: Not together
"Not together" $=$ total $-$ together. For the example above: $6! - 240 = 480$.

## Repeated objects

If some objects are identical, swapping them does not make a new arrangement, so divide.

$$\text{arrangements} = \frac{n!}{p!\,q!\cdots}$$

where $p$, $q$, ... count the repeats.

### Worked example: BANANA
6 letters with A three times and N twice:
$$\frac{6!}{3!\,2!} = \frac{720}{12} = 60$$

### Calculator tip: nPr, factorials and brackets
Use `nPr` for ordered selections: $\,{}^{8}\mathrm{P}_{3}$ is `8 nPr 3` = 336. For arrangements with repeated letters, put the whole bottom in brackets: `10! ÷ ( 3! × 3! × 2! )`. Without the brackets, the calculator divides by 3! and then multiplies by the rest.

**Mind map**

```text
Counting and permutations
  Principles
    And means multiply
    Or means add
  Factorial
    n! = n × (n - 1) × ... × 1
    0! = 1
  Permutations
    Order matters
    nPr = n!/(n - r)!
  Restrictions
    Do the restriction first
    Together: make a block
    Not together: total minus together
  Repeats
    Divide by p! q!
  Circles
    (n - 1)!
```

### Year 11: Arrangements in a circle
Around a round table there is no "first seat", so turning everyone one place gives the same seating. Fix one person and arrange the rest:
$$\text{ways} = (n - 1)!$$
**Example.** 7 people at a round table: $6! = 720$ ways. If two particular people must sit together: block them, so 6 items in a circle, $5! \times 2! = 240$.

**Someone in the middle.** For Ana to sit between Ben and Cal at a round table of 8, glue B, A, C into a block (2 orders, since only B and C can swap). Then 6 items around the circle: $5! \times 2 = 240$.

**Two tables.** For 9 people at round tables of 5 and 4, first choose who sits at the table of 5, $\binom{9}{5}$, then arrange each table: $\binom{9}{5} \times 4! \times 3! = 18\,144$.

**Alternating.** 4 boys and 4 girls in a row, alternating: the row starts with a boy or a girl (2 ways), then $4!$ for the boys and $4!$ for the girls: $2 \times 4! \times 4! = 1152$.

**Apart, with repeated letters.** In LETTER the T's are together in $\dfrac{5!}{2!} = 60$ arrangements (glue TT, the two E's still repeat). All arrangements: $\dfrac{6!}{2!\,2!} = 180$. So the T's are apart in 120.

**Numbers with conditions.** For 4-digit numbers above 7000 from the digits 1 to 9 without repeats, the first digit is 7, 8 or 9: $3 \times 8 \times 7 \times 6 = 1008$.

**Cases.** When one rule does not cover everything, split into cases that do not overlap and add them. For example, count numbers whose repeated digit is the first digit, then those whose repeated digit is a different digit.

**Every box used.** To put 4 different objects into 3 different boxes with none empty, count all ways ($3^4 = 81$), subtract those missing a chosen box ($3 \times 2^4 = 48$), then add back those missing two boxes ($3 \times 1 = 3$): $81 - 48 + 3 = 36$.

### Year 11: Probability with counting
A 4 digit code uses digits 1 to 9 with no repeats. Find the probability it is even.

Total: $^9P_4 = 3024$. Even: last digit from $\{2, 4, 6, 8\}$ in 4 ways, then $8 \times 7 \times 6$ for the rest: $4 \times 336 = 1344$.
$$P = \frac{1344}{3024} = \frac{4}{9}$$

### Year 12: Counting meets probability distributions
In Year 12, the binomial distribution counts the number of ways to get $k$ successes in $n$ trials. That count comes straight from this chapter and the next.

## Quick summary

- Multiply the number of choices for each step. "And" multiplies, "or" adds.
- $n! = n(n - 1)\cdots 1$ and $0! = 1$.
- Order matters: $^nP_r = \dfrac{n!}{(n - r)!}$.
- Handle restrictions first. Use blocks for "together" and subtract for "not together".
- Identical objects: divide by the factorial of each repeat count. Circles: $(n - 1)!$.

## 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. Multiply: $4 \times 3$.

2. A 3 digit code uses digits 0 to 9, and digits can repeat. How many codes?
   - A. 30
   - B. 720
   - C. 1000
   - D. 999
   - Answer: C. 1000. $10 \times 10 \times 10$.

3. Evaluate $5!$.
   - A. 25
   - B. 60
   - C. 120
   - D. 720
   - Answer: C. 120. $5 \times 4 \times 3 \times 2 \times 1$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/permutations/#quiz