NSWMaths Extension 1Syllabus dot point

How do we count the number of ordered arrangements of a set of objects?

Use the multiplication principle and the permutation formula to count ordered arrangements, including restrictions and repeated elements

A focused answer to the HSC Maths Extension 1 dot point on permutations. The multiplication principle, the formula $\frac{n!}{(n - r)!}$ for arrangements of $r$ from $n$ , permutations of objects with repeats, circular permutations, and counting with restrictions, with worked examples.

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

Have a quick question? Jump to the Q&A page

What this dot point is asking

NESA wants you to count ordered arrangements (permutations) of objects, with or without repeats, in circular or linear settings, and to apply common restrictions (objects must or must not be adjacent, must be at fixed positions, etc.).

The answer

The multiplication principle

If a procedure can be performed in $n_1$ ways at step 1, and (independent of step 1) in $n_2$ ways at step 2, $\dots$ , and $n_k$ ways at step $k$ , then the total number of ways to complete the procedure is $n_1 \cdot n_2 \cdot \dots \cdot n_k$ .

This is the foundation of every counting problem.

Permutations of $n$ distinct objects

The number of ways to arrange all $n$ distinct objects in a row is

n! = n \cdot (n - 1) \cdot (n - 2) \cdots 2 \cdot 1.

Reasoning: $n$ choices for the first position, $n - 1$ for the second (one used up), $n - 2$ for the third, and so on.

By convention $0! = 1$ .

Permutations of $r$ from IMATH_18

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

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

The notation is sometimes $P(n, r)$ or $P^n_r$ . NESA tends to use $^n P_r$ .

Permutations with repeats

If you have $n$ objects of which $n_1$ are alike, $n_2$ are alike, $\dots$ , $n_k$ are alike (with $n_1 + n_2 + \dots + n_k = n$ ), the number of distinct arrangements is

\frac{n!}{n_1! \, n_2! \cdots n_k!}.

This divides out the over-count from treating identical objects as distinguishable.

Circular permutations

The number of distinct circular arrangements of $n$ distinct objects is $(n - 1)!$ . Reasoning: fix one object to break the rotational symmetry, then arrange the remaining $n - 1$ linearly.

If reflections are also considered the same (necklace problems), divide by another $2$ .

Restrictions

Two objects must be together: glue them together as a single block, arrange as if $n - 1$ objects, then multiply by $2!$ for the internal arrangement of the block.

Two objects must not be together: count total arrangements minus the "together" count.

Vowels (or some other type) together: glue all vowels into a single block, treat as one object.

Particular object in a fixed position: lock that object, count the arrangements of the rest.

Summary recipe

Identify whether order matters (yes for permutations, no for combinations).
Identify whether repetition is allowed (with-repeat formulas are $n^r$ ).
Identify whether you are arranging all $n$ or only $r$ of them.
Apply restrictions by gluing, locking or subtracting.

Worked example

All distinct

In how many ways can $6$ people sit in a row of $6$ seats? $6! = 720$ .

IMATH_3 from IMATH_4

How many four-digit codes using digits $1$ to $9$ with no repeated digit? ${}^9 P_4 = 9 \cdot 8 \cdot 7 \cdot 6 = 3024$ .

Word with repeats

How many arrangements of the letters in MISSISSIPPI? Letters: M, I (4), S (4), P (2). Total $11$ letters.

$\frac{11!}{1! \, 4! \, 4! \, 2!} = \frac{39 \, 916 \, 800}{24 \cdot 24 \cdot 2} = \frac{39 \, 916 \, 800}{1152} = 34 \, 650$ .

Circular

In how many ways can $7$ people sit at a round table? $(7 - 1)! = 720$ .

Together restriction

How many arrangements of $5$ people in a row if Alice and Bob must sit together?

Glue them: $4$ "objects", $4! = 24$ arrangements. Internal arrangement of Alice and Bob: $2$ . Total: $24 \cdot 2 = 48$ .

Not together restriction

Same setup but Alice and Bob must not sit together.

Total arrangements: $5! = 120$ . Minus together arrangements ( $48$ ): $120 - 48 = 72$ .

Fixed position

How many arrangements of the letters MATHS such that the M is in the first position? Lock the M: arrange the remaining $4$ letters in $4! = 24$ ways.

Common traps

Forgetting the over-count from repeats: "Arrangements of BANANAS" with $7! = 5040$ over-counts by ignoring the three A's and two N's. Divide by $3! \, 2!$ .
Circular without fixing: Treating circular like linear over-counts by a factor of $n$ (each rotation looks like a "new" arrangement but it is the same). Always fix one object.
Together restriction missing internal arrangement: Gluing two objects gives the position; you must also multiply by the number of ways to arrange them inside the block.
Confusing permutation with combination: Permutation counts ordered arrangements; combination counts unordered subsets. Same problem in two different "frames".
Misapplying the multiplication principle: Steps must be independent. If choosing seat 1 affects how many options are available for seat 2 (because the same person cannot sit twice), the second factor is $n - 1$ , not $n$ .

Past exam questions, worked

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

2021 HSC Q62 marksHow many five-letter arrangements of the letters in the word "CHAIR" are there?

Show worked answer →

All five letters are distinct. The number of arrangements of $5$ distinct letters is $5! = 120$ .

Markers reward identifying distinct objects and applying $n!$ directly.

2020 HSC Q233 marksIn how many different ways can the letters of the word "BANANAS" be arranged?

Show worked answer →

Letters: B, A, N, A, N, A, S. Total $7$ letters. Repeats: $3$ A's, $2$ N's.

Number of distinct arrangements: $\frac{7!}{3! \cdot 2!} = \frac{5040}{12} = 420$ .

Markers reward identifying the multiset structure, applying the formula for permutations of objects with repeats, and clean arithmetic.