What is permutations with repeats (identical objects)?

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

What is two objects must be together?

Glue them into a single block and arrange the block among the other objects: if there were n objects, you now arrange n - 1 items, giving (n - 1)!, then multiply by the internal arrangements of the block (2! for two glued objects, m! for a block of m). "All the vowels together" is the same move: glue every vowel into one block.

What is two objects must not be together?

Do not try to count this directly. Use the complement: (not together) = (total) - (together). This is almost always faster and is what markers expect.

What is a particular object in a fixed position?

Lock that object in place and arrange the rest. With one object fixed, you arrange the remaining n - 1 in (n - 1)! ways.

What is "Together" without the internal arrangement?

Gluing two objects into a block fixes their position relative to the rest, but you must still multiply by 2! for the two orders inside the block (or m! for a block of m).

§-Syllabus dot point

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 stepped diagrams and worked examples.

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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 the common restrictions: objects that must or must not be adjacent, objects fixed at particular positions, and so on. The whole topic rests on one idea, the multiplication principle, so the real skill is learning to break a counting problem into a sequence of independent choices and then deciding which standard pattern it matches.

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 in the course. The word "independent" is doing real work: the number of options at step 2 must not depend on which option you took at step 1 (the options themselves can change, only the count must stay fixed). When you arrange distinct people in seats, choosing person A for seat 1 versus person B for seat 1 leaves a different set of people for seat 2, but the same number ( $n - 1$ ) of them, so the principle applies.

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 straight from the multiplication principle: $n$ choices for the first position, $n - 1$ for the second (one object is used up), $n - 2$ for the third, and so on down to $1$ for the last. By convention $0! = 1$ , which is what makes the $r$ from $n$ formula below work when $r = n$ .

Permutations of $r$ from $n$

The number of ways to choose and arrange $r$ objects from $n$ distinct objects (order matters, no repetition) is

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

The product form on the right is usually faster by hand: it is just $r$ descending factors starting at $n$ . For example ${}^{9}P_4 = 9 \cdot 8 \cdot 7 \cdot 6$ , four factors, with no need to compute $9!$ and $5!$ separately. The notation is sometimes $P(n, r)$ or $P^n_r$ ; NESA tends to write $^n P_r$ .

Watch the multiplication principle fill the positions, stage by stage

Counting ${}^{5}P_3$ (arrange $3$ of the $5$ distinct letters $A, B, C, D, E$ ) is the clearest way to see why the formula is a product of descending factors. Fill the positions left to right; each time a letter is used it leaves the available pool, so the choice count drops by one.

Stage 1, choose the first position. All $5$ letters are available, so there are $5$ ways to fill position 1.

Stage 2, choose the second position. Say $A$ took position 1. It is now used, so only $B, C, D, E$ remain: $4$ ways to fill position 2.

Stage 3, choose the third position. Two letters are now used, so $3$ remain for position 3. By the multiplication principle the total is the product of the choice counts: $5 \times 4 \times 3 = 60 = {}^{5}P_3$ . Notice this is exactly $\frac{5!}{(5 - 3)!} = \frac{5!}{2!}$ , the descending product stopping after $3$ factors.

Permutations with repeats (identical objects)

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!}.

Here is why you divide. Pretend for a moment the identical objects are labelled and distinct: then there are $n!$ arrangements. But swapping the $n_1$ identical objects among themselves does not produce a new arrangement, and there are $n_1!$ such swaps; likewise $n_2!$ for the second group, and so on. So $n!$ over-counts each genuinely different arrangement by a factor of $n_1!\, n_2! \cdots n_k!$ , and dividing corrects it. This "divide out the over-count" move is exactly the same idea that turns permutations into combinations on the next dot point.

Circular permutations

The number of distinct circular arrangements of $n$ distinct objects is

(n - 1)!.

Reasoning: around a round table there is no fixed "first seat", so every arrangement looks the same under rotation. There are $n$ rotations of each seating, so the $n!$ linear arrangements over-count by a factor of $n$ , giving $\frac{n!}{n} = (n - 1)!$ . The standard way to do a circular problem is to fix one person in a seat to kill the rotational symmetry, then arrange the remaining $n - 1$ people linearly in the other seats. If reflections are also counted as the same (turning a bracelet or necklace over), divide by a further $2$ to get $\frac{(n - 1)!}{2}$ .

Restrictions: the standard moves

Two (or more) objects must be together: Glue them into a single block and arrange the block among the other objects: if there were $n$ objects, you now arrange $n - 1$ items, giving $(n - 1)!$ , then multiply by the internal arrangements of the block ( $2!$ for two glued objects, $m!$ for a block of $m$ ). "All the vowels together" is the same move: glue every vowel into one block.
Two objects must not be together: Do not try to count this directly. Use the complement: (not together) $=$ (total) $-$ (together). This is almost always faster and is what markers expect.
A particular object in a fixed position: Lock that object in place and arrange the rest. With one object fixed, you arrange the remaining $n - 1$ in $(n - 1)!$ ways.

Some objects in particular kinds of slots (for example, an arrangement must start with a vowel, or the sexes must alternate). Count the restricted positions first, then fill the rest, multiplying the stage counts together.

Summary recipe

Decide whether order matters. Yes for permutations; no for combinations (the next dot point).
Decide whether repetition is allowed. With unlimited repeats and $r$ slots the count is $n^r$ (each slot independently has $n$ options); without repeats use ${}^{n}P_r$ .
Decide whether you arrange all $n$ or only $r$ of them.
Apply restrictions by gluing (together), subtracting (not together) or locking (fixed position), and handle circles by fixing one object.

How exam questions ask about permutations

"In how many ways can ... be arranged in a row / a line / on a shelf?": a straight $n!$ (all distinct) or ${}^{n}P_r$ (only $r$ of them).
"How many arrangements of the letters of the word ...?": check for repeated letters; if any, use $\dfrac{n!}{n_1! \cdots n_k!}$ .
"... seated around a circular / round table?": circular, answer $(n - 1)!$ (fix one person).
"... if [two named people] sit together / are next to each other?": glue into a block, $(n-1)! \times 2!$ .
"... if [two named people] do not sit together / must be separated?": complement, total minus together.
"... if it must begin with [a vowel / a specific object] / [object] is in a fixed seat?": lock that position, then arrange the rest.
"How many [codes / numbers / arrangements] with no repeated [digit / letter]?": ${}^{n}P_r$ . If repeats are allowed it is $n^r$ instead.
The word "different" or "distinct" is the cue to divide out repeats; its absence when identical objects are present is the classic trap.

Exam technique

Lay the working out as a product of clearly labelled choices, exactly like the stage diagrams above: write the boxes or the descending factors so the marker sees the multiplication principle in action, not just a final number. For an "arrangements of a word" question, list the letters and their frequencies first (" $7$ letters: $3$ A, $2$ N, $1$ B, $1$ S") so the denominator is justified. For "together" glue and state the internal $2!$ ; for "not together" write " $=$ total $-$ together" explicitly. For circular, write "fix one person, then arrange the other $n - 1$ ", which earns the method mark even if the arithmetic slips. Always sanity-check the size of the answer: a permutation count should be larger than the corresponding combination count by a factor of $r!$ .

Worked example

Arrange all distinct objects

Six students line up for a photo. How many orders are possible? All six are distinct and all are used, so this is a straight factorial:

6! = 720.

Arrange only $r$ of $n$ objects

A four-digit PIN is made from the digits $1$ to $9$ with no digit repeated. How many PINs? Order matters and repetition is banned, so use ${}^{9}P_4$ as four descending factors:

{}^{9}P_4 = 9 \cdot 8 \cdot 7 \cdot 6 = 3024.

Count arrangements of a word with repeated letters

How many distinct arrangements of the letters in MISSISSIPPI? List the letters and frequencies: M $(1)$ , I $(4)$ , S $(4)$ , P $(2)$ , for $11$ letters in total. Divide the $11!$ by the factorial of each repeat count:

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

Seat people around a circular table

In how many ways can $7$ guests sit at a round table, if only the order around the table matters? Fix one guest to remove the rotational symmetry, then arrange the remaining $6$ :

(7 - 1)! = 6! = 720.

A "must sit together" restriction

Five friends sit in a row, but Anh and Bilal insist on sitting together. Glue Anh and Bilal into one block, leaving $4$ items to arrange, then multiply by the $2!$ orders inside the block:

4! \times 2! = 24 \times 2 = 48.

A "must not sit together" restriction by complement

Same five friends, but now Anh and Bilal refuse to sit together. Count the complement: all arrangements minus the "together" count found above:

5! - 48 = 120 - 48 = 72.

Lock an object into a fixed position

How many arrangements of the letters of MATHS begin with M? Lock M in the first seat and arrange the other $4$ letters freely:

4! = 24.

Combine restrictions with the multiplication principle

A committee photo seats $3$ teachers and $3$ students in a row, alternating teacher, student, teacher, student, teacher, student (so a teacher is at each end). Fill the three teacher seats first, then the three student seats, and multiply the two counts:

3! \times 3! = 6 \times 6 = 36.

Common traps

Forgetting the over-count from repeats: "Arrangements of BANANAS" is not $7! = 5040$ ; that treats the three A's and two N's as distinct. Divide by $3!\, 2!$ to get $420$ .
Treating a circle like a line: A round table has no fixed start, so $n!$ over-counts by the $n$ rotations. Fix one object and use $(n - 1)!$ .
"Together" without the internal arrangement: Gluing two objects into a block fixes their position relative to the rest, but you must still multiply by $2!$ for the two orders inside the block (or $m!$ for a block of $m$ ).
Confusing permutation with combination: A permutation counts ordered arrangements; a combination counts unordered selections. The same scenario can be either, depending on whether order matters. They differ by a factor of $r!$ .
Misusing the multiplication principle: The number of options at each step must be independent of earlier choices. If the same person cannot sit twice, the second factor is $n - 1$ , not $n$ . If steps overlap or interact, the principle does not apply directly.
Adding when you should multiply, or the reverse: Independent stages of one arrangement multiply; genuinely separate, mutually exclusive cases add. "First seat AND second seat" multiplies; "starts with a vowel OR ends with a vowel" needs cases (and usually the complement).

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.

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?