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Permutations: counting ordered arrangements with the multiplication principle

Syllabus dot point

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

HSCMaths Extension 1Combinatorics (ME-A1)13 min read

Quick answer

Permutations count ordered arrangements; the basic formulas are n!n! for all nn distinct, nPr=n!(n−r)!{}^n P_r = \frac{n!}{(n - r)!} for rr from nn, n!n1!⋯nk!\frac{n!}{n_1! \cdots n_k!} for objects with repeats, and (n−1)!(n - 1)! for circular arrangements, all flowing from the multiplication principle. Handle restrictions by gluing (together), subtracting the complement (not together) or locking a fixed position.

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  1. What this dot point is asking
  2. The answer
  3. Exam-style questions

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 n1n_1 ways at step 1, and (independent of step 1) in n2n_2 ways at step 2, …\dots, and nkn_k ways at step kk, then the total number of ways to complete the procedure is

n1⋅n2⋅⋯⋅nk.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−1n - 1) of them, so the principle applies.

Permutations of nn distinct objects

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

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

Reasoning straight from the multiplication principle: nn choices for the first position, n−1n - 1 for the second (one object is used up), n−2n - 2 for the third, and so on down to 11 for the last. By convention 0!=10! = 1, which is what makes the rr from nn formula below work when r=nr = n.

Permutations of rr from nn

The number of ways to choose and arrange rr objects from nn distinct objects (order matters, no repetition) is

nPr=n!(n−r)!=n(n−1)(n−2)⋯(n−r+1).{}^{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 rr descending factors starting at nn. For example 9P4=9⋅8⋅7⋅6{}^{9}P_4 = 9 \cdot 8 \cdot 7 \cdot 6, four factors, with no need to compute 9!9! and 5!5! separately. The notation is sometimes P(n,r)P(n, r) or PrnP^n_r; NESA tends to write nPr^n P_r.

Watch the multiplication principle fill the positions, stage by stage

Counting 5P3{}^{5}P_3 (arrange 33 of the 55 distinct letters A,B,C,D,EA, 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 55 letters are available, so there are 55 ways to fill position 1.

Permutation stage 1: five choices for the first positionThree empty slots for arranging three of the five letters A B C D E. The first position has five letters available, so there are five choices.Position 1: 5 choicesposition 15choicesposition 2position 3letters still available:ABCDEAny of the 5 letters can take position 1.

Stage 2, choose the second position. Say AA took position 1. It is now used, so only B,C,D,EB, C, D, E remain: 44 ways to fill position 2.

Permutation stage 2: four choices remain for the second positionThe first slot holds A. Four letters B C D E remain, so the second position has four choices.Position 2: 4 choices leftAposition 15choicesposition 24choicesposition 3×letters still available:BCDEA is used, so 4 letters remain for position 2.

Stage 3, choose the third position. Two letters are now used, so 33 remain for position 3. By the multiplication principle the total is the product of the choice counts: 5×4×3=60=5P35 \times 4 \times 3 = 60 = {}^{5}P_3. Notice this is exactly 5!(5−3)!=5!2!\frac{5!}{(5 - 3)!} = \frac{5!}{2!}, the descending product stopping after 33 factors.

Permutation stage 3: three choices remain for the third positionThe first two slots hold A and B. Three letters C D E remain, so the third position has three choices. The product five times four times three equals sixty.Position 3: 3 choices leftAposition 15choicesBposition 24choicesposition 33choices××letters still available:CDETotal arrangements = 5 × 4 × 3 = 60 = ⁵P₃.

Key fact

The multiplication principle is the master tool: a sequence of independent steps with n1,n2,…,nkn_1, n_2, \dots, n_k options multiplies to n1n2⋯nkn_1 n_2 \cdots n_k arrangements. From it follow n!n! (arrange all nn distinct), nPr=n!(n−r)!{}^{n}P_r = \dfrac{n!}{(n - r)!} (ordered rr from nn), n!n1! n2!⋯nk!\dfrac{n!}{n_1!\, n_2! \cdots n_k!} (objects with repeats) and (n−1)!(n - 1)! (distinct objects in a circle).

Permutations with repeats (identical objects)

If you have nn objects of which n1n_1 are alike, n2n_2 are alike, …\dots, nkn_k are alike (with n1+n2+⋯+nk=nn_1 + n_2 + \dots + n_k = n), the number of distinct arrangements is

n!n1! n2!⋯nk!.\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!n! arrangements. But swapping the n1n_1 identical objects among themselves does not produce a new arrangement, and there are n1!n_1! such swaps; likewise n2!n_2! for the second group, and so on. So n!n! over-counts each genuinely different arrangement by a factor of n1! n2!⋯nk!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 nn distinct objects is

(n−1)!.(n - 1)!.

Reasoning: around a round table there is no fixed "first seat", so every arrangement looks the same under rotation. There are nn rotations of each seating, so the n!n! linear arrangements over-count by a factor of nn, giving n!n=(n−1)!\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−1n - 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 22 to get (n−1)!2\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 nn objects, you now arrange n−1n - 1 items, giving (n−1)!(n - 1)!, then multiply by the internal arrangements of the block (2!2! for two glued objects, m!m! for a block of mm). "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−1n - 1 in (n−1)!(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

  1. Decide whether order matters. Yes for permutations; no for combinations (the next dot point).
  2. Decide whether repetition is allowed. With unlimited repeats and rr slots the count is nrn^r (each slot independently has nn options); without repeats use nPr{}^{n}P_r.
  3. Decide whether you arrange all nn or only rr of them.
  4. 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!n! (all distinct) or nPr{}^{n}P_r (only rr of them).
  • "How many arrangements of the letters of the word ...?": check for repeated letters; if any, use n!n1!⋯nk!\dfrac{n!}{n_1! \cdots n_k!}.
  • "... seated around a circular / round table?": circular, answer (n−1)!(n - 1)! (fix one person).
  • "... if [two named people] sit together / are next to each other?": glue into a block, (n−1)!×2!(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]?": nPr{}^{n}P_r. If repeats are allowed it is nrn^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 ("77 letters: 33 A, 22 N, 11 B, 11 S") so the denominator is justified. For "together" glue and state the internal 2!2!; for "not together" write "== total −- together" explicitly. For circular, write "fix one person, then arrange the other n−1n - 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!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!=720.6! = 720.

Arrange only rr of nn objects

A four-digit PIN is made from the digits 11 to 99 with no digit repeated. How many PINs? Order matters and repetition is banned, so use 9P4{}^{9}P_4 as four descending factors:

9P4=9⋅8⋅7⋅6=3024. {}^{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)(1), I (4)(4), S (4)(4), P (2)(2), for 1111 letters in total. Divide the 11!11! by the factorial of each repeat count:

11!1! 4! 4! 2!=39 916 80024⋅24⋅2=39 916 8001152=34 650.\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 77 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 66:

(7−1)!=6!=720. (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 44 items to arrange, then multiply by the 2!2! orders inside the block:

4!×2!=24×2=48.4! \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=72.5! - 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 44 letters freely:

4!=24.4! = 24.

Combine restrictions with the multiplication principle

A committee photo seats 33 teachers and 33 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!×3!=6×6=36.3! \times 3! = 6 \times 6 = 36.

Common traps
Forgetting the over-count from repeats
"Arrangements of BANANAS" is not 7!=50407! = 5040; that treats the three A's and two N's as distinct. Divide by 3! 2!3!\, 2! to get 420420.
Treating a circle like a line
A round table has no fixed start, so n!n! over-counts by the nn rotations. Fix one object and use (n−1)!(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!2! for the two orders inside the block (or m!m! for a block of mm).
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!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−1n - 1, not nn. 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 questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style2 marks
How 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 55 distinct letters is 5!=1205! = 120.

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

HSC-style3 marks
In 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 77 letters. Repeats: 33 A's, 22 N's.

Number of distinct arrangements: 7!3!⋅2!=504012=420\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.

Practise this

Sources & how we know this

Based on the official NESA syllabus and published NESA past papers and marking guidelines. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

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