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Permutations


The Logic behind permutations
Types of Permutations
Permutation (Full)
Permutation with Identical Items
Partial Permutation without Repetition
Permutation Notation
Permutation with Repetition
Circular Permutation
Derangement
Permutations FAQ


Permutations: Understanding Ordered Arrangements

Permutations are arrangements where order matters. As we discussed on the main combinatorics page, this is what distinguishes permutations from combinations.

On this page, we'll explore four key permutation scenarios: Full Permutation (arranging all n distinct elements), Permutation with Repetition (arrangements where certain elements can repeat), Permutation without Repetition (selecting and arranging r elements from n options), and Circular Permutation (arrangements around a circle where rotations are identical).

Each scenario has specific formulas and real-world applications. Ready to dive into the details, or want to compare with combinations where order doesn't matter?

Mastering these permutation types gives you the tools to solve ordering problems across mathematics, computer science, and everyday scenarios. The key is recognizing which type fits your specific situation.




The Logic behind permutations

In combinatorics, we distinguish between several types of permutations based on different criteria, but they all share one key property: in every type, the order of arrangement matters ( and that is exactly what makes the difference between permutations and variations).
In all different scenarios, the mechanism for creating a permutation is similar, and it is important to understand it.
When creating an arrangement, we must make two decisions:
1. Which item to pick?
2. Where to place it in the arrangement?

Let us examine the simplest case. Imagine we need to arrange nn different items (all distinct) in a row, without repetition. We can break it down into steps:

Step 1: Choose an item for the first position. There are nn options.
Step 2: Choose an item for the second position. Now there are n−1n-1 options.
Step 3: Choose an item for the third position. Now there are n−2n-2 options.
… and so on, until
Step nn: Only one item remains, so there is just 1 option.

Each step reduces the number of choices (since repetition is not allowed). Because we must make all these choices in sequence, we apply the multiplication principle.

Thus, to calculate the total number of permutations, we multiply:

n×(n−1)×(n−2)×⋯×1,n \times (n-1) \times (n-2) \times \dots \times 1,

which is the definition of a factorial.

For nn distinct objects, the number of permutations is:
                                                 n!=n×(n−1)×(n−2)×⋯×2×1.n! = n \times (n−1) \times (n−2) \times \dots \times 2 \times 1.
To summarize:

* We start with nn options.
* Each choice reduces the number of remaining options by one.
* We multiply the number of choices at each step.

This gives us n!n! total permutations.

Here is an example of simple permutation: creating 3 letter string.

Permutations of ABC: Step-by-Step A B C First choice: A B C 3 options to choose first ball Second choice: B C A C A B 2 options to choose second ball Third choice: C B C A B A 1 option left for the last ball Final permutations: ABC ACB BAC BCA CAB CBA 3! = 3 × 2 × 1 = 6 total permutations
You can see how the principle explained before is applied here.
First, one of 3 letters is picked and for each scenario there are 2 different choices to select the second letter.Once the choice is made, only one letter left.

The result is:
                                                 3×2×1=6(n!)3\times 2 \times 1 = 6 (n!)
Where nn is total number of items.

Original set 2! = 2 ways All possible arrangements Original set 3! = 6 ways All possible arrangements Original set 4! = 24 ways All possible arrangements Permutation: Arrange ALL items, order matters Take n distinct items → arrange them in every possible order → get n! arrangements

Types of Permutations

Although in all types of permutations the order of arrangement matters, the key differences between them arise from other considerations: Can items be repeated? Are some items identical? Is the arrangement linear or circular? The table below summarizes these distinctions, showing how different scenarios lead to different counting formulas.

Permutation Scenarios

ScenarioOrder Matters?Repetition?Identical Items?Arrangement TypeFormulaExample
Full permutationYesNoNoLinearn!n!Arrange 6 books in a row
Partial permutationYesNoNoLinearn!(n−r)!\frac{n!}{(n-r)!}Arrange 3 of 10 runners
Permutation with repetitionYesYesNoLinearnrn^r4-digit PIN code
Permutation with identical itemsYesN/AYesLinearn!n1! n2!…nk!\frac{n!}{n_1! \, n_2! \dots n_k!}Rearrange "BALLOON"
Circular permutationYesNoNoCircular(n−1)!(n-1)!Seat 5 people around a table


By addressing these additional questions, we can classify permutations into several types, each with its own counting method. Understanding these distinctions helps in correctly applying the appropriate formula for a given arrangement problem.

Permutation (Full)

Full Permutation applies when you arrange all distinct items in a specific order with no repetitions allowed.

Property Full Permutation
Use all items ✓
Order matters ✓
Identical items ✗
Repetitions ✗
Linear arrangement ✓

Full Permutation — Examples:
Arranging all books on a shelf, deciding the lineup of all players in a team, organizing all photos in an album, setting the order of speakers in a presentation, ordering all tasks in a workflow, ranking all competitors in a contest, scheduling all meetings in a day, sorting all files in a folder.


The intuition and logic we use for this kind of permutation has been explained pretty well in previous section and it works basically for all types of permutations.
Just in case of full permutations (unlike in other types) no further modifications needed and the calculation is simple.

Notation:
In combinatorics, we use specific notation to represent different types of permutations.
For arranging rr objects selected from nn distinct objects, the standard notation is P(n,r)P(n,r) or nPr_nP_r.
When applied to full permutations, apparently r=nr=n (because we use all the items ), and the notation turns into :

P(n,n)P(n,n)

or nPn_nP_n

or simply P(n)P(n).

Learn more about combinatorial notations and symbols here →

Formula:
The number of possible results is calculated by the formula :
                                                  P(n)=n!P(n)=n!
Product of all integers from nn down to 11.
Where nn is the number of items to arrange.

Use combinatorics calculator →


Permutation with Identical Items

Permutation with Identical Items applies when arranging all items where some identical elements appear multiple times in your collection. Classic examples include arranging all letters in words like "MISSISSIPPI" or organizing all colored balls when you have multiple balls of the same color.

Property Permutation with Identical Items
Use all items ✓
Order matters ✓
Identical items ✓
Repetitions ✗
Linear arrangement ✓

Permutation with Identical Items — Examples:
Rearranging letters in the word “BALLOON,” organizing colored balls where some colors repeat, sequencing identical files and unique ones in storage, arranging identical chairs and distinct tables in a layout, ordering repeated ingredients in a recipe list.

Notation:

(nn1,n2,…,nk)\binom{n}{n_1, n_2, \ldots, n_k}

or P(n;n1,n2,n3,…,nk)P(n; n_1, n_2, n_3, \ldots, n_k)

or Pnn1,n2,n3,…,nkP_n^{n_1, n_2, n_3, \ldots, n_k}

Learn more about combinatorial notations and symbols here →

Formula:

To calculate the number of possible arrangements
                                                 (nn1,n2,…,nk)=n!n1!n2!…nk!\binom{n}{n_{1},n_{2},\dots,n_{k}} = \frac{n!}{n_{1}!n_{2}!\dots n_{k}!}
Where multinomial coefficient: counts distinct arrangements of nn items split into groups of sizes n1,n2,…,nkn_{1},n_{2},\dots,n_{k}.

Use combinatorics calculator →



Permutations: All Different vs. Some Identical Items Case 1: All Different Items 3! = 6 permutations Initial Set: Case 2: Some Identical Items 3!/2! = 3 permutations Initial Set: RGB RBG GRB GBR BRG BGR RRB RBR - Canceled BRR RBR BRR - Canceled RRB - Canceled All Different Items: • Each ball is unique (Red, Green, Blue) • Every arrangement creates a distinct permutation • Formula: n! = 3! = 6 permutations • All 6 arrangements are counted Some Identical Items: • Two red balls are identical, one blue ball is unique • Swapping identical red balls gives same arrangement • Formula: n!/k! = 3!/2! = 3 permutations • Red X marks duplicate arrangements Identical items reduce total permutations Legend: Valid permutation (no marking) Duplicate arrangement - Canceled
Permutations: All Different vs. Some Identical Items Case 1: All Different Items 3! = 6 permutations Initial Set: Case 2: All Identical Items 3!/3! = 1 permutation Initial Set: RGB RBG GRB GBR BRG BGR RRR RRR - Canceled RRR - Canceled RRR - Canceled RRR - Canceled RRR - Canceled All Different Items: • Each ball is unique (Red, Green, Blue) • Every arrangement creates a distinct permutation • Formula: n! = 3! = 6 permutations • All 6 arrangements are counted All Identical Items: • All three red balls are identical • Any rearrangement gives the exact same result • Formula: n!/n! = 3!/3! = 1 permutation • Only 1 distinct arrangement possible Identical items reduce total permutations Legend: Valid permutation (no marking) Duplicate arrangement - Canceled
2 red, 1 blue 3!/(2!×1!) = 3 Identical items reduce count 2 red, 2 blue 4!/(2!×2!) = 6 Two types of repetition 3 red, 1 blue, 1 green ⋮ 5!/(3!×1!×1!) = 20 Only 10 shown above Permutations with Identical Items Formula: n! / (n₁! × n₂! × ... × nₖ!) where nᵢ = count of each identical type

Partial Permutation without Repetition

Partial Permutation without Repetition applies when selecting and arranging only some items from a larger collection, with all items being distinct. Classic examples include choosing and ordering contestants from a group or selecting and arranging books from a library shelf.

Property Partial Permutation without Repetition
Use all items ✗
Order matters ✓
Identical items ✗
Repetitions ✗
Linear arrangement ✓

Partial Permutation (without repetition) — Examples:
Selecting and ordering finalists from a group of contestants, arranging a subset of books on a display, picking and sequencing players for a relay team, choosing and ordering questions for an interview, organizing a limited set of tasks for a project phase.

Notation:

Most common:

P(n,r)P(n,r)

Alternative notations:

nPr {}^nP_r

AnrA_n^r

nPr\text{nPr}

Learn more about combinatorial notations and symbols here →

Formula:
The number of arrangements for Partial Permutation without Repetition is
                                                 P(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}
Where n!n! counts all arrangements, dividing by (n−r)!(n-r)! removes the unused positions.

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Partial Permutations: P(4,2) Selecting and arranging 2 items from 4 available items P(4,2) = 4!/(4-2)! = 4!/2! = 24/2 = 12 permutations Available Items: Choose 2 and arrange them All 12 Partial Permutations: Starting with Red: Starting with Green: Starting with Blue: Starting with Yellow: Partial Permutations Explained: Step 1: Choose the first position - 4 choices (any of the 4 colored balls) Step 2: Choose the second position - 3 remaining choices Total: 4 × 3 = 12 different ordered pairs Key point: Order matters! Red-Green is different from Green-Red Formula: P(n,r) = n!/(n-r)! where n=4 items available, r=2 items selected
n = 3 items choose r = 2 P(3,2) = 6 3!/(3-2)! = 6 n = 4 items choose r = 2 n = 4 items choose r = 2 P(4,2) = 12 4!/(4-2)! = 12 n = 5 items choose r = 2 P(5,2) = 20 5!/(5-2)! = 20 Permutation without Repetition: Select r from n, order matters Formula: P(n,r) = n!/(n-r)! - Choose r items from n, arrange them in order

Permutation Notation

Notation

Permutation Notation

Three marks carry every count on this page: the exclamation that multiplies downward, the two-letter family that orders selections, and the same exclamation moved out front to count the restless. Each scenario section keeps its own Notation block for the variant spellings.
The choose-family marks (nr)\binom{n}{r} and C(n,r)C(n, r) belong to the binomial coefficient; the set-size bars behind the counting arguments to cardinality.
n!n!
n factorial
The postfix count of complete rearrangements: n!=n×(n−1)×⋯×1n! = n \times (n-1) \times \cdots \times 1, introduced in The Logic behind permutations above. The convention 0!=10! = 1 is load-bearing — it keeps every counting formula honest at the edges.
CasesThe mark binds tight — 2n!2n! means 2⋅(n!)2 \cdot (n!), and (2n)!(2n)! needs its brackets; stacked marks are different animals: (n!)!(n!)! iterates, while n!!n!! skips every second factor (the double factorial), not a repeat.
Do not confusePunctuation. A sentence ending on 55 with an exclamation and the operator in 5!5! share a glyph — and in programming ! negates; position and context carry the meaning.
P(n,r)P(n, r) · nPr_nP_r
P n r; n permute r
The arrangement-count family: P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!} orders rr items chosen from nn — functional P(n,r)P(n, r) in texts, subscripted nPr_nP_r on calculator keys, as the Notation blocks of Permutation (Full) and Partial Permutation without Repetition above list.
CasesAnrA_n^r — the Arrangements letter of French and Russian textbooks, same count in a different alphabet; the full case collapses the second argument: P(n,n)=P(n)=n!P(n, n) = P(n) = n!.
Do not confuseThe choose family. P(n,r)P(n, r) counts ordered selections, (nr)\binom{n}{r} unordered ones — they differ by exactly the factor r!r!, and the binomial coefficient owns the other side of that ledger.
!n!n · DnD_n
subfactorial n; derangements of n
The factorial's mirror image: !n!n counts permutations that leave nothing in place — the prefix position is the entire distinction, as Derangement below explains. DnD_n is the older spelling, still common.
CasesThe two interchange freely — !5=D5=44!5 = D_5 = 44; the prefix form reads badly mid-sentence, which is precisely what keeps the subscripted DnD_n alive in running text.
Do not confusen!n! read backwards. !4=9!4 = 9 while 4!=244! = 24 — position flips the meaning entirely, and a hasty transcription trades all arrangements for only the restless ones.

Permutation with Repetition

Permutation with Repetition applies when arranging items where you can reuse the same element multiple times during the arrangement process. Classic examples include creating PIN codes where digits can repeat, or forming passwords where letters can be used multiple times.

Property Permutation with Repetition
Use all items ✓
Order matters ✓
Identical items ✗
Repetitions ✓
Linear arrangement ✓

Permutation with Repetition — Examples:
Generating PIN codes from digits, creating letter sequences for passwords, arranging beads in a bracelet where colors can repeat, composing license plate numbers, assigning seats with multiple people allowed in the same category.

Notation:

Permutation with repetition (positions)

Prep(n,r)P_{\text{rep}}(n,r)

Alternative notations:

P‾(n,r)\overline{P}(n,r)

P(r)(n)P^{(r)}(n)

nPr(r) {}^nP^{(r)}_r

Learn more about combinatorial notations and symbols here →

Formula:
Total number of permutations with repetitions possible and all items used is:
                                                 Prep(n,r)=nrP_{\text{rep}}(n,r)=n^{r}
Where each of the rr positions can be filled in nn ways.


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Permutations with Repetition Selecting 2 items from 4 available items (repetition allowed) nr = 42 = 16 permutations Available Items (can be reused): Choose 2 (repetition allowed) All 16 Permutations with Repetition: Starting with Red: Starting with Green: Starting with Blue: Starting with Yellow: Permutations with Repetition Explained: Step 1: Choose the first position - 4 choices (any color, can be reused) Step 2: Choose the second position - 4 choices (still all colors available) Total: 4 × 4 = 16 different ordered pairs Formula: nr where n=4 items available, r=2 positions to fill Note: The diagonal (RR, GG, BB, YY) shows repetition is allowed

Circular Permutation

Circular Permutation applies when arranging items around a circle where rotations are considered identical arrangements. Classic examples include seating people around a round table or arranging objects in a circular pattern.

Property Circular Permutation
Use all items ✓
Order matters ✓
Identical items ✗
Repetitions ✗
Linear arrangement ✗

Circular Permutation — Examples:
Arranging guests around a round table, seating participants in a circular panel, ordering beads in a circular necklace, organizing tasks in a repeating cycle, setting positions for players in a circular game arrangement.

Notation:
Circular permutation

Pcirc(n)P_{\text{circ}}(n)

Alternative notations

CnC_n

Pcyclic(n)P_{\text{cyclic}}(n)

n!n\frac{n!}{n}

Learn more about combinatorial notations and symbols here →

Formula:

The total number of possible circular arrangements is
                                                 Pcirc(n)=(n−1)!P_{\text{circ}}(n) = (n-1)!
If nn is a total number of items to arrange then fixing one item reduces circular arrangements to (n−1)!(n-1)! linear ones.

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Original set (2-1)! = 1 way Rotations are same Original set (3-1)! = 2 ways Fix one position Original set (4-1)! = 6 ways Red fixed at top Circular Permutation: Arrange in circle, rotations are identical Fix one item's position → arrange remaining (n-1) items → get (n-1)! arrangements
Circular Permutations of ABC with Connected Balls ABC group: A B C B C A C A B These three arrangements are identical (just rotated clockwise) ACB group: A C B C B A B A C These three arrangements are identical (just rotated clockwise) In circular permutations, rotations are considered the same arrangement.
A B C A B C A C B ABC = BCA = CAB ACB = CBA = BAC These three arrangements are identical because in circular permutations, only relative order matters These three arrangements are identical because rotations of the same circular order are equivalent

Derangement


A derangement is a permutation with an added constraint: no element appears in its original position. The other five permutation types vary the question of which arrangements to count — all items or a subset, with or without repetition, linear or circular. The derangement varies the constraint on the arrangement itself: same items, same order-matters property, same linear shape, with the extra requirement that every item must have moved.

The classical motivation is the hat-check problem. nn guests check nn hats, and the attendant returns them in a random order. In how many of those orderings does no guest receive their own hat? Each such ordering is a derangement of the nn hats.

Notation


Two notations are in common use:

• !n!n — the subfactorial, the modern standard. The exclamation precedes the operand, mirroring the factorial in reverse.
• DnD_n — an alternative notation, common in older texts.

The two are interchangeable.

Formula


The explicit formula is

Derangement
!n=n!∑k=0n(−1)kk!.!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}.
Learn more about this formula: Derangement →


It is derived by applying the inclusion–exclusion principle to the set of all permutations: start with n!n! total, subtract those that fix at least one specific position, add back those that fix at least two, and continue with alternating signs through all nn positions. The full derivation lives on the inclusion–exclusion page; the permutations page uses the result.

A recursive form is also useful for computation:

Derangement Recurrence
!n=(n−1)(!(n−1)+!(n−2)),!0=1,!1=0.!n = (n-1) \bigl( !(n-1) + !(n-2) \bigr), \quad !0 = 1, \quad !1 = 0.
Learn more about this formula: Derangement Recurrence →


The asymptotic behavior of the formula is striking. As n→∞n \to \infty,

Derangement Asymptotic Limit
lim⁡n→∞!nn!=1e≈0.367879.\lim_{n \to \infty} \frac{!n}{n!} = \frac{1}{e} \approx 0.367879.
Learn more about this formula: Derangement Asymptotic Limit →


The proportion of permutations that are derangements approaches a fixed limit rather than vanishing or filling the space — for any sufficiently large nn, roughly 36.8%36.8\% of all permutations of nn items are derangements.

Small Values


The first few subfactorials:

• !1=0!1 = 0 — with one item, the only permutation leaves it in place
• !2=1!2 = 1 — the swap
• !3=2!3 = 2
• !4=9!4 = 9
• !5=44!5 = 44
• !6=265!6 = 265

Example


In the hat-check problem with 4 guests, the number of orderings in which no guest receives their own hat is

!4=4!∑k=04(−1)kk!=24(1−1+12−16+124)=9.!4 = 4! \sum_{k=0}^{4} \frac{(-1)^k}{k!} = 24 \left(1 - 1 + \tfrac{1}{2} - \tfrac{1}{6} + \tfrac{1}{24}\right) = 9.


Out of the 4!=244! = 24 possible orderings of the four hats, exactly 9 satisfy the no-fixed-point condition.

Permutations FAQ

What is a permutation?

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An arrangement of items where the order counts. Listing three books on a shelf in a different sequence produces a different permutation, even though the same books are involved. This is the defining contrast with selections, where rearranging the chosen items changes nothing about the result.Read more →

What is the difference between permutations with and without repetition?

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Without repetition each item can be used once, so the pool shrinks with every position filled and the count is a descending product. With repetition the full pool is available every time, so the count is one number raised to the power of the number of positions. A PIN allows repeats; seating people does not. Compare both on the permutation calculator.Read more →

How do you count arrangements when some items are identical?

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Count as if every item were distinct, then divide by the factorial of each repeated group's size. The division removes the rearrangements among identical items, which produce no visibly different result. Arranging the letters of a word with repeated letters is the standard case where this correction is needed.Read more →

Why do you divide by n when counting circular permutations?

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Because rotating the whole arrangement produces the same circular seating. Every distinct circle corresponds to n straight-line arrangements, one for each starting position, so the linear count overcounts by exactly that factor. Fixing one person in place and arranging the rest gives the same answer more directly.Read more →

What is the difference between nPr and n factorial?

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Factorial arranges every item, while nPr arranges only r of them chosen from n. Factorial is the special case where r equals n, so all positions are filled. Reading nPr as a factorial of something is a common slip; it is a descending product that stops after r terms rather than continuing to one.Read more →