In short

A permutation of r objects chosen from n distinct objects is an arrangement in a specific order. The number of such permutations is {}^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1). When you arrange all n objects (r = n), the count is simply n!. The formula comes directly from applying the multiplication principle to r sequential choices with shrinking options.

A cricket team of 11 players has won a tournament. The organiser wants to line them up on the podium for a photograph — but the order matters, because the captain should stand in the centre, the player of the tournament next to the captain, and so on. The photographer asks: in how many different ways can these 11 players stand in a row?

For the first position, you have 11 choices. Once one player stands there, 10 remain for the second position. Then 9, then 8, down to 1. By the multiplication principle:

11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 11! = 39{,}916{,}800

Nearly forty million arrangements. That number is large enough that even if the photographer took one photo per second, it would take more than a year to exhaust them all.

But what if only 3 of the 11 players receive awards (Player of the Tournament, Best Bowler, Best Fielder) and you need to arrange just those 3 on a smaller podium? Now you have 11 choices for the first award recipient, 10 for the second, 9 for the third:

11 \times 10 \times 9 = 990

This is the key question of permutations: how many ways can you pick r objects from n and arrange them in order?

What is a permutation?

Permutation

A permutation of r objects from a collection of n distinct objects is an ordered arrangement of r of those objects. Two arrangements are considered different if they contain different objects or the same objects in a different order.

The number of such permutations is denoted {}^nP_r (also written P(n, r) or P^n_r).

The word "ordered" is the crucial distinction. Picking players A, B, C for the three awards is a different permutation from picking A, C, B — because the awards are different (Player of the Tournament is not the same honour as Best Fielder). If you do not care about order — if A, B, C and A, C, B count as the same selection — that is a combination, a different concept covered in combinations.

Permutations versus combinations for choosing 2 from 3 objectsOn the left, the heading Permutations with six arrangements of objects A B C taken two at a time: AB, AC, BA, BC, CA, CB. On the right, the heading Combinations with three groupings: AB, AC, BC. The permutations distinguish order while the combinations do not. Permutations of 2 from {A,B,C} Combinations of 2 from {A,B,C} A, B A, C B, A B, C C, A C, B {A,B} {A,C} {B,C} 6 = ³P₂ 3 = C(3,2) order matters order does not matter
Choosing $2$ from $\{A, B, C\}$: permutations (left) count each ordering separately, giving $6$; combinations (right) merge orderings that contain the same elements, giving $3$. The ratio is $2! = 2$ — each combination corresponds to $2!$ permutations.

Deriving the {}^nP_r formula

The derivation is a direct application of the multiplication principle, step by step.

You have n distinct objects and you want to fill r ordered positions.

By the multiplication principle, the total number of ordered arrangements is:

{}^nP_r = n \times (n-1) \times (n-2) \times \dots \times (n-r+1)

This is a product of r consecutive descending integers, starting from n.

Now express this in terms of factorials. The product n \times (n-1) \times \dots \times (n-r+1) is the "top part" of n!, with the "bottom part" (n-r) \times (n-r-1) \times \dots \times 1 missing. So:

n \times (n-1) \times \dots \times (n-r+1) = \frac{n!}{(n-r)!}

Check: the numerator is n! = n \times (n-1) \times \dots \times (n-r+1) \times (n-r)!, and dividing by (n-r)! cancels the tail, leaving exactly the r factors you want.

The permutation formula

{}^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1) \qquad (0 \leq r \leq n)
Derivation of nPr: r slots with decreasing choicesA row of r boxes. The first box is labelled n choices, the second n minus 1, the third n minus 2, with dots indicating continuation, and the last box is labelled n minus r plus 1 choices. Below, the product is shown equalling n factorial over n minus r factorial. slot 1 slot 2 slot 3 ⋯ slot r n n−1 n−2 n−r+1 choices choices choices choices ⁿPᵣ = n × (n−1) × (n−2) × ⋯ × (n−r+1) = n! / (n−r)!
The $r$ slots are filled left to right. Each slot has one fewer choice than the previous one, because one object has been placed. The product of these $r$ descending integers is ${}^nP_r = n!/(n-r)!$.

Special case: r = n. When you arrange all n objects:

{}^nP_n = \frac{n!}{(n-n)!} = \frac{n!}{0!} = \frac{n!}{1} = n!

This confirms what you already know from factorial notation: the number of ways to arrange n distinct objects in a row is n!.

Special case: r = 1. Choosing one object from n (where order "matters" trivially):

{}^nP_1 = \frac{n!}{(n-1)!} = n

There are n ways to pick one item — no surprise.

Special case: r = 0. Choosing zero objects:

{}^nP_0 = \frac{n!}{n!} = 1

There is exactly one way to choose nothing: do nothing. This mirrors the convention 0! = 1.

Permutations of n different objects

The most common permutation problem in school mathematics is: how many ways can n distinct objects be arranged in a row? The answer is n! = {}^nP_n, and the derivation is the multiplication-principle staircase from slot 1 (with n choices) to slot n (with 1 choice).

Here are some concrete instances:

All 6 permutations of 3 objects A B CSix rows, each showing a different arrangement of the letters A, B, and C. The arrangements are ABC, ACB, BAC, BCA, CAB, CBA. Each row has three boxes holding one letter each. All permutations of {A, B, C} A B C ① A B C A C B ② A C B B A C ③ B A C B C A ④ B C A C A B ⑤ C A B C B A ⑥ C B A 3! = 6 permutations
All $6$ permutations of $\{A, B, C\}$. Each row is one arrangement — the same three letters in a different order. The count $3! = 3 \times 2 \times 1 = 6$ matches the enumeration exactly.

A natural question: what happens when some objects are identical? If you have the letters \{A, A, B\}, swapping the two A's does not produce a new arrangement. The formula for this case — permutations with repetition — is covered in permutations with constraints. For now, every formula in this article assumes all n objects are distinct.

Two worked examples

Example 1: How many 4-letter "words" can be formed from the letters of PENCIL (no letter used twice)?

The word PENCIL has 6 distinct letters: P, E, N, C, I, L. A "word" here means any sequence of 4 letters chosen from these 6, regardless of whether it makes dictionary sense.

Step 1. Identify n and r.

n = 6 (the pool of letters), r = 4 (the length of each word).

Why: each "word" is an ordered arrangement of 4 letters chosen from 6 distinct letters, with no repetition. That is exactly the setup for {}^nP_r.

Step 2. Apply the formula.

{}^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{720}{2} = 360

Why: 6! = 720 and 2! = 2. The division by (n-r)! = 2! removes the tail of the factorial that corresponds to the 2 unused letters.

Step 3. Verify by the multiplication principle.

Position 1: 6 choices. Position 2: 5. Position 3: 4. Position 4: 3.

6 \times 5 \times 4 \times 3 = 360

Why: this is the expanded form of {}^6P_4 — four consecutive descending integers starting from 6. The two approaches give the same answer, as they must.

Step 4. Sanity check.

If you used all 6 letters, the count would be 6! = 720. Using only 4 gives 360 = 720/2, which is exactly half — because the 2 unused letters can be arranged in 2! = 2 ways among themselves, and those arrangements are "absorbed" by the division. The ratio makes sense.

Result. {}^6P_4 = 360 four-letter words.

Four slots filled from six letters PENCILFour boxes in a row representing the four letter positions. Below each box is the number of available choices: 6, 5, 4, 3. The six available letters P E N C I L are shown above. The product 6 times 5 times 4 times 3 equals 360 is displayed at the bottom. Letters: P E N C I L pos 1 pos 2 pos 3 pos 4 6 5 4 3 choices choices choices choices ⁶P₄ = 6 × 5 × 4 × 3 = 360
Four positions, each with one fewer choice than the last. The multiplication principle gives $6 \times 5 \times 4 \times 3 = 360$, matching the formula ${}^6P_4 = 6!/(6-4)! = 720/2 = 360$.

Each of the 360 "words" is a distinct sequence of four letters. PENC and PENL are different words; PENC and CNEP are different words (same letters, different order). Order matters in every one of these 360 arrangements — that is what makes them permutations.

Example 2: In how many ways can a president, vice-president, and secretary be chosen from a club of $10$ members?

Step 1. Identify n and r.

n = 10 (the members), r = 3 (the positions: president, vice-president, secretary).

Why: the three positions are distinct — being president is not the same as being secretary — so the order of selection matters. This is a permutation problem, not a combination problem.

Step 2. Apply the formula.

{}^{10}P_3 = \frac{10!}{(10-3)!} = \frac{10!}{7!}

Why: you are choosing 3 people from 10 and assigning them to 3 distinct roles. The formula {}^nP_r = n!/(n-r)! handles exactly this.

Step 3. Cancel and compute.

\frac{10!}{7!} = 10 \times 9 \times 8 = 720

Why: 10! = 10 \times 9 \times 8 \times 7!, and dividing by 7! leaves the three top factors. This cancellation is the standard factorial simplification from factorial notation.

Step 4. Interpret.

There are 720 ways to fill the three officer positions. If you only wanted to know how many groups of 3 members could be chosen (ignoring which role each gets), you would divide by 3! = 6, getting 720/6 = 120 — that is the combination \binom{10}{3}. The permutation count is larger by a factor of r! = 3! = 6 because each group of 3 members corresponds to 6 different role assignments.

Result. {}^{10}P_3 = 720 ways to choose the three officers.

Choosing 3 officers from 10 membersTen circles representing club members numbered 1 through 10 arranged in a row. Three of them (members 3, 7, and 5) are highlighted in red and connected by arrows to three labelled boxes below: President, Vice-President, Secretary. The arrangement shows one specific permutation out of 720 possible. 10 club members 1 2 3 4 5 6 7 8 9 10 President Vice-Pres. Secretary member 3 member 7 member 5 One of ¹⁰P₃ = 720 possible assignments swapping any two officers gives a different permutation
One of the $720$ ways to assign $3$ officer roles to $10$ members. Member $3$ is president, member $7$ is vice-president, member $5$ is secretary. Swapping any two of these — say, making $7$ the president and $3$ the vice-president — is a *different* permutation, even though the same three people hold office.

The distinction between permutations and combinations is visible in the diagram: if you ignored the role labels (president, vice-president, secretary) and cared only about which three members were chosen, the count would drop by a factor of 3! = 6. Permutations track both the selection and the arrangement; combinations track only the selection.

Common confusions

Going deeper

If you have the {}^nP_r formula and can apply it to arrangement problems, you are ready for permutations with constraints and combinations. The rest of this section is for readers interested in the algebraic structure of permutations and their connection to group theory.

Permutations as functions

A permutation of n objects can be thought of as a bijection — a one-to-one and onto function from the set \{1, 2, \dots, n\} to itself. The permutation "move object 1 to position 3, object 2 to position 1, object 3 to position 2" is the function \sigma where \sigma(1) = 3, \sigma(2) = 1, \sigma(3) = 2.

This functional viewpoint lets you compose permutations: apply one, then apply another. If \sigma and \tau are both permutations of \{1, 2, 3\}, the composition \tau \circ \sigma is also a permutation. The set of all n! permutations of \{1, \dots, n\}, together with composition, forms a group — the symmetric group S_n. Group theory, which studies these structures abstractly, is one of the central branches of modern mathematics, and the symmetric group is its most concrete and most important example.

Derangements: permutations with no fixed points

A derangement is a permutation where no object ends up in its original position. If you have n letters and n envelopes (each letter belongs to a specific envelope), a derangement is a way to put every letter in the wrong envelope.

The number of derangements of n objects is:

D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} = n!\left(1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!}\right)

For n = 3: D_3 = 3!(1 - 1 + 1/2 - 1/6) = 6 \times 1/3 = 2. The two derangements of \{1, 2, 3\} are (2, 3, 1) and (3, 1, 2) — the only arrangements where no element occupies its original position.

As n grows, the ratio D_n / n! approaches 1/e \approx 0.3679. So roughly 36.8\% of all permutations are derangements — a fact that is useful in probability.

Cayley's formula

How many labelled trees can be built on n vertices? The answer is n^{n-2}, a result known as Cayley's formula. The proof uses a beautiful bijection between labelled trees and sequences of n - 2 numbers from \{1, \dots, n\} (Prufer sequences), turning a graph-theory question into a counting question solvable by the multiplication principle: (n-2) positions, each with n choices, giving n^{n-2}. It is a striking example of how permutation-style counting reaches far beyond the "arrange objects in a row" problems of this chapter.

Where this leads next

Permutations are the gateway to the full combinatorics sequence.