Permutation

Definitionnoun

What is a permutation in math?

A permutation is an arrangement of objects in a specific order, so ABC and CBA count as different. The number of ways to arrange r objects chosen from n different objects is P(n, r) = n!(n − r)!. For example, picking a president and vice president from 5 people gives 5 × 4 = 20 permutations.

In Spanish
permutación
Also called
arrangement, ordering, nPr, P(n, r)
Grade level
Precalculus
Algebra 2PrecalculusStatisticsCCSS HSS-CP.B.9SATACTASVAB

Permutation in a picture

Permutations: arrangements where order mattersThe six orders of A, B and C are listed. Choosing a president and vice president from 5 people gives 5 times 4, which is 20 permutations.All orders of A, B, C: 3! = 6 permutationsABCACBBACBCACABCBAABC and CBA use the same letters but count as different.Choose a president and a vice president from 5 people5president4vice president×= 20P(5, 2) = 5! ÷ (5 − 2)! = 120 ÷ 6 = 20Order matters: Ana-then-Ben is not Ben-then-Ana.If order did not matter, it would be a combination (10).
In a permutation the same objects in a different order make a new arrangement. Filling ordered slots one at a time gives 5 × 4 = 20, which matches P(5, 2) = 5!3!.

Formal definition and notation

A permutation of r objects from a set of n distinct objects (0 ≤ r ≤ n) is an ordered selection of r of them. Their number is P(n, r) = n(n − 1) ⋯ (n − r + 1), which equals n!(n − r)! using factorials. A permutation of all n objects is an ordering of the whole set, and there are n! of them. If some objects are identical, the count of distinct orderings is n!k1! k2! ⋯.

Symbol Read it as Meaning
P(n, r) P of n, r; the permutations of n things taken r at a time the number of ordered arrangements of r objects from n
nPr n P r the same number, as written on calculators; also printed with small n and r beside the P
n! n factorial n × (n − 1) × ⋯ × 1, the number of orderings of all n objects

Key facts

  • Order matters in a permutation. If order does not matter (a team, a committee, a hand of cards), use a combination instead.
  • P(n, n) = n! and P(n, 1) = n. Also P(n, 0) = 1: there is one way to arrange nothing.
  • The counting principle is behind the formula: n choices for the first spot, n − 1 for the second, and so on.
  • Permutations count is always at least the combinations count: P(n, r) = C(n, r) × r!. For 5 people choose 2: 20 = 10 × 2.
  • Identical objects reduce the count: the letters of BOOK can be arranged in 4!2! = 12 distinct ways, not 24.

Permutation formula

Permutation formula
P(n, r) = n!(n − r)!
n
the number of different objects available
r
how many are chosen and placed in order
n!
n factorial

Shortcut: multiply r numbers counting down from n. P(8, 3) = 8 × 7 × 6 = 336. With repeated letters, divide by the factorial of each repeat count.

Permutation examples, step by step

Example 1: Use the formula

Warm-up

Find P(6, 2).

  1. P(6, 2) = 6!(6 − 2)! = 6!4!.
  2. Cancel 4!: 6 × 5 × 4!4! = 6 × 5.
  3. 6 × 5 = 30.

Answer: P(6, 2) = 30.

Example 2: Medals in a race

Core

Eight runners are in a final. In how many ways can the gold, silver and bronze medals be awarded?

  1. Order matters: gold, silver and bronze are different prizes, so this is a permutation.
  2. n = 8 runners, r = 3 medals.
  3. P(8, 3) = 8 × 7 × 6.
  4. 8 × 7 × 6 = 336.

Answer: The medals can be awarded in 336 ways.

Example 3: Arrangements with repeated letters

Test-style

How many different arrangements of the letters in the word ALGEBRA are there?

  1. ALGEBRA has 7 letters, so 7! = 5040 if every letter were different.
  2. The letter A appears twice. Swapping the two A’s gives the same word, so each arrangement was counted 2! = 2 times.
  3. Divide: 50402 = 2520.

Answer: There are 2,520 different arrangements.

Common mistakes

✗ MistakeUsing a permutation to count 3-person committees from 8 people: P(8, 3) = 336.
✓ InsteadA committee has no order, so use a combination: 3363! = 56.

Ask whether swapping two chosen people creates a new outcome. For medals it does (permutation); for a committee it does not (combination).

✗ MistakeP(8, 3) = 8!3! = 6720.
✓ InsteadP(8, 3) = 8!(8 − 3)! = 8!5! = 336.

The bottom of the formula is (n − r)!, the factorial of the number of objects not chosen.

✗ MistakeThe letters of BOOK can be arranged in 4! = 24 ways.
✓ Instead4!2! = 12 distinct arrangements.

The two O’s look the same, so swapping them does not make a new word. Divide by the factorial of each repeat count.

Not to be confused with

Permutationvs.Combination

A combination is a selection where order does not matter. From A, B, C, D, choosing 2 gives 6 combinations (AB = BA) but 12 permutations (AB and BA both count).

Permutationvs.Factorial

n! counts the orders of all n objects. A permutation count P(n, r) orders only r of them; P(n, n) = n! is the special case.

Permutationvs.Probability

Permutations count outcomes; probability compares counts. The chance that a random order of A, B, C is exactly ABC is 13! = 16.

Quick check: 3 questions

Question 1 of 3

In how many ways can 4 different books be arranged on a shelf?

Show answer and explanation

Answer: D. All 4 books are placed in order: 4! = 4 × 3 × 2 × 1 = 24. 4 counts only the first spot, 16 is 42, and 10 adds 4 + 3 + 2 + 1.

Question 2 of 3

Which situation is a permutation?

Show answer and explanation

Answer: C. In a PIN, 1234 and 4321 are different codes, so order matters. Toppings, a card hand and invited friends are the same group in any order, so those are combinations.

Question 3 of 3

What is P(7, 3)?

Show answer and explanation

Answer: A. P(7, 3) = 7 × 6 × 5 = 210. 35 is the combination C(7, 3), 21 is 7 × 3, and 840 is 7 × 6 × 5 × 4, one factor too many.

Permutation: frequently asked questions

What is the difference between a permutation and a combination?

In a permutation order matters; in a combination it does not. Choosing a president and a vice president from 5 people is a permutation (20 ways), because Ana-president-Ben-vice differs from the reverse. Choosing a 2-person committee from the same 5 is a combination (10 ways). Ask: does rearranging the chosen items create a new outcome?

How do you calculate a permutation?

Use P(n, r) = n!(n − r)!, where n is the number of objects and r is how many you arrange. A faster way is to multiply r numbers counting down from n: P(9, 3) = 9 × 8 × 7 = 504. Calculators usually have an nPr key that does this directly.

What if some objects are the same?

Divide by the factorial of each group of identical objects. The word LEVEL has 5 letters with two L’s and two E’s, so it has 5!2! × 2! = 1204 = 30 distinct arrangements. Without dividing, every arrangement would be counted once for each way of swapping the identical letters.

Practice and tools

Sources

  1. Common Core State Standards, High School Statistics and Probability (HSS-CP.B.9)
  2. OpenStax, College Algebra, 9.5 Counting Principles

Updated November 2026·Reviewed by the Effortless Math team·Standards: CCSS HSS-CP.B.9

Cite this entry

Effortless Math Team. “Permutation.” Effortless Math Dictionary. Updated November 2026. https://www.effortlessmath.com/math-dictionary/permutation/

Flashcard

TermPermutationP(n, r) = n!(n − r)!
DefinitionAn arrangement where order matters. Gold, silver and bronze among 8 runners: 8 × 7 × 6 = 336 permutations.
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