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Permutation Calculator — nPr with Steps

Calculate permutations with step-by-step factorial expansion

P(n, r)

720

n

10

r

3

C(n,r)

120

Each item can only be selected once. Formula: n! / (n-r)!

Permutation Result

P(10, 3)

720

10ʳ (with rep.)

1,000

Step-by-Step

P(10, 3) = 10! / (10 - 3)!

= 10! / 7!

= 3,628,800 / 5,040

= 720

Permutation vs Combination

P(10, 3) — order matters720
C(10, 3) — order ignored120
P / C = r! = 3!6

Each combination of 3 items can be arranged in 6 different orders.

Formulas

P(n, r) = n! / (n − r)!
With repetition = nʳ
P(n, r) = C(n, r) × r!
P(n, n) = n!
P(n, 0) = 1

Formulas Used

Permutation (without repetition)

P(n, r) = n! / (n - r)!

Count the number of ordered arrangements of r items chosen from n distinct items, where each item is used at most once.

Where:

n= Total number of items to choose from
r= Number of items being arranged
n!= n factorial = n × (n-1) × ... × 1

Permutation (with repetition)

P_rep = n^r

Count the number of ordered arrangements when items can be reused. Each of the r positions can be filled by any of the n items.

Where:

n= Number of available items
r= Number of positions to fill

Relationship to Combinations

P(n, r) = C(n, r) × r!

Each combination of r items can be arranged in r! different orders, so permutations equal combinations times r factorial.

Where:

C(n,r)= Number of combinations (unordered selections)
r!= Number of ways to arrange r items

Example Calculations

1P(10, 3) — Top 3 from 10 Contestants

Inputs

n (Total Items)10
r (Items Chosen)3

Result

P(10, 3)720
With Repetition (10³)1,000
C(10, 3)120

P(10,3) = 10!/(10-3)! = 10!/7! = 10 × 9 × 8 = 720. There are 720 ways to award gold, silver, and bronze to 3 of 10 contestants.

2P(5, 5) — Arrange All 5 Items

Inputs

n (Total Items)5
r (Items Chosen)5

Result

P(5, 5)120
With Repetition (5⁵)3,125
C(5, 5)1

P(5,5) = 5! = 5 × 4 × 3 × 2 × 1 = 120. There are 120 ways to arrange all 5 books on a shelf.

3P(8, 2) — President and VP from 8 Candidates

Inputs

n (Total Items)8
r (Items Chosen)2

Result

P(8, 2)56
With Repetition (8²)64
C(8, 2)28

P(8,2) = 8!/6! = 8 × 7 = 56. There are 56 ways to choose a president and vice-president from 8 candidates.

Frequently Asked Questions

Q

What is a permutation and how is it different from a combination?

A permutation counts the number of ways to arrange items where order matters. In permutations, ABC and CBA are different arrangements. Combinations ignore order, so ABC and CBA count as the same selection. P(n,r) is always greater than or equal to C(n,r).

  • P(10,3) = 720 arrangements (assigning 1st, 2nd, 3rd place from 10 people)
  • C(10,3) = 120 selections (choosing a 3-person committee from 10 people)
  • P(n,r) = C(n,r) × r! — each combination generates r! permutations
  • A 4-digit PIN from digits 0–9 without repeats: P(10,4) = 5,040 arrangements
  • Choosing 4 digits (order irrelevant): C(10,4) = 210 selections
ScenarioTypeFormulaResult
3 winners from 10 (ranked)PermutationP(10,3)720
3-person committee from 10CombinationC(10,3)120
5-letter codes from 26 lettersPermutationP(26,5)7,893,600
5 cards from 52-card deckCombinationC(52,5)2,598,960
Q

How do you calculate permutations with repetition?

Permutations with repetition use the formula n^r, where n is the number of items and r is the number of positions. Each position can use any of the n items, even if that item has already been used. For example, a 4-digit PIN with digits 0-9 has 10^4 = 10,000 possibilities.

  • 4-digit PIN (0–9): 10⁴ = 10,000 possible PINs
  • 3-letter code (A–Z): 26³ = 17,576 possible codes
  • Binary string of length 8: 2⁸ = 256 possible bytes
  • Without repetition: P(10,4) = 5,040 (much fewer)
  • With repetition: order still matters, but reuse is allowed
ScenarioWithout RepetitionWith Repetition
4 from 10 digitsP(10,4) = 5,04010⁴ = 10,000
3 from 26 lettersP(26,3) = 15,60026³ = 17,576
2 from 6 dice facesP(6,2) = 306² = 36
Q

What is the factorial formula for permutations?

The permutation formula is P(n,r) = n! / (n-r)!, where n! (n factorial) means n × (n-1) × (n-2) × ... × 1. For P(10,3): 10! / 7! = (10 × 9 × 8 × 7!) / 7! = 10 × 9 × 8 = 720.

  • P(10,3) = 10!/7! = 10 × 9 × 8 = 720
  • P(5,5) = 5!/0! = 120 (0! = 1 by definition)
  • P(n,1) = n (just picking one item from n)
  • P(n,0) = 1 (there is exactly one way to choose nothing)
ExpressionExpansionResult
P(10,3)10 × 9 × 8720
P(5,5)5 × 4 × 3 × 2 × 1120
P(8,2)8 × 756
P(20,3)20 × 19 × 186,840
Q

What are real-world examples of permutations?

Permutations appear whenever order matters in a selection. Examples include ranking contestants in a competition, creating passwords, arranging books on a shelf, scheduling tasks in a specific order, and generating license plate numbers.

  • Race results: 10 runners competing for gold, silver, bronze = P(10,3) = 720 outcomes
  • Phone unlock pattern: connecting 4 of 9 dots = P(9,4) = 3,024 patterns
  • Seating arrangement: 6 people at 6 chairs = P(6,6) = 6! = 720 ways
  • License plates: 3 letters then 4 digits = 26³ × 10⁴ = 175,760,000
  • Playlist order: shuffling 20 songs = 20! ≈ 2.43 × 10¹⁸ orders
Q

Why does P(n,0) = 1 and why is 0! = 1?

P(n,0) = n!/n! = 1 because there is exactly one way to arrange zero items: do nothing. Similarly, 0! = 1 by convention, which keeps the factorial formula consistent. Without 0! = 1, formulas like C(n,n) = n!/(n! × 0!) would break.

  • P(n,0) = n!/n! = 1 for any non-negative n
  • 0! = 1 by mathematical convention (empty product)
  • This makes C(n,0) = 1 and C(n,n) = 1 work correctly
  • The gamma function Γ(1) = 0! = 1 confirms this in continuous math

Understanding Permutations in Combinatorics

Permutations are a fundamental concept in combinatorics that count the number of ways to arrange items in a specific order. Unlike combinations, permutations treat different orderings as distinct outcomes. The classic formula P(n,r) = n!/(n-r)! gives the number of ways to arrange r items chosen from a set of n items.

Permutations have wide applications in computer science (algorithm analysis, cryptography), probability theory (counting sample spaces), and everyday problems like password generation and tournament scheduling. Understanding when to use permutations vs combinations is one of the most important skills in discrete mathematics.

When repetition is allowed, the formula simplifies to n^r, which grows much faster. A 4-digit PIN from 10 digits has 10,000 possibilities with repetition but only 5,040 without. This distinction is critical in fields like information security and coding theory.

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Last Updated: Mar 9, 2026

This calculator is provided for informational and educational purposes only. Results are estimates and should not be considered professional financial, medical, legal, or other advice. Always consult a qualified professional before making important decisions. UseCalcPro is not responsible for any actions taken based on calculator results.

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