Calculate Permutations

Calculate the number of possible arrangements (orders) without repetition

Factorial n!: Number of ways to arrange n different objects

Permutations Calculator

Permutations without Repetition

Calculates P(n) = n! - the number of ways to arrange n different objects in different orders.

Number of different objects to arrange
Calculation Result
n! =

Permutations Example

Default Example: 4! = 24
Objects: n = 4
Calculation: 4! = 4×3×2×1
Result: 4! = 24
24 different arrangements of 4 objects
Concrete Example: Objects {A, B, C}

All possible arrangements of A, B, C:

ABC ACB BAC BCA CAB CBA

3! = 6 permutations

Important Properties
  • Order is crucial: ABC ≠ BAC
  • All objects must be used
  • No repetition: Each object exactly once
  • n! grows very quickly with n

Mathematical Foundations of Factorial

Permutations without repetition are based on the concept of factorial:

Factorial Definition
\[n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1\]

Product of all natural numbers from 1 to n

Permutation Principle
\[\text{Position 1: n choices, Position 2: (n-1) choices, etc.}\]

Step-by-step reduction of choice options

Permutation Formulas and Examples

General Permutation Formula
\[P(n) = n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1\]

Number of arrangements of n different objects

Step-by-Step Calculation: 4!

Given: n = 4 different objects {A, B, C, D}

1. Calculate factorial:

\[4! = 4 \times 3 \times 2 \times 1\]

2. Step-by-step multiplication:

4 × 3 = 12

12 × 2 = 24

24 × 1 = 24

3. Result:

\[4! = 24 \text{ different arrangements}\]
Intuitive Explanation with Positions

4 objects {A, B, C, D} in 4 positions:

Position 1: 4 choices (A, B, C, or D)

Position 2: 3 choices (3 remaining objects)

Position 3: 2 choices (2 remaining objects)

Position 4: 1 choice (1 remaining object)

Total: 4 × 3 × 2 × 1 = 24 arrangements

Additional Calculation Examples
Small factorials:

0! = 1 (by definition)

1! = 1

2! = 2

3! = 6

Large factorials:

5! = 120

6! = 720

7! = 5,040

8! = 40,320

Complete arrangements for n=3:
ABC ACB BAC BCA CAB CBA

3! = 6 different permutations

Factorial Growth
Small Values

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

Large Values

10! = 3,628,800

12! = 479,001,600

15! ≈ 1.3 × 10¹²

20! ≈ 2.4 × 10¹⁸

25! ≈ 1.6 × 10²⁵

Factorial grows extremely fast - even 13! exceeds 6 billion!

Permutations Reference

Default Example
4! = 24 4×3×2×1 24 arrangements
Special Values

0! = 1: Empty set (by definition)

1! = 1: One object

n! = n × (n-1)!: Recursion

Stirling: n! ≈ √(2πn)(n/e)ⁿ

Properties

Order: ABC ≠ BAC ≠ CBA

Use all: Each object exactly once

Without repetition: All different

Exponential growth: Very large numbers

Applications

Sequences: People in a line

Anagrams: Rearranging letters

Scheduling: Task ordering

Codes: Password generation

Permutations - Detailed Description

Fundamentals of Permutations

Permutations are fundamental objects in combinatorics and describe the number of possible arrangements of n different objects. The order is crucial - ABC is a different permutation than BAC.

Characteristics:
• Order is crucial: ABC ≠ ACB
• All objects must be used
• No repetition: Each object exactly once
• Calculation via factorial: n!

The Factorial

The factorial n! is the mathematical tool for calculating permutations. It arises from the principle of step-by-step choice reduction: For the first position there are n possibilities, for the second n-1, etc.

Recursive Definition

n! = n × (n-1)! with 0! = 1 as base
This recursive structure enables efficient calculations

Practical Applications

Permutations find applications in many areas: from arranging people in a queue to generating anagrams to complex scheduling problems in computer science.

Typical Scenarios:
• Seating arrangements at events
• Order of tasks/processes
• Anagram generation
• Password combinations

Exponential Growth

A fascinating aspect of the factorial is its extreme growth. Even with relatively small values of n, n! reaches astronomical magnitudes, which is of great importance in cryptography and complexity theory.

Stirling Approximation

For large n: n! ≈ √(2πn) × (n/e)ⁿ
This approximation helps estimate large factorials

Practical Examples and Applications

Seating Arrangement

Problem: 6 people at round table

Solution: (6-1)! = 5! = 120

Reason: Rotational symmetry

Application: Dinner planning

Anagram Calculation

Word: "MATH" (4 letters)

Anagrams: 4! = 24

Examples: MATH, MTHA, AMTH...

Application: Word games, puzzles

Task Scheduling

Problem: 5 tasks sequentially

Possibilities: 5! = 120

Optimization: Find best order

Application: Project management

Permutation Variants
  • Regular permutations: n! - All objects, without repetition
  • k-permutations: P(n,k) = n!/(n-k)! - Select k from n
  • Circular permutations: (n-1)! - Round arrangements
  • With repetition: Multinomial coefficients
  • Derangements: Permutations without fixed points
  • Involutions: Self-inverse permutations


Other Combinatorics Functions

Combinations with Repetition  •  Combinations without Repetition  •  Permutations  •  Rule of Product  •  Variations with Repetition  •  Variations without Repetition  •  Activity Selection Problem  •