Permutations: A Comprehensive Guide

Permutations count the ways to arrange objects where order matters, a cornerstone of combinatorics. Essential in sequences, rankings, and algorithms, permutations differ from combinations by emphasizing position. This MathMultiverse guide covers their formulas, interactive examples, visualizations, and applications.

From ancient puzzles to modern computing, permutations quantify ordered arrangements, adapting to constraints like repetition or circular setups. Try our interactive permutations calculator to explore these concepts hands-on.

Permutation Formulas

Distinct Objects

Arrange \( n \) distinct objects in a sequence:

\[ P(n) = n! \]

Partial Selection

Arrange \( k \) items from \( n \) distinct items:

\[ P(n, k) = \frac{n!}{(n - k)!} \]

Repetition Allowed

Choose \( k \) items from \( n \) choices with repetition:

\[ n^k \]

Identical Objects

Arrange items with repetitions (\( n_1, n_2, \ldots \)):

\[ \frac{n!}{n_1! \cdot n_2! \cdot \ldots} \]

Circular Permutations

Arrange \( n \) distinct objects in a circle:

\[ (n - 1)! \]

Derangements

Arrange \( n \) objects so none are in their original position:

\[ !n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} \]

Permutation Examples

Basic Arrangement

Arrange 3 distinct letters (A, B, C):

\[ 3! = 3 \times 2 \times 1 = 6 \]

Sequences: ABC, ACB, BAC, BCA, CAB, CBA.

Steps: Calculate \( 3! = 3 \times 2 \times 1 = 6 \), representing all possible orderings of 3 distinct items.

Try it: n =

Partial Selection

Arrange 2 books from 4 distinct books:

\[ P(4, 2) = \frac{4!}{(4-2)!} = \frac{24}{2} = 12 \]

Steps: \( 4! = 24 \), \( (4-2)! = 2! = 2 \), so \( P(4, 2) = \frac{24}{2} = 12 \). Examples include AB, BA, AC, CA, etc.

n = k =

Repetition

Create 3-digit codes using digits 0-9 (repetition allowed):

\[ 10^3 = 10 \times 10 \times 10 = 1,000 \]

Steps: Each of the 3 positions can be any of 10 digits, so \( 10^3 = 1,000 \). Examples: 000, 001, 010, etc.

n = k =

Identical Objects

Arrange the letters in "BOOK" (B, O, O, K):

\[ \frac{4!}{1! \cdot 2! \cdot 1!} = \frac{24}{1 \cdot 2 \cdot 1} = 12 \]

Steps: Total letters = 4, with O repeated twice. Calculate \( \frac{4!}{1! \cdot 2! \cdot 1!} = \frac{24}{2} = 12 \). Examples: BOOK, BOKO, OBOK, etc.

Circular Arrangement

Arrange 6 people around a circular table:

\[ (6 - 1)! = 5! = 120 \]

Steps: For circular arrangements, use \( (n-1)! \). Here, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

n =

Permutation Counts

Number of permutations for 3 to 6 distinct objects.

Applications of Permutations

Scheduling

Order 8 cars in a race:

\[ 8! = 40,320 \]

Steps: Calculate \( 8! = 8 \times 7 \times \ldots \times 1 = 40,320 \), representing all possible race orderings.

Cryptography

Generate 5-character passwords using letters A-Z (repetition allowed):

\[ 26^5 = 11,881,376 \]

Steps: Each of 5 positions can be any of 26 letters, so \( 26^5 = 26 \times 26 \times 26 \times 26 \times 26 = 11,881,376 \).

Optimization

Arrange 5 cities for a traveling salesman (circular):

\[ (5 - 1)! = 24 \]

Steps: For a circular route, use \( (n-1)! \). Here, \( 4! = 24 \).

Statistics

Rank 6 candidates:

\[ 6! = 720 \]

Steps: Calculate \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \), representing all possible rankings.

Try Our Permutations Calculator

Use our interactive permutations calculator to compute permutations and permutations with repetition with step-by-step solutions.