Permutations Calculator

Permutations is evaluated from Total Items, Items to Arrange and Allow Repetition?. The calculation reports Permutations P, Combinations C and Circular Permutations.

Results

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About the Permutations Calculator

Permutations is treated here as a quantitative relation between Total Items, Items to Arrange and Allow Repetition? and Permutations P, Combinations C, Circular Permutations and Probability of Specific Order.

The calculator uses a multi formula configuration. Each reported value is read as a direct evaluation of the stored rules with the declared field formats and units.

Formula basis:
P(n,r) = n! / (n - r)!
P with repetition = nʳ
Circular P(n) = (n - 1)!

Interpret the outputs in the order shown by the result fields. Optional inputs affect only the outputs that depend on those variables.

Formula & How It Works

The calculation applies the following relations exactly as recorded in the metadata:

P(n,r) = n! / (n - r)!
P with repetition = nʳ
Circular P(n) = (n - 1)!

Each output field is produced by substituting the supplied inputs into the relevant relation and then applying the declared rounding or text format.

Worked Examples

Example 1: 4-Digit PIN (with repetition)

Inputs

n_val: 10 r_val: 4 repetition: yes

Permutations P: 10,000. Combinations C: 210. Circular Permutations: 0. Probability of Specific Order: 0.0001

With Total Items = 10, Items to Arrange = 4 and Allow Repetition? = yes as the stated inputs, the result is Permutations P = 10,000, Combinations C = 210 and Circular Permutations = 0. Each value corresponds to the declared output fields.

Example 2: Race Finishing Order — Top 3 of 8

Inputs

n_val: 8 r_val: 3 repetition: no

Permutations P: 336. Combinations C: 56. Circular Permutations: 0. Probability of Specific Order: 0.0029761905

With Total Items = 8, Items to Arrange = 3 and Allow Repetition? = no as the stated inputs, the result is Permutations P = 336, Combinations C = 56 and Circular Permutations = 0. Each value corresponds to the declared output fields.

Example 3: Seating 6 People at a Round Table

Inputs

n_val: 6 r_val: 6 repetition: no

Permutations P: 720. Combinations C: 1. Circular Permutations: 120. Probability of Specific Order: 0.0013888889

With Total Items = 6, Items to Arrange = 6 and Allow Repetition? = no as the stated inputs, the result is Permutations P = 720, Combinations C = 1 and Circular Permutations = 120. Each value corresponds to the declared output fields.

Example 4: License Plate — Letters and Digits

Inputs

n_val: 26 r_val: 3 repetition: yes

Permutations P: 17,576. Combinations C: 2,600. Circular Permutations: 0. Probability of Specific Order: 0.0000568958

With Total Items = 26, Items to Arrange = 3 and Allow Repetition? = yes as the stated inputs, the result is Permutations P = 17,576, Combinations C = 2,600 and Circular Permutations = 0. Each value corresponds to the declared output fields.

Common Use Cases

Count ways to arrange letters in a word
Find number of race finishing orders
Calculate PIN and password combinations
Count ways to assign seats in a row