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K-Permutations

Tool to count and generate partial k-permutations, ie combinations where the order matters (A,B distinct with B,A).

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K-Permutations -

Tag(s) : Combinatorics

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K-Permutations

k-Permutation Generator







Permutation Generator

⮞ Go to: Permutations

Combinations Generator

k-Permutation with Repetition Generator

Counting Partial Permutations




Answers to Questions (FAQ)

What is a partial permutation? (Definition)

In mathematics, a partial permutation (also called k-permutation or arrangements) is an ordered list of items without repetition. These are the permutations of each combination.

How to generate k-permutations?

A partial permutation of k items out of n consists in the list of all possible permutations of each combination of k out of n.

Example: 2 items among 3 (A,B,C) can be shuffled in 6 ways: A,B A,C B,A B,C C,A C,B

In a k-permutation the notion of order is important A,B is different from B,A, contrary to combinations.

How to count k-permutations?

The counting of the permutations uses combinatorics and factorials

Example: For k items out of n, the number of k-permutations is equal to $$ n!/(n-k)! $$

When to use k-permutations?

In counting/combinatorial, the k-permutations are used to count the cases where $ k $ elements are selected from among $ n $ when the order matters.

Example: Number of podium of 3 horses among 20 participants in a horse race: 20 * 19 * 18 = 6840 possibilities

Example: Number of routes passing through 5 French cities (35,000 municipalities): 52506870250563750840000 separate routes

Example: Number of sentences of 10 words in the dictionary (100,000 words): 99955008699055063270306822366827310265723712000000 possible sentences

Chat is the list of arrangements?

The list is infinite, here are some examples:

A(2,3)6 arranged subsets(1,2)(2,1)(1,3)(3,1)(2,3)(3,2)
A(2,4)12 arranged subsets(1,2)(2,1)(1,3)(3,1)(1,4)(4,1)(2,3)(3,2)(2,4)(4,2)(3,4)(4,3)
A(2,5)20 arranged subsets(1,2)(2,1)(1,3)(3,1)(1,4)(4,1)(1,5)(5,1)(2,3)(3,2)(2,4)(4,2)(2,5)(5,2)(3,4)(4,3)(3,5)(5,3)(4,5)(5,4)
A(2,6)30 arranged subsets(1,2)(2,1)(1,3)(3,1)(1,4)(4,1)(1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,4)(4,2)(2,5)(5,2)(2,6)(6,2)(3,4)(4,3)(3,5)(5,3)(3,6)(6,3)(4,5)(5,4)(4,6)(6,4)(5,6)(6,5)
A(3,4)24 arranged subsets(1,2,3)(2,1,3)(3,1,2)(1,3,2)(2,3,1)(3,2,1)(1,2,4)(2,1,4)(4,1,2)(1,4,2)(2,4,1)(4,2,1)(1,3,4)(3,1,4)(4,1,3)(1,4,3)(3,4,1)(4,3,1)(2,3,4)(3,2,4)(4,2,3)(2,4,3)(3,4,2)(4,3,2)

How to remove the limit when computing k-permutations?

Partial permutations need exponential resources of computer time, so the generation must be paid.

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K-Permutations on dCode.fr [online website], retrieved on 2024-12-21, https://www.dcode.fr/partial-k-permutations

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