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

K-Permutations - dCode

Tag(s) : Combinatorics

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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.

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.

The counting of the permutations uses combinatorics and factorials

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

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

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) |

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

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Cite as source (bibliography):

*K-Permutations* on dCode.fr [online website], retrieved on 2024-10-04,

arrangement,permutation,combination,counting,order,combinatorics

https://www.dcode.fr/partial-k-permutations

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