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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K-Permutations
k-Permutation 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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