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The usual way to prove that there are n! different permutations of n objects is to observe that the first object can be chosen in n different ways, the next object in n − 1 different ways (because choosing the same number as the first is forbidden), the next in n − 2 different ways (because there are now 2 forbidden values), and so forth.
A main problem in permutation codes is to determine the value of (,), where (,) is defined to be the maximum number of codewords in a permutation code of length and minimum distance . There has been little progress made for 4 ≤ d ≤ n − 1 {\displaystyle 4\leq d\leq n-1} , except for small lengths.
(3.a) We may choose one of the f(k − 1, j − 1) permutations with k − 1 elements and j − 1 fixed points and add element k as a new fixed point. (3.b) We may choose one of the f(k − 1, j) permutations with k − 1 elements and j fixed points and insert element k in an existing cycle of length > 1 in front of one of the (k − 1) − j ...
This means in 3d it is sufficient to take cyclic or anticyclic permutations of (1, 2, 3) and easily obtain all the even or odd permutations. Analogous to 2-dimensional matrices, the values of the 3-dimensional Levi-Civita symbol can be arranged into a 3 × 3 × 3 array:
A map of the 24 permutations and the 23 swaps used in Heap's algorithm permuting the four letters A (amber), B (blue), C (cyan) and D (dark red) Wheel diagram of all permutations of length = generated by Heap's algorithm, where each permutation is color-coded (1=blue, 2=green, 3=yellow, 4=red).
As for the equal probability of the permutations, it suffices to observe that the modified algorithm involves (n−1)! distinct possible sequences of random numbers produced, each of which clearly produces a different permutation, and each of which occurs—assuming the random number source is unbiased—with equal probability.
The ! permutations of the numbers from 1 to may be placed in one-to-one correspondence with the ! numbers from 0 to ! by pairing each permutation with the sequence of numbers that count the number of positions in the permutation that are to the right of value and that contain a value less than (that is, the number of inversions for which is the ...
From a permutations perspective, let the event A be the probability of finding a group of 23 people without any repeated birthdays. Where the event B is the probability of finding a group of 23 people with at least two people sharing same birthday, P(B) = 1 − P(A).