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Permutations without repetition on the left, with repetition to their right. If M is a finite multiset, then a multiset permutation is an ordered arrangement of elements of M in which each element appears a number of times equal exactly to its multiplicity in M. An anagram of a word having some repeated letters is an example of a multiset ...
Any Hamiltonian path through the created graph is a superpermutation, and the problem of finding the path with the smallest weight becomes a form of the traveling salesman problem. The first instance of a superpermutation smaller than length 1 ! + 2 ! + … + n ! {\displaystyle 1!+2!+\ldots +n!} was found using a computer search on this method ...
The twelve problems of counting equivalence classes of functions do not involve the same difficulties, and there is not one systematic method for solving them. Two of the problems are trivial (the number of equivalence classes is 0 or 1), five problems have an answer in terms of a multiplicative formula of n and x, and the remaining five ...
Of course, this problem is solvable by finitely many trials. Rules which would push the number of trials below the number of permutations of the given points, are not known. The rule that one first should go from the starting point to the closest point, then to the point closest to this, etc., in general does not yield the shortest route. [5]
In computer science and mathematics, the Josephus problem (or Josephus permutation) is a theoretical problem related to a certain counting-out game. Such games are used to pick out a person from a group, e.g. eeny, meeny, miny, moe. A drawing for the Josephus problem sequence for 500 people and skipping value of 6.
Considering the symmetric group S n of all permutations of the set {1, ..., n}, we can conclude that the map sgn: S n → {−1, 1} that assigns to every permutation its signature is a group homomorphism. [2] Furthermore, we see that the even permutations form a subgroup of S n. [1] This is the alternating group on n letters, denoted by A n. [3]
It can be used to solve a variety of counting problems, such as how many ways there are to put n indistinguishable balls into k distinguishable bins. [4] The solution to this particular problem is given by the binomial coefficient ( n + k − 1 k − 1 ) {\displaystyle {\tbinom {n+k-1}{k-1}}} , which is the number of subsets of size k − 1 ...
The transformations of the 15 puzzle form a groupoid (not a group, as not all moves can be composed); [12] [13] [14] this groupoid acts on configurations.. Because the combinations of the 15 puzzle can be generated by 3-cycles, it can be proved that the 15 puzzle can be represented by the alternating group. [15]