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Then the word problem in is solvable: given two words , in the generators of , write them as words in and compare them using the solution to the word problem in . It is easy to think that this demonstrates a uniform solution of the word problem for the class K {\displaystyle K} (say) of finitely generated groups that can be embedded in G ...
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 ...
In group theory, a word is any written product of group elements and their inverses. For example, if x , y and z are elements of a group G , then xy , z −1 xzz and y −1 zxx −1 yz −1 are words in the set { x , y , z }.
The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
For example, the word "encyclopedia" is a sequence of symbols in the English alphabet, a finite set of twenty-six letters. Since a word can be described as a sequence, other basic mathematical descriptions can be applied. The alphabet is a set, so as one would expect, the empty set is a subset. In other words, there exists a unique word of ...
Combinations and permutations in the mathematical sense are described in several articles. Described together, in-depth: Twelvefold way; Explained separately in a more accessible way: Combination; Permutation; For meanings outside of mathematics, please see both words’ disambiguation pages: Combination (disambiguation) Permutation ...
In the given example, there are 12 = 2(3!) permutations with property P 1, 6 = 3! permutations with property P 2 and no permutations have properties P 3 or P 4 as there are no restrictions for these two elements. The number of permutations satisfying the restrictions is thus: 4! − (12 + 6 + 0 + 0) + (4) = 24 − 18 + 4 = 10.
In combinatorics, the twelvefold way is a systematic classification of 12 related enumerative problems concerning two finite sets, which include the classical problems of counting permutations, combinations, multisets, and partitions either of a set or of a number.