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An example of legal and illegal permutations can be better demonstrated by a specific problem such as balanced brackets (see Dyck language). A general problem is to count the number of balanced brackets (or legal permutations) that a string of m open and m closed brackets forms (total of 2m brackets). By legally arranged, the following rules apply:
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.
C n is the number of permutations of {1, ..., n} that avoid the permutation pattern 123 (or, alternatively, any of the other patterns of length 3); that is, the number of permutations with no three-term increasing subsequence. For n = 3, these permutations are 132, 213, 231, 312 and 321.
The sum of the numbers in the factorial number system representation gives the number of inversions of the permutation, and the parity of that sum gives the signature of the permutation. Moreover, the positions of the zeroes in the inversion table give the values of left-to-right maxima of the permutation (in the example 6, 8, 9) while the ...
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.
The types of permutations presented in the preceding two sections, i.e. permutations containing an even number of even cycles and permutations that are squares, are examples of so-called odd cycle invariants, studied by Sung and Zhang (see external links). The term odd cycle invariant simply means that membership in the respective combinatorial ...
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.
For n ≥ 0 and 0 ≤ k ≤ n, the rencontres number D n, k is the number of permutations of { 1, ..., n } that have exactly k fixed points. For example, if seven presents are given to seven different people, but only two are destined to get the right present, there are D 7, 2 = 924 ways this could happen.