Let’s say, that the question was to calculate all the possible permutations of the word baboon but by picking 4 letters each time (instead of 6). When we are in a position to get all the possible permutations, we will be able to calculate the permutations of more complicated problems. If we want to get the number of rows of the table, which are actually our permutations: dim(my_matrix)Īs expected we got 180 rows (the permutations) and 6 columns (the number of letters) n and r are dictated by the limiting factor in question: which people get to be seated in each of the limited number of chairs (n of people, r of. This is less important when the two groups are the same size, but much more important when one is limited. My_list<-combinat::permn(c("b","a","b","o","o","n")) The general permutation can be thought of in two ways: who ends up seated in each chair, or which chair each person chooses to sit in. Keine Ahnung, was das bedeuten soll Keine Sorge. Working with combinat package library(combinat) Permutation ist die Gesamtheit der möglichen Kombinationen von Elementen einer gegebenen Menge miteinander. Hence the number of permutations is \(P=\frac = 180\) \(n_4\) is the number of objects of type 4, for example, the number of n which is 1.\(n_3\) is the number of objects of type 3, for example, the number of o which is 2.\(n_2\) is the number of objects of type 2, for example, the number of a which is 1.
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