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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.
C n is the number of noncrossing partitions of the set {1, ..., n}. A fortiori, C n never exceeds the n-th Bell number. C n is also the number of noncrossing partitions of the set {1, ..., 2n} in which every block is of size 2. C n is the number of ways to tile a stairstep shape of height n with n rectangles. Cutting across the anti-diagonal ...
Sorting a set of unlabelled weights by weight using only a balance scale requires a comparison sort algorithm. A comparison sort is a type of sorting algorithm that only reads the list elements through a single abstract comparison operation (often a "less than or equal to" operator or a three-way comparison) that determines which of two elements should occur first in the final sorted list.
Shuffling can also be implemented by a sorting algorithm, namely by a random sort: assigning a random number to each element of the list and then sorting based on the random numbers. This is generally not done in practice, however, and there is a well-known simple and efficient algorithm for shuffling: the Fisher–Yates shuffle .
For a given set of n beads, all distinct, the number of distinct necklaces made from these beads, counting rotated necklaces as the same, is n! / n = (n − 1)!. This is because the beads can be linearly ordered in n ! ways, and the n circular shifts of such an ordering all give the same necklace.
The permutation graphs of stack-sortable permutations are trivially perfect. [4] For each element i of a permutation p, define b i to be the number of other elements that are to the left of and greater than i. Then p is stack-sortable if and only if, for all i, b i − b i + 1 ≤ 1. [1]
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 ...
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.