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In mathematics, a combination is a selection of items from a set that has distinct members, such that the order of selection does not matter (unlike permutations).For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.
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 problems have an answer in terms of combinatorial functions (Stirling numbers and the partition function for a given number of parts).
By the definition of the lexicographic ordering, two k-combinations that differ in their largest element c k will be ordered according to the comparison of those largest elements, from which it follows that all combinations with a fixed value of their largest element are contiguous in the list.
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 ...
Two examples of this type of problem are counting combinations and counting permutations. More generally, given an infinite collection of finite sets S i indexed by the natural numbers , enumerative combinatorics seeks to describe a counting function which counts the number of objects in S n for each n .
A combination of numbers is usually presented in ascending order. An eventual 7th drawn number, the reserve or bonus, is presented at the end. An alternative method of calculating the odds is to note that the probability of the first ball corresponding to one of the six chosen is 6/49; the probability of the second ball corresponding to one of ...
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In this example, the rule says: multiply 3 by 2, getting 6. The sets {A, B, C} and {X, Y} in this example are disjoint sets, but that is not necessary.The number of ways to choose a member of {A, B, C}, and then to do so again, in effect choosing an ordered pair each of whose components are in {A, B, C}, is 3 × 3 = 9.