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The number of k-combinations for all k is the number of subsets of a set of n elements. There are several ways to see that this number is 2 n. In terms of combinations, () =, which is the sum of the nth row (counting from 0) of the binomial coefficients in Pascal's triangle.
The number associated in the combinatorial number system of degree k to a k-combination C is the number of k-combinations strictly less than C in the given ordering. This number can be computed from C = {c k, ..., c 2, c 1} with c k > ... > c 2 > c 1 as follows.
The Indian mathematician Mahāvīra (c. 850) provided formulae for the number of permutations and combinations, [13] [14] and these formulas may have been familiar to Indian mathematicians as early as the 6th century CE. [15]
One must divide the number of combinations producing the given result by the total number of possible combinations (for example, () =,,).The numerator equates to the number of ways to select the winning numbers multiplied by the number of ways to select the losing numbers.
For any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of (k − 1)-element subsets of a set with n − 1 elements. For example, if n = 10 and k = 4, the theorem gives the number of solutions to x 1 + x 2 + x 3 + x 4 = 10 (with x 1, x 2, x 3, x 4 > 0) as the binomial coefficient
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 ...
An archetypal double counting proof is for the well known formula for the number () of k-combinations (i.e., subsets of size k) of an n-element set: = (+) ().Here a direct bijective proof is not possible: because the right-hand side of the identity is a fraction, there is no set obviously counted by it (it even takes some thought to see that the denominator always evenly divides the numerator).
The formula counting all functions N → X is not useful here, because the number of them grouped together by permutations of N varies from one function to another. Rather, as explained under combinations , the number of n -multicombinations from a set with x elements can be seen to be the same as the number of n -combinations from a set with x ...