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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.
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.
These combinations (subsets) are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to 2 n − 1, where each digit position is an item from the set of n. Given 3 cards numbered 1 to 3, there are 8 distinct combinations , including the empty set:
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
The same argument shows that the number of compositions of n into exactly k parts (a k-composition) is given by the binomial coefficient (). Note that by summing over all possible numbers of parts we recover 2 n−1 as the total number of compositions of n:
Frobenius coin problem with 2-pence and 5-pence coins visualised as graphs: Sloping lines denote graphs of 2x+5y=n where n is the total in pence, and x and y are the non-negative number of 2p and 5p coins, respectively.
For each of the n − 1 hats that P 1 may receive, the number of ways that P 2, ..., P n may all receive hats is the sum of the counts for the two cases. This gives us the solution to the hat-check problem: Stated algebraically, the number ! n of derangements of an n -element set is ! n = ( n − 1 ) ( !
For instance, the number 25 in column k = 3 and row n = 5 is given by 25 = 7 + (3×6), where 7 is the number above and to the left of 25, 6 is the number above 25 and 3 is the column containing the 6.