Consider the integer set S = {1, 2, ... , n}. A combination is a subset of S. We emphasize that a combination differs from a permutation in that elements in a combination have no specific ordering. The 2-element subsets of S = {1, 2, 3, 4} are {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} , whereas the 2-permutations of S are more numerous with (1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3) . There are therefore fewer 2-element subsets of S than 2-permutations of S.

We can compute the number of k-element combinations of S = {1, 2, ... , n} as follows. First, we note that a k-permutation can be formed by first selecting k objects from S and then ordering them. There are k! distinct ways of ordering k objects. The number of k-permutations must then equal the number of k-element combinations multiplied by k!. Since the total number of k-permutations of S is n! / (n-k)!, the number of k-element combinations is found to be

This expression is called a binomial coefficient. Observe that selecting a k-element subset of S is equivalent to choosing the n-k elements that belong to its complement.

Example - Sampling without Replacement or Ordering: An urn contains n balls numbered one through n. A ball is drawn from the urn and placed in a jar. This process is repeated until the jar contains k balls, where . We wish to compute the number of distinct combinations the jar can hold after the completion of this experiment. Because there is no ordering in the jar, this amounts to counting the number of k-element subsets of a given n-element set, which is given by

Again, let S = {1, 2, ... , n}. Since a combination is also a subset and the number of k-element combinations of S is , the sum of the binomial coefficients over all values of k must be equal to the number of elements in the power set of S. Thus, we get