> c2 > c1 as follows. From the definition of the ordering it follows that for each k-combination S strictly less than C, there is a unique index i such that ci is absent from S, while ck, ..., ci+1 are present in S, and no other value larger than ci is. One can therefore group those S according to the possible values 1, 2, ..., k of i, and count each group separately. For a given value of i one must include ck, ..., ci+1 in S, and the remaining i elements of S must be chosen from the ci non-negative integers strictly less than ci; moreover any such choice will result in a S strictly less than C. The number of possible choices is , which is therefore the number of combinations in group i; the total number of k-combinations strictly less than C then is and this is the index (starting from 0) of C in the ordered list of k-combinations. Obviously there is for every N ∈ N exactly one k-combination at index N in the list (supposing k ≥ 1, since the list is then infinite), so the above argument proves that every N can be written in exactly one way as a sum of k binomial coefficients of the given form. Finding the k-combination for a given number The given formula allows finding the place in the lexicographic ordering of a given k-combination immediately. The reverse process of finding the k-combination at a given place N requires somewhat more work, but is straightforward nonetheless. By the definition of the lexicographic ordering, two k-combinations that differ in their largest element ck 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. Moreover the smallest combination with ck as the largest element is , and it has ci = i − 1 for all i < k (for this combination all terms in the expression except are zero). Therefore ck is the largest number such that . If k > 1 the remaining elements of the k-combination form the -combination corresponding to the number in the combinatorial number system of degree , and can therefore be found by continuing in the same way for and instead of N and k. Example Suppose one wants to