# Combinatorial number system

In mathematics, and in particular in combinatorics, the combinatorial number system of degree k (for some positive integer k), also referred to as combinadics, or the Macaulay representation of an integer, is a correspondence between natural numbers (taken to include 0) N and k-combinations. The combinations are represented as strictly decreasing sequences ck > ... > c2 > c1 ≥ 0 where each ci corresponds to the index of a chosen element in a given k-combination. Distinct numbers correspond to distinct k-combinations, and produce them in lexicographic order. The numbers less than ${\displaystyle {\tbinom {n}{k}}}$ correspond to all k-combinations of {0, 1, ..., n − 1}. The correspondence does not depend on the size n of the set that the k-combinations are taken from, so it can be interpreted as a map from N to the k-combinations taken from N; in this view the correspondence is a bijection.

The number N corresponding to (ck, ..., c2, c1) is given by

${\displaystyle N={\binom {c_{k}}{k}}+\cdots +{\binom {c_{2}}{2}}+{\binom {c_{1}}{1}}}$.

The fact that a unique sequence corresponds to any non-negative number N was first observed by D. H. Lehmer.[1] Indeed, a greedy algorithm finds the k-combination corresponding to N: take ck maximal with ${\displaystyle {\tbinom {c_{k}}{k}}\leq N}$, then take ck−1 maximal with ${\displaystyle {\tbinom {c_{k-1}}{k-1}}\leq N-{\tbinom {c_{k}}{k}}}$, and so forth. Finding the number N, using the formula above, from the k-combination (ck, ..., c2, c1) is also known as "ranking", and the opposite operation (given by the greedy algorithm) as "unranking"; the operations are known by these names in most computer algebra systems, and in computational mathematics.[2][3]

The originally used term "combinatorial representation of integers" was shortened to "combinatorial number system" by Knuth,[4] who also gives a much older reference;[5] the term "combinadic" is introduced by James McCaffrey[6] (without reference to previous terminology or work).

Unlike the factorial number system, the combinatorial number system of degree k is not a mixed radix system: the part ${\displaystyle {\tbinom {c_{i}}{i}}}$ of the number N represented by a "digit" ci is not obtained from it by simply multiplying by a place value.

The main application of the combinatorial number system is that it allows rapid computation of the k-combination that is at a given position in the lexicographic ordering, without having to explicitly list the k-combinations preceding it; this allows for instance random generation of k-combinations of a given set. Enumeration of k-combinations has many applications, among which are software testing, sampling, quality control, and the analysis of lottery games.

## Ordering combinations

A k-combination of a set S is a subset of S with k (distinct) elements. The main purpose of the combinatorial number system is to provide a representation, each by a single number, of all ${\displaystyle {\tbinom {n}{k}}}$ possible k-combinations of a set S of n elements. Choosing, for any n, {0, 1, ..., n − 1} as such a set, it can be arranged that the representation of a given k-combination C is independent of the value of n (although n must of course be sufficiently large); in other words considering C as a subset of a larger set by increasing n will not change the number that represents C. Thus for the combinatorial number system one just considers C as a k-combination of the set N of all natural numbers, without explicitly mentioning n.

In order to ensure that the numbers representing the k-combinations of {0, 1, ..., n − 1} are less than those representing k-combinations not contained in {0, 1, ..., n − 1}, the k-combinations must be ordered in such a way that their largest elements are compared first. The most natural ordering that has this property is lexicographic ordering of the decreasing sequence of their elements. So comparing the 5-combinations C = {0,3,4,6,9} and C′ = {0,1,3,7,9}, one has that C comes before C′, since they have the same largest part 9, but the next largest part 6 of C is less than the next largest part 7 of C′; the sequences compared lexicographically are (9,6,4,3,0) and (9,7,3,1,0).

Another way to describe this ordering is view combinations as describing the k raised bits in the binary representation of a number, so that C = {c1, ..., ck} describes the number

${\displaystyle 2^{c_{1}}+2^{c_{2}}+\cdots +2^{c_{k}}}$

(this associates distinct numbers to all finite sets of natural numbers); then comparison of k-combinations can be done by comparing the associated binary numbers. In the example C and C′ correspond to numbers 10010110012 = 60110 and 10100010112 = 65110, which again shows that C comes before C′. This number is not however the one one wants to represent the k-combination with, since many binary numbers have a number of raised bits different from k; one wants to find the relative position of C in the ordered list of (only) k-combinations.

## Place of a combination in the ordering

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 = {ck, ..., c2, c1} with ck > ... > 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 k-combinations 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 k-combinations S strictly less than C. The number of possible choices is ${\displaystyle {\tbinom {c_{i}}{i}}}$, which is therefore the number of combinations in group i; the total number of k-combinations strictly less than C then is

${\displaystyle {\binom {c_{1}}{1}}+{\binom {c_{2}}{2}}+\cdots +{\binom {c_{k}}{k}},}$

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 ${\displaystyle {\tbinom {c_{k}}{k}}}$, and it has ci = i − 1 for all i < k (for this combination all terms in the expression except ${\displaystyle {\tbinom {c_{k}}{k}}}$ are zero). Therefore ck is the largest number such that ${\displaystyle {\tbinom {c_{k}}{k}}\leq N}$. If k > 1 the remaining elements of the k-combination form the k − 1-combination corresponding to the number ${\displaystyle N-{\tbinom {c_{k}}{k}}}$ in the combinatorial number system of degree k − 1, and can therefore be found by continuing in the same way for ${\displaystyle N-{\tbinom {c_{k}}{k}}}$ and k − 1 instead of N and k.

### Example

Suppose one wants to determine the 5-combination at position 72. The successive values of ${\displaystyle {\tbinom {n}{5}}}$ for n = 4, 5, 6, ... are 0, 1, 6, 21, 56, 126, 252, ..., of which the largest one not exceeding 72 is 56, for n = 8. Therefore c5 = 8, and the remaining elements form the 4-combination at position 72 − 56 = 16. The successive values of ${\displaystyle {\tbinom {n}{4}}}$ for n = 3, 4, 5, ... are 0, 1, 5, 15, 35, ..., of which the largest one not exceeding 16 is 15, for n = 6, so c4 = 6. Continuing similarly to search for a 3-combination at position 16 − 15 = 1 one finds c3 = 3, which uses up the final unit; this establishes ${\displaystyle 72={\tbinom {8}{5}}+{\tbinom {6}{4}}+{\tbinom {3}{3}}}$, and the remaining values ci will be the maximal ones with ${\displaystyle {\tbinom {c_{i}}{i}}=0}$, namely ci = i − 1. Thus we have found the 5-combination {8, 6, 3, 1, 0}.

## National Lottery example

For each of the ${\displaystyle {\binom {49}{6}}}$ lottery combinations c1 < c2 < c3 < c4 < c5 < c6 , there is a list number N between 0 and ${\displaystyle {\binom {49}{6}}-1}$ which can be found by adding

${\displaystyle {\binom {49-c_{1}}{6}}+{\binom {49-c_{2}}{5}}+{\binom {49-c_{3}}{4}}+{\binom {49-c_{4}}{3}}+{\binom {49-c_{5}}{2}}+{\binom {49-c_{6}}{1}}.}$