Question

UPDATE:
Combinatorics and unranking was eventually what I needed.
The links below helped alot:

http://msdn.microsoft.com/en-us/library/aa289166(v=vs.71).aspx

http://www.codeproject.com/Articles/21335/Combinations-in-C-Part-2

The Problem
Given a list of N symbols say {0,1,2,3,4…}
And NCr combinations of these

eg. NC3 will generate:

0 1 2  
0 1 3  
0 1 4  
...  
...  
1 2 3  
1 2 4  
etc...  

For the ith combination (i = [1 .. NCr]) I want to determine Whether a symbol (s) is part of it.
Func(N, r, i, s) = True/False or 0/1
eg. Continuing from above
The 1st combination contains 0 1 2 but not 3

F(N,3,1,"0") = TRUE  
F(N,3,1,"1") = TRUE  
F(N,3,1,"2") = TRUE  
F(N,3,1,"3") = FALSE  

Current approaches and tibits that might help or be related.
Relation to matrices
For r = 2 eg. 4C2 the combinations are the upper (or lower) half of a 2D matrix

    1,2 1,3 1,4  
    ----2,3 2,4  
    --------3,4  

For r = 3 its the corner of a 3D matrix or cube
for r = 4 Its the “corner” of a 4D matrix and so on.

Another relation
Ideally the solution would be of a form something like the answer to this:
Calculate Combination based on position

The nth combination in the list of combinations of length r (with repitition allowed), the ith symbol can be calculated
Using integer division and remainder:

n/r^i % r = (0 for 0th symbol, 1 for 1st symbol….etc)

eg for the 6th comb of 3 symbols the 0th 1st and 2nd symbols are:

i = 0 => 6 / 3^0 % 3 = 0   
i = 1 => 6 / 3^1 % 3 = 2   
i = 2 => 6 / 3^2 % 3 = 0   

The 6th comb would then be 0 2 0

I need something similar but with repition not allowed.

Thank you for following this question this far :]
Kevin.

Answer 1

I believe your problem is that of unranking combinations or subsets.

I will give you an implementation in Mathematica, from the package Combinatorica, but the Google link above is probably a better place to start, unless you are familiar with the semantics.

UnrankKSubset::usage = "UnrankKSubset[m, k, l] gives the mth k-subset of set l, listed in lexicographic order."

UnrankKSubset[m_Integer, 1, s_List] := {s[[m + 1]]}
UnrankKSubset[0, k_Integer, s_List] := Take[s, k]
UnrankKSubset[m_Integer, k_Integer, s_List] := 
       Block[{i = 1, n = Length[s], x1, u, $RecursionLimit = Infinity}, 
             u = Binomial[n, k]; 
             While[Binomial[i, k] < u - m, i++]; 
             x1 = n - (i - 1); 
             Prepend[UnrankKSubset[m - u + Binomial[i, k], k-1, Drop[s, x1]], s[[x1]]]
       ]

Usage is like:

UnrankKSubset[5, 3, {0, 1, 2, 3, 4}]

   {0, 3, 4}

Yielding the 6th (indexing from 0) length-3 combination of set {0, 1, 2, 3, 4}.