If you just want to know the place of the combination, then you could do the following:
let n be the 0 based length of your set (for ABCDE n = 4)
let x be the 0 based position of the first element (for 'CD' x = 2)
let y be the 0 based position of the second element (for 'CD' y = 3)
The position in the outputed combination would be:
[n(n+1)-x(x+1)]/2 + y-x
so for 'CD': [4(4+1) - 2(2+1)]/2 + 3 - 2
==> [20 - 6]/2 + 1 = 8 (solution is not 0 based)
not sure if that is what you wanted, but fun to think about.
*Updated* added the code tags per L~R's request.