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)