Yes, that's what I meant by requiring that the checksum is 3. The only valid numbers for you are ones in which the sum of the digits is 3. That is why you have to do a little bit of combinatorics to convert from the binary number to the enumeration.

If the first 1 is in digit x (counting from the right), second 1 in digit y (again counting from the right) and third 1 is in digit z, then I think that the number in the enumeration is (x-1 choose 3) + (y-1 choose 2)+ (z-1)+1