Assign the bucket to the number (instead of the number to the bucket), and you have a much easier to comprehend solution. As a sanity check, you will have Y^x results.

One approach, with 8 objects in 2 bins, would be to treat it as binary. A digit of '0' in position X would put item X into bucket 0. A digit of '1' in position X would put item X into bucket 1. If you iterate over all of the values for that binary number, you have just iterated over all of the solutions - in this case 256.

00000000  => (87654321), ()
00000001  => (8765432), (1)
00000010  => (8765431), (2)
00000011  => (876543), (21)
00000100  => (8765421), (3)
...
11111111  => (), (87654321)
[download]

Extrapolate this to any value of X (8 above) and Y (2 above), and you have, I believe, your solution.

Update: Added example data

Update: You could also handle the case where all of the items do not need to be used by assigning a bin Y+1 as a hidden bin.