We seem to be missing something (concept-wise) for #21.
The way we understand it, the number of blue balls that should fall should correspond with the number indicated by the sum of the values of bits that face to the right.
1st bit: 1
2nd bit: 2
3rd bit: 4
4th bit: 8
So if, for instance, only the top bit faces right, only one ball should drop.
My son and I were unable to come up with a way to divert the balls to the right (red) output with the allowed parts. This was the only way that we thought we would be able to control the number of balls that would drop, since the red input was empty.
When we checked the solution, it did not solve the puzzle as we expected it to be solved. Instead, all of the blue balls went to the blue output regardless of the bit orientation.
Can someone explain what we are missing?