Hi. I'm happy to provide the solution for this.
This indeed is a difficult counting problem, one more difficult that what one might expect to see on the real GMAT.
I'll consider four slot arrangements, with the first two having the major voting rights, and the last two having the minor voting rights.
The first two slots can have eight combinations, four of which involve E (the red ones):
AC,
AD,
AE,
BC,
BD,
BE,
CE,
DE
If E is not chosen, then that means there's one from Division X and one from Division Y in the first pair. That leaves three people from three different divisions. Any combination would work, so there are 3C2 = 3 possibilities. For those four green pairs, there are 3 possible second pairs, for a total of 12.
If E is chosen in the first pair (e.g. AE), that means two of the remaining three people are from the same division and can't both be picked. Instead of three possible pairs from those remaining people, there would only be two --- for example, if AE is the first pair, then either BC or BD could be the second pair, but not CD. For the four red pairs, there are 2 possible second pairs, for a total of 8.
That gives 12+8=20 possibilities total. Since there are only 20, I'll list them (first pair has major voting rights, second pair has minor).
(AC BD), (AC BE), (AC DE), (AD BC), (AD BE),
(AD CE), (AE BC), (AE BD), (BC AD), (BC AE),
(BC DE), (BD AC), (BD AE), (BD AE), (BE AC),
(BE AD), (CE AD), (CE BD), (DE AC), (DE BC)
Does all that make sense? Please let me know if anyone reading this has any questions.
Mike
