This is a binary grouping game with a 2 sets of information, so you can think of this as 3D binary in a way.
This game tells us up front that we are selecting 5 of 9, which means 4 will be out.
After diagramming the rules, that is the setup I had initially.
I know that we cannot have 3 A's. I know that I have 1 D. Of the 4 that are out, I knew that we had to have 2 D's out and at least 1 A out. There is only 1 more spot left in the out column. So this means that at the very least we must have 2 P's. We can have at most 1 P out.
This is a very restrictive idea. Either we have 3 P's in or 2 P's in. What a great framing opportunity with a clear split.
I know that the 3 P's being in frame will really give me limited options, I expect to have only 1 hypothetical dealing with this idea.
However, with the 2 P's being in, I know that I have a variety of options: Either UV, UW, or VW.
I am prepared to move on to the questions if I feel that this part of the frame gets too involved. I think it is a great thing to explore, however.
The entire framing below took me about 3 minutes. You can then answer the questions in literally 2 minutes.
To accomplish this framing, do the following:
Show the 3 P's frame: UVW being in.
For the next frame, draw out 3 hypothetical sketches.
Show one with UW, one with UV, one with VW.
I also feel obligated to tell you that in all instances, I would approach the typical first hypothetical question by applying the rules 1 by 1 until one choice remains. This is always the most efficient approach.