Q22

[Virg.elizabeth.gordon]

How can a red ball be in both boxes 2 and 3? Wouldn't a WG block need to go before any red ball, thus making it impossible for a red ball to be in position 2?

[ohthatpatrick]

No, the WG block can be after (higher) than the reds in 2 and 3 without breaking the rule.

The rule is that "there is a green ball that is lower than all the reds". That rule is NOT saying that "all greens are lower than all reds".

With reds in 2 and 3, we HAVE to put a G in spot 1, so that we can comply with the rule that there is a green lower than all reds.

Then we have put the WG chunk somewhere in the last three spots.

So we could have something like
G R R W G _
or
G R R _ W G

The missing box can NOT be another White, or else there would be 2 W's and 2 R's, breaking rule 1.

It could be another G or another R.

[CharlesS800]

I began this problem by diagramming the new conditions of this game:

6 _
5 _
4 _
3 R
2 R
1 _

I realized then that to heed the rules of the game, in particular the second rule, there must be a G in slot one, so I was left with:

6 _
5 _
4 _
3 R
2 R
1 G

To comply with the third rule, I determined that there must be a W/G chunk in slots 4, 5, or 6. Because this only occupied two slots, there was a still a bit of fungibility in my diagram but I figured it would be enough to at least work through most of the game.

With my (incomplete) diagram, I was able to eliminate answer choices A and B immediately. I glanced over answer choice C and it seemed to work so I jumped down to D and E.

Answer choice D was also out of the question as placing a red ball in box five would mean that the W/G chunk couldn't be placed.

Answer choice E would also not work as even if you placed a white ball in box six, you would have to have another white ball in box four and a green ball in box five to fulfill the third rule.

This elimination left me with C as the correct answer.