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Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13302)

The correct answer choice is (C)

he question stem establishes that there are red balls in boxes 2 and 3:

PT63_Game_#4_#22_diagram 1.png
In order to conform with the second rule, a green ball must then be in box 1, and a GW vertical block must exist to satisfy the third rule:

PT63_Game_#4_#22_diagram 2.png
The GW vertical block can be placed in boxes 5-4 or 6-5, leaving two basic templates in this question.

PT63_Game_#4_#22_diagram 3.png
Because there are at least 2 green balls and at least two red balls, the distribution in this question is either 3-2-1 or 2-3-1. Thus, the remaining ball can be either green or red, and in both solutions there is no restriction on color based on position. Thus, a G/R dual-option can be placed in the remaining empty spaces:

PT63_Game_#4_#22_diagram 4.png
As is apparent in template #2, box 4 could contain a green ball. Accordingly, answer choice (C) could be true and is correct.
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Hi Team -

Why we can't place White Ball in position one as long as we place Red Ball in Position 6, Green Ball in Position 5 and White Ball in Position 4 - we can still meet Rule #2 and #3?

Many thanks!
 Luke Haqq
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Hi christinecwt!

Happy to address this question.

The hypothetical you suggest would actually violate the second rule: "There is a box containing a green ball that is lower in the stack than any box that contains a red ball." Your suggestion was:

6: R
5: G
4: W
3: R
2: R
1: W
That setup of the variables doesn't meet the requirement that there be a green ball below "any box that contains a red ball." Rather, there are red balls in boxes 2 and 3, with no green ball below them.

In fact, the second rule allows us to infer a global rule from the outset. Namely, there can't be a red ball in the first box. If there were a red ball in the first box, this would prevent a green ball from being "lower in the stack than any box that contains a red ball."
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Thanks a lot, Luke :)

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