A is CORRECT. I’ll prove it must be true. Green must be before red. And there are always at least two reds. If there are three or more reds, we can’t put green fourth. Green would go third at earliest.
So let’s try with just two reds. We could put them in five and six, and put a green in four:
So far so good – but what color are the other balls? We can only have one white ball, because we need more red balls than white balls. So the other balls must be green. Here’s one example:
With so many greens, there’s no way to avoid putting them earlier than four!
My scenario above disproves B and C.
This scenario disproves D and E.
Remember, you can do anything as long as the rules don’t specifically ban it.
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Steph says
Just adding onto what I have posted, as long as one G is before R, I thought G could go anywhere in the stack. Is this correct?
Steph says
Hi, I have a question for the answer choice A. I have R-R-G-W-R-W (from 6 to 1) placing only one G on 4. Why can’t we have just one G and two Ws and three Rs as a result?
TutorRosalie (LSATHacks) says
In the sequence you cited (R-R-G-W-R-W), there isn’t a G below the lowest R. If you had only one G, then to be in accordance with Rule 2, it has to be below/before the first R. If you want a situation where you only have one G, then it would have to look like this: W-R-R-R-G-W (from 6 to 1; the W-R-R-R part doesn’t have to be in that order).
Yes, so long as there is a G before/below R, then the other G’s can go anywhere.