Great question. This is definitely a tough one, and it helps if you have the four scenarios of ball distribution down first.
We know there are more red balls than white ones. This is huge, as we also know that there is at least 1 ball of each color. This means there are four possibilities of ball color distribution
1) 4 red, 1 white, and 1 green (with the white immediately below the green to satisfy the final rule)
2) 3 red, 2 green, and 1 white
3) 2 red, 3 green, and 1 white
4) 3 red, 2 white, and 1 green
One way to answer this question is to simply try the answer choices out, starting with (A).
The way you check these answer choices is to see if you can make a scenario where the box stated in the answer choice is the only box with a given color. If that scenario works, then you know that that answer choice is out.
Looking at (A), we have box 2. If we put green in box 2, white in boxes 1 and 3, and red in 4, 5, and 6, then we're good to go, so (A) is out.
With (B), if we put green in 3, white in 1 and 2, and red in 4, 5, and 6, we're also good.
With (C), we could have green in 1, red in 2 and 3, and then white in 4, followed by green in 5 and 6, so (C) is out.
With (D), we could put green in 1, red in 2 and 3, green in 4, white in 5, and green in 6, and we'd be good, so (D) is out.
By process of elimination, we're left with (E), which is correct.
Now, there is another way to solve this problem more quickly, but it requires your ability to perceive why box 6 can't be a solo color.
Why can't 6 be solitary? It's because whatever the solitary thing is, it's going to be green or white because there are at least 2 reds in every scenario. It can't be green because there is a green that's lower than any of the reds, and it can't be white because there is a white that's lower than at least one green. Thus, we know for sure that box 6 can't be solitary. This is the fastest way to solve the problem, but it requires having a really strong grasp of how the rules work together.
Does this make sense? Let us know if you have any other questions!
