#10 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=351&t=3899)

The correct answer choice is (B)

If no green bulbs are selected, then each bulb is either purple or yellow:

Because the question asks for how many different color sequences are possible, the best approach is to test light 1, first as purple, and then as yellow.

When light 1 is purple:

When light 1 is purple, according to the first rule light 2 must be yellow. But, when light 2 is yellow, from the contrapositive of the third rule, light 3 cannot be purple or yellow, and must be green. But, as this violates the condition in this question stem, this does not allow for a workable solution. Thus, there are no viable solutions when light 1 is purple.

When light 1 is yellow:

When light 1 is yellow, then light 2 can be purple or yellow:

When light 2 is yellow, from the contrapositive of the third rule, light 3 cannot be purple or yellow, and must be green. But, as this violates the condition in this question stem, this does not allow for a workable solution. Thus, there are no possible solutions when light 1 is yellow, and light 2 is yellow.

When light 2 is purple, then light three can be purple or yellow:

When light 3 is purple or yellow, no violations occur, meaning that there are two possible solutions when light 1 is yellow and light 2 is purple. Thus, two is the correct answer, and answer choice (B) is correct.

[theTRSlegend]

I had a little trouble with this question but reviewed and think I understand:

If we cannot use G, then we only have P & Y.

Due to Rule 1, if P is first then Y is second. However, Due to Rule 3, we cannot put Y second because P must maintain that spot (assuming no G). Thus the only option would be to put Y first only, or Y first and third, leading to YPP or YPY as the only two possible unique combos. Is this line of thinking correct?

Thanks!

[Robert Carroll]

TRS,

That's exactly the correct way of thinking. I got the same conclusion a little differently at the end, or maybe I just thought it using different words: 1 is definitely yellow, so 1 and 2 are completely fixed, and only 3 can be different (although not green), so two different total sequences. Thus, answer choice (B) is correct.

Robert Carroll