This is a linear game that requires us to determine the order of six obstacles - R S T V W Z.
___ ___ ___ ___ ___ ___ 1 2 3 4 5 6
The following rules apply: (1) S must be 3d or 4th.
___ ___ ___ ___ ___ ___ 1 2 3 4 5 6 /S /S
(2) W must be placed just before Z.
This rule tells us that we must have a WZ block in any valid scenario.
___ ___ ___ ___ ___ ___ 1 2 3 4 5 6 ~Z /S /S ~W
(3) R cannot be placed just before or after V.
This rule tells us we cannot have RV or VR combination.
Since rule (1) is fairly restrictive, let's consider both of these scenarios:
A. __ __ S __ __ __
Here we could place WZ first and second resulting in:
W Z S R/V T V/R scenario
or we could place WZ after S which would require T as well as either V or R to be first or second since V & R cannot be adjacent to each other.
V/R + T S V/R + WZ
Let's consider the second scenario:
B. ___ ___ __ S __ __
If we place WZ after S, V & R must be separated by T
V/R T R/V S WZ
If we place WZ before S, then again either V or R as well as T must be placed after S
WZ + V/R S T + V/R.
The question asks us which of the following a complete list of all positions in which tunnel could be placed. We see that every possible answer choice includes 1 and 2, so there is no reason to check that T could possibly take one of these spots. From our scenarios above we see that when S is #3, T must be either #5 or 1 or 2. And when S is #4, T could be either #2 or #5 or #6. Thus, we can conclude that T could be placed 1, 2, 5, or 6 (B).
Let me know if this makes sense or if you have any other questions.
