Linear Setup Practice Drill #11

[username1991]

N.11 on p.3-36
if we only know that:
Z_T
Y3 ->>>S8 (and contrapositive if not S8 then not Y3)
3rd is Y/Z
5th is T/V

Here is my reasoning:
1) to establish a clearer picture of the diagram we have the following:
1.1 if Y 3rd then a)Z is not 3rd, b) S is 8th which creates the picture:

___ ___ _Y_ ___ ___ ___ ___ _S_
_1___2____3___ 4____5____6___7__8__
wouldn't that be all that we could possibly know based on that?
the author suggests that we would know the following(based on the same information):

_________Y___Z__V___T
___ _Z_ _Y_ _T_ _V_ ___ ___ _S_
_1__2___3__ 4__5__6___7__8__

I am trying to figure out if V/T was supposed to be placed on 5th position how do we know that Z for example becomes second?
Thanks

[Ron Gore]

Hi 91!

Thanks for your question. I'm going to restate what I understand to be your question, so please let me know if I misunderstood.

You're asking about the scenario in which we're told that Y is 3rd. In that scenario, you ask how we know where Z would go. The answer is that we don't know exactly where Z goes, but we have it down to an either/or option.

So, if Y is 3, then we know from Rule 4 that S is 8. And, we know from the combination of Rules 3 and 5 that Z and T must be in this relationship to each other: Z _ T. As you point out, since Y is 3, then Z cannot be. Since Z cannot be in 3, then T cannot be in 5, which means that V must be in 5 (Rule 1). That gets us to this situation:

___ ___ Y ___ V ___ ___ S

Consider all the places that V cannot go:

can't go in 1 because Y is in 3;
can't go in 3 because Y is there;
can't go in 5 because V is there;
can't go in 6 because S is in 8;
can't go in 7 because there must be two spaces left after Z is placed for _ T;
can't go in 8 for the same reason, not to mention that S is there.

So, the only two spaces left for Z are 2 and 4. We don't know which one Z will actually occupy, but it must be one of the two.

Let me know if that helps.

Ron