This is an explanation of the fouth logic game from Section IV of LSAT 34, the June 2001 LSAT.

There are six doctors, namely: Juarez, Kudrow, Longtree, Nance, Onawa, and Palermo (J, K, L, N, O, P). Each of them must be in one of two clinics: Souderton or Randsborough (S, R). You must decide which clinic the doctors are in.

### Game Setup

This is a grouping game. Every doctor is at one of two clinics. This means you can treat the game as an in-out grouping game, like the birds in the forest game from test 33. “You’re either in or out” is the same as “you’re either at S or R”.

We’re going to draw all the rules, then recombine them into two chains of sufficient and necessary conditions. This will let us figure out a *lot*.

The first two rules can be easily confused if you don’t read them carefully. LSAC** **placed the sufficient condition second. The first rule would mean the same thing if it read:

“If Juarez is at Souderton, Kudrow is at Randsborough.”

Always focus on words such as “if” “only if” etc. Word order doesn’t always change much. Like Yoda be.

Here are the rules drawn out. The subscripts indicate the clinics:

When you’ve finished drawing the rules on this sort of game, look back over the game’s written rules to make sure you haven’t made any mistakes. I *always *do. Any mistake is a disaster, especially when you’re combining rules.

I’ve also drawn the contrapositives below. Remember that you have to change “and” to “or.” For example, since L being at S requires *both *N and P to be at R then *either *one of them being at S is sufficient to show that L must be at R.

In a two group game, the negation of L being at S is L being at R. There’s no need to put a line through the variables. There are only two groups, so if you aren’t at one then you must be at the other.

Now it’s time for a game of mix and match. If the same variable is both a necessary and a sufficient condition for two separate statements, you can join the two. It’s like lego!

For example, Or leads to Js. And Js leads to Kr. So we get:

For in-out or two group games, it’s often the case you can connect *everything*. So let’s build on that and see if we can fit all of the variables.

For example, Nr leads to Or.

Next, Kr leads to Ps. And Ps leads to Lr. Done!

(Note: we also know Or leads directly to Ps. You can draw that separately if you want, but it’s not necessary since Or leads to Ps indirectly through Kr.)

Now we can take the contrapositive. Be very careful when you do this, you don’t want to misinterpret your AND/OR statements. I always reread my rules once I’ve drawn a contrapositive chain, to make sure I didn’t make a mistake.

### L actually can’t be in S

I start with Ls leads to Pr. Then Pr leads to Ks. Ks leads to Jr.

Next, Jr leads to Os, and Os leads to Ns.

(I took a shortcut. Pr leads to both Ks and Os. I didn’t draw Pr ➞ Os, because Ks leads to Os.)

Lastly, Ns leads to Lr.

(The diagram is smaller only because it’s very long and too big for the page)

Huh? Yes, you read that diagram right. Ls leads to Lr.

Obviously, that doesn’t work, L can’t be in two places at once. A hidden deduction is that L always *has *to be at R.

This is very weird, since they gave us a rule that tells us what happens if L is at S. But it’s true: L can’t ever be at S.

If you’re not convinced, here’s a diagram of what happens when we take the Ls rule and combine it with other rules:

Clearly, we can’t have Ps and Pr at the same time. L has to be in R.

And we’re done! Combining rules really is just like lego. Or dominoes.

We can set up the deduction about L in a table.

There’s a bit more we can add. Consider these rules:

If J is at S, K is at R, and vice versa. J and K can’t both be at S; one of them has to be at R.

Both *could *be in R. This may look confusing. Try putting J in R.

Jr is the *necessary condition* in the rule above. Knowing that J is in R can’t tell us *anything* about K. You can only read the diagrams left to right, not backwards.

This is the most confusing relationship on in-out grouping games, so make sure you understand it.*

O and J have a similar relationship. One has to be in S. You can add both relationships to the table.

* Here’s an example of the “at least one” relationship. If you’re a parent, you have a boy or a girl. You could also have both. So we no that if someone is a parent, and:

no boy ➞ girl

no girl ➞ boy

Parents always have at least one, and maybe both.

Whenever the sufficient condition is negated, and the necessary condition is positive, you must have at least one, and *could have both.*

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Why are we showing J/K in the R column if we can have both? I see that you wrote “You can only read the diagrams left to right, not backwards.” Can you elaborate?

Because we need at least one of J/K. We could have both, but we *need* to have one. The columns only show what is needed, at minimum.