Ok, so I just finished it. It didn't take me long, but I see how this could have give you some issues. The key here is to recognize that the first rule seriously limits the possibilities of this game. So I'll give you some of my thoughts as I worked through this:
STEP ONE: Setup the Game Board, Make Inferences
Three variables (M, S, T), we're assigning them to three different days. Hey this seems really easy. Oh, wait, there's three sessions (H, I, R) per day. So at this moment, I'm thinking that we have three variables going into nine slots, at three slots per day. The big question I have at this point is whether we can have multiple variables (e.g. MS) in one slot.
Now let's look at the rules. Sweet, we actually have six variables, two of each of M, S, and T. There are some restrictions on M and S for one of these sessions. A restriction on T for one day. And there it is! Two, but not more than two variables can be assigned to any one session. So now I'm going to set up the game board and make some inferences:
A couple things to point out here. First off, you'll notice that I have split some sessions into two positions to reflect the rule that we can have two people in one session. But I didn't do that for session I, because we can't have M or S there ever; so if someone is assigned to I, it must be T. Next, you'll notice that I diagrammed the first rule with X's, but that's just because it's a random variable that I use to reflect rules that say things can't be in the same horizontal group, or vertical group.
As far as inferences go, one thing you might have noticed is that there must be a T on day 2 and one on day 1. To see why this must be true, do a quick hypo where you assume T can't be on day 2----because you already have a rule that says T can't be on day three, under this scenario, both T variables would be forced into day one, which would violate the first rule.
Another inference is that there's at least two variables by itself in a session. You could infer this by doing a hypothetical with T in I2, which would result in three variables in H or R, and since you can have two at the most in one position, one of M, S, and T, would be floating. On the other hand, what if there's no T in group I? Then you would have three variables assigned to H, and three variables assigned to R, here you would have two variables alone (assuming two variables are together in both H and R), and possibly six alone (assuming no variables go to the same session as each other on the same day) in a space.
Side note: I never diagrammed this the same way as my initial diagram. I just remembered that sessions in group H and R could have two, but not more than two variables in the same session. So I dropped the session divider to make things faster. Still conceptually, it was pretty easy to look at my master diagram for the global questions and think through it.
Now let's go to the questions:
Question Six: Max Sessions
Let's max out the sessions by not combining any variables into a session on the same day. So my diagram looks like this:
H-----M, /, S
I------/, T, /
R-----T, S, M
I count six sessions. There's no way we could have seven, because we only have six variables. Six is the next highest number, so we've solved this one.
Question Seven: Must be False
We don't have much to go on here, so let's see how far our inferences take us. The first answer is easy, we could split M into day one and day two in groups H and R, no problem. The second answer has the same effect as the first, so all good so far. The third answer is also easy to eliminate, we can make that work. The fourth answer works, we could have T in group I in position one, S in group H in position one, MT in group H in position two, and SM in group R in position three. So that leaves us with answer choice (E), let's see what it does:
Yup, no way we can do that. Three sessions on day three? That would require T to be on day three, and that won't work.
Question Eight: Session Two, Day Three
Well, if two sessions on day three are occupied, then they must be filled by M and S in groups H and R, because T can't be on day three, and we can't have the same variable in the same vertical group. So where does that leave us? Well we have taken away one M and one S, we don't know which group either is in, but we know both are used on day three. That leaves us with two T variables, and one M and one S variable. We know that there's a T on day one and on day two, we have no idea which group though. We know that our remaining M and our remaining S are floating, they could both have T (See the hypothetical below), but they can't be with each other because of the condition in the question stem.
H-----MT, /, S
I------/, /, /
R-----/, ST, M
We're looking for what must necessarily be true, that makes every answer containing the word "exactly," easy to eliminate, and we know that the last two answers aren't necessarily true.
Question Nine: Session Two, Day Three
Another question that doesn't really give us much to work with. Let's skim and see what jumps out. We can make (A) work, we just split the T variables and throw M and S around. There's no way that the second answer works!----we inferred that at the beginning.
Question Ten: Day One, all Sessions
So here we have something like this: Note that M and S are interchangeable, you could have written M/S instead. The things in parenthesis are floating variables within the groups they're next to.
H-----S, ?, ? (M, maybe T)
I------T, /, /
R-----M, ?, ? (S, maybe T)
We know that there is a T to be assigned to day 2 in group H or group R. We also know that we have one S and one M floating just about anywhere in groups H and R, but we know they can't be together. And that's the first answer! Nice.
Question Eleven: Day One, MT
So here we have MT together on day one, and nobody else is on day one. So we're left with something like this:
H-----MT, ?, ? (S)
I------/, ?, / (maybe T)
R-----/, ?, ? (S, M, maybe T)
All that we really know is that we have things floating. Right off the bat, answer choice (A) should stick out because M is restricted in this question. There's no way M could be with S twice because nobody except MT can be on day one.