PT31.S1.Q01 (G1) - four boys

Rigid Designator

Hey, hope this is the right place for this to go. I had a search of the forum and didn't see this question discussed before, so thought I'd throw out my 2 cents.

Doing some drills before this week's test I noticed a much faster way of getting to the right answer choice on Question 1 of Game 1 in this PT, compared to the video explanation. In the video JY skips Q1 because by the end of G1 you have more points of reference with which to brute force it. But I think it's perfectly do-able with just the rules, and I think it's a fast inference.

I set the game up the same as JY, except instead of representing his rule 1 with two crossed out and stacked boxes of BB (boy, boy) and GG (girl, girl), I just used the notation 1+ ---> BG. I think this notation helped me more quickly spot the inference I'm about to explain here.

So from the initial setup and from the first indented prompt we know that the game is going to have 3 lockers with one person in, and 2 lockers with two people in.

Combine this rule with JY's (or my) rule 1 and we then know that there is going to be two lockers with BG in there.

For the next two rules (rules 2 and 3) I used the exact same notation as JY.

As JY explains, from rule 2 and 3 we know J will have to share with N or T, since R must be alone. This allows for a further, vey simple inference which I didn't see JY make with respect to Q1.

We know we have two shared lockers with 1B and 1G in there, and we know that one of those two shared lockers is J and N/T. Since we know that the other shared locker has to have 1B and 1G in it, we then know that the other shared locker is going to have a girl in it. But there's only three girls. And one of them is going with J, and the other is always on her own (R). So the girl that goes in the other shared locker is just the one left over from our choice in the J, N/T locker. (T/N)

Thus, just from this basic inference from the setup and rules 2 and 3 we know at the very least that in the two shared lockers we will have J and N/T and, in the other, T/N. This allows us to completely solve question 1.

All we need to do is look for an answer choice that has J, N and T in it. Since only one answer choice has all these three in it, we know E is right straight away. If there was another answer choice that was, say, J N T and F, then we would have to check to see if F needs to share. But luckily there isn't, so you can answer it right away.

Hope this makes sense!
https://7sage.com/lsat_explanations/lsat-31-section-1-game-1/