Hi, this is a game where the setup is almost the entire game. We have four spots in (ordered) and two which must be out
_ _ _ _ (_ _)
We also know that since both of our females are in, both our out group are males. We also know that since one lab is in and one is out, we have the breakdown of who is in and out. In: 2 f, 2 M, 1L, 3 G Out 2M 1 g 1 L Since we know that S is in, and S is a female lab, we know that U is out, and U is a male lab. So, we have a rough order to the six
1: P/R (greyhound, sex unknown) 2 P/r (G, sex unknown) 3: S (L, F) 4: Q/T (G, sex unknown)
Outs: Q/T (GM), U(LM)
With this understanding, we know that all of the answer choices here except for A must be true.
Jalvarezon October 16 at 11:58PM
How are we sure S is in? Why can't U be in? Also how do we know the second out one is a greyhound how come it can't be a labrador as well?
Jalvarezon October 17 at 12:02AM
Is it because S places ahead of Q and T, and since Q and T can't both be out because then two labradors would be in? I am still stumped on why can't there be a third labrador out (One in and two out)?
Emil-Kunkinon October 26 at 03:24AM
We know that there always must be two that come after S, so the lowest position S can be in is fourth. So, it must always be in, you're spot on in that respect! U can never be in because S, a lab must always be in, and we're only allowed to have one lab in.
I'm actually pretty sure that the last out could be a lab, I think I misremembered the 2-4 rule for the dogs sex and applied it to the breed wrongly. So, I do think we could have two labs out.
