Numerical Distribution ID Drill - Seven animals

[freddythepup]

Hi,

Can you please explain more in depth number 4 in this drill set? I'm not sure how you can make an inference that there will be 4 cages? Thanks.

[Adam Tyson]

Hey Freddy, if you are talking about the drill with the 7 animals being placed into cages, you cannot infer that there are only 4 cages! Our answer key shows that as a possibility (a 3-2-1-1 or 2-2-2-1 distribution), but it also shows another possibility with only 3 cages (a 3-2-2 distribution, which obeys all the rules).

We can't do it in more than 4 cages, because we would have to break a rule there (more than 2 cages with just 1 animal in them), and we cannot do it in fewer than 3 cages because we would have to break at least one other rule (max 3 per cage). So, the distributions here only allow for either 3 or 4 cages, but we cannot be sure which it is. In a real game scenario, that would probably mean setting up three templates based around those three distributions!

I hope that clarifies it for you!

[ncg293]

Hi,

Can you please explain why there are max 3 animals per cage? I understood the rule "At most one of the cages contains three animals," to mean that no more than one cage can contain three animals. I used the same interpretation for the rule "At most two of the cages contain one animal each." With this interpretation, I identified 2-5 and 2-4-1 as possible numerical distributions in addition to those listed in the answer key. I would understand the max 3 interpretation if the rule stated something along the lines of "A cage can contain at most three animals." I'd greatly appreciate any clarification. Thanks.

[Robert Carroll]

ncg,

The first part of the drill states: "Seven Animals are placed into cages containing anywhere from one to three animals..."

I think that should clarify. If not, let us know!

Robert Carroll