LSAT and Law School Admissions Forum

Get expert LSAT preparation and law school admissions advice from PowerScore Test Preparation.

 josuecarolina
  • Posts: 24
  • Joined: Jul 20, 2012
|
#5719
online course, p 9-46 #4

Seven animals placed in cages. I only got one group 3-2-1-1

I don't understand how the other groups were found.

Thanks in advance!
User avatar
 Dave Killoran
PowerScore Staff
  • PowerScore Staff
  • Posts: 4105
  • Joined: Mar 25, 2011
|
#5722
Hey Josue,

Good to hear from you. That one distribution you found is definitely correct, so that's a great start! But, as you noted, there are other distributions, so let's go over how we found those.

The first thing to note is that there is no specified number of cages. That usually means there will be multiple solutions. The second thing is that all the rules do not specify an exact number of animals, but leave it slightly open:
  • 1. At most one of the cages contains three animals.

    2. At least one of the cages contains exactly two animals.

    3. At most two of the cages contain exactly one animal.
I added the italics because they show that the following can also occur:
  • 1. You don't have to have a cage with three animals.

    2. You can have more than one cage with two animals,

    3. You can have zero or one cage with just one animal.
Because of all these options, more distributions exist than just the one you found:
  • 3-2-2

    In this distribution, all seven animals are accounted for and each of the rules is still met. Here we first maxed out a cage with three animals, then made sure we had a cage with two animals. That leaves two animals. One option is to split them into cages with just one animal each (creating a 3-2-1-1, which you found) or to put them both into one cage, which produces this 3-2-2 distribution.

    2-2-2-1

    Here we avoid have a cage with three animals, and create a cage with two animals (to meet the second rule). That leaves five animals; with the remaining rule, the only way to split them up is 2-2-1, which when combined with our first cage creates a 2-2-2-1 distribution.
Please let me know if that helps. Thanks!
 josuecarolina
  • Posts: 24
  • Joined: Jul 20, 2012
|
#5735
It does help (Please tell me that a lot of people have toruble with this one... :cry: )

However, how does:
3. At most two of the cages contain exactly one animal.

translate to:

3. You can have zero or one cage with just one animal.
?

Also, doesn't the second rule mean that at least one cage has to have 2 animals? you said it translates to: 2. You can have more than one cage with two animals,

sorry if it's obvious. I ripped through the rest of the examples, but this one has me stumped.
User avatar
 Dave Killoran
PowerScore Staff
  • PowerScore Staff
  • Posts: 4105
  • Joined: Mar 25, 2011
|
#5737
Sure, let me clarify. In my second part, I'm listing the other possibilities (aside from the one clearly contained in the rule). So, when it says you "At most two of the cages contain exactly one animal," I figure that having two cages with one animal is clear, but that it is also possible that you can have zero or one cage with just one animal. See, from the distribution you found, it appeared that you just took each rule at exact value: one cage with 3, one with 2, and two with 1 (3-2-1-1). But, my point was that there are other possibilities besides those. Does that help?

It is a tough one, because of the undefined number of cages--that makes it tricky because you have to play around more with the numbers.

Thanks!
 Law1
  • Posts: 4
  • Joined: Oct 29, 2019
|
#71574
"Seven Animals are placed into cages containing anywhere from one to three animals..."

Two part question: 3-2-2 can be a distribution being that it is not explicitly stated that there are 4 cages, thus - there could be 3 or 4 cages to place the animals within, correct?

If so (or not), how can I consistently work on getting into the habit of identifying this subtly consistently as I practice?

Best
 Rachael Wilkenfeld
PowerScore Staff
  • PowerScore Staff
  • Posts: 488
  • Joined: Dec 15, 2011
|
#71592
Hi Law1,

You are absolutely correct that 3-2-2 is a solution to this drill. We have an unknown number of cages here, so we are limited by the distribution restrictions, but not by any other cage maximums or minimums. We need a total of 7 animals, in cages of 1-3 animals. There is at most one cage with three animals, at least one with two animals, and at most two of one animal. That's consistent with a 3-2-2 distribution.

I think you are right on target. Don't read in rules that aren't stated, and stick to the information in scenario/rules.

Keep up the great work!
Rachael

Get the most out of your LSAT Prep Plus subscription.

Analyze and track your performance with our Testing and Analytics Package.