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!
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