Hi Tim,
Thanks for the question! You are partially right but not completely because there are more distributions, namely these three: 3-2-2, 3-2-1-1, 2-2-2-1. Let's go through them:
- We start with: Seven boxes distributed into an unknown number of rooms, or, written out: 7
?
The reason this feels different is because it's rare to have one of the sets be "unknown." It's never specified as to exactly how many rooms there are, so that makes this much harder and difficult to be systematic in the normal manner we typically can use. It's also why it's the last one in the set—we often save the hardest for last!
So, how do we do break this one down? First, start with the known rules:
"At most one of the rooms contains three boxes." — Okay, so at most we can have a single 3. We don't have to have it, but we can have it if needed. This also means a bunch of possible distributions are eliminated immediately, such as 4-3, 4-2-1, 4-1-1-1, 5-2, and 5-1-1.
"At least one of the rooms contains exactly two boxes." — Good, you have to love a minimum when they give it to you in this situation! Now we have a minimum of at least one 2. This sets us up as:
- 7 ?
But one room has at least 2, meaning our distribution is:
2-??
and we have 5 remaining boxes to fill in the ?? or to add to the 2.
"At most two of the rooms contain exactly one box." — This means that a 1-1 arrangement is our max for singles. Thus, 1-1-1 is impossible, and any extended version of that such as 1-1-1-1, etc. This kills off distributions like 2-1-1-1-1-1, and it tells us that we're not going to have a bunch of rooms here; it's going to be limited since we can have only a max of two 1s.
With those rules in mind, how many variations do we have? This is harder since we can't just meet the minimums and then distribute the extras, like usual. Instead, let's look at each rule for it's max or min, depending on what is specified:
First rule: max 3
Second rule: min 2
Third rule: max 1-1
This actually works out as a full combination and is 7 boxes total. Thus, this is one of the distributions: 3-2-1-1.
Of course, we don't have to have two 1s at the end, and could combine those, making for a 3-3-2.
And, returning to the first distribution, we don't have to have a room with 3, so let's take one away there, giving us 2-2-1-1. We still have an extra box, but we can't have another room with a 1 as that would violate the third rule, so let's instead add that extra box to one of the 1, resulting in: 2-2-2-1.
All of these distributions are Unfixed since they aren't tied to any specific room or room number.
By the way, the answer key for this is on page 9-27.
Please let me know if this helps. Thanks!
