Pigeonhole Principle Problem - Circle I'm working on a homework problem with the following instruction:
Suppose there is a class of 2n students with different names sat in a large show and tell circle. The teacher returns their homework, but each student in the circle is given another student's homework. Therefore, the class agrees to pass the homework to the student sitting to their right. Once again, the names are all wrong. For the first n times they do this, they all have the wrong named homework. So in all there are n + 1 configurations in which every student is matched with someone else's homework). Show that if they continue passing the homeworks to their right, they will eventually reach a configuration where 3 students have the homework with their own names.
I've done a few problems with the pigeonhole principle, but this one is giving me exceptionally more trouble. I've illustrated two separate cases, both where n = 2, which both result in a class size of 4. The uppercase letters are the children and the lowercase letters are the corresponding homeworks.
Test with class of 4
With the above starting configuration, I could not get it to be where three students concurrently have the correct homework, only two at once. I also didn't break any of the apparent contraints:

*

*Each student does not recieve their homework initally

*Each student does not have their homework after 1 round of passing

The second case in the drawings is to exemplify what would happen if the student to one's left had their homework, i.e. necessitating only one pass. However, I'm fairly certain the question says this is not allowed since

Therefore, the class agrees to pass the homework to the student sitting to their right. Once again, the names are all wrong.

Meaning that the round after the first pass must also ensure no one has their homework. I trust the question is correct in what it is proposing, but I am having trouble understanding it and applying the Pigeonhole principle. Are the "holes" the kids and the "pigeons" he homework and if so, how would one calculate in the case where it's not as simple as a pigeon in the hole, but the correct pigeon in the correct hole? Any help would be appreciated.
 A: If we perform a full $2n - 1$ passes, then everyone has seen their homework exactly once. We know that the first $n$ passes, the first $n + 1$ configurations, no one has seen their homework. If we continue on into the remaining $n - 1$ configurations, and at most 2 students see their homework in every configuration, at most how many students have seen their homework? How does this contradict what we've already said?
A: Here's how you can approach it. Fill in the gaps yourself.
Generally when applying PP, it's very important to be clear on what the holes and pigeons are. Defining them often a great start, so:
Let the holes be "Configurations where the students are given homework rotated $k$ times".
Let the pigeons be "Student has the correct homework".
Assuming that these definitions are the correct ones to use, let's pursue how PP could work:
How many pigeons are there?
What is the maximum number of holes we can have to ensure that there is some hole with 3 pigeons?
Can we show that we have at most this many holes? If no, how can we tweak the holes (or pigeons)?

There is some uncertainty about what the exact number of holes are.

*

*If the number of holes is $n-1$ (meaning that there are $n+1$ configurations which are compeltely mismatched), then PP works directly.

*If the number of holes is $n$ (meaning that there are $n$ configurations which are compeltely mismatched), then the question is wrong. In particular, the statement is true for even $n$ and false for odd $n$. This can be proved by 1) Getting a contradiction for "exactly 2 papers match for $n$ is even", and 2) Constructing a case for "exactly 2 papers match for $n$ is odd".

