I'm not following: how does this equate to finding a set of boxes that have 50 balls total?

If there is a set of boxes with total of $50$ balls, it should be a partition of $50$.

How do we know that we get a partition of 50 though?

We have $51$ non-empty boxes. So from $1+1+...+1$ to $50$, we can place the balls in a way that one of these partitions occur.

I am still not seeing why this would give you a set of boxes with exactly 50 balls in it though. What set of boxes would that be? You put 50 balls in a set of boxes, but what about the remaining 50 balls, wouldn't where you put them matter?

First, we are trying to put balls to boxes in a way that we can get each partition of $50$. But we don't know whether there might be a placement that does not have any partition of $50$ in it. We can clearly get each partition of $50$ in an algorithmic way. First put $50$ balls to first box and $1$ balls to each other. then we can take balls from the first box and put them to other boxes. Then we can get $49+1,48+2,48+1+1$, etc. Then, we calculate total number of placements, which is $p(49)$. Then a partition of $50$ must occur one of these placements. Take that partition as a set, we're done.

OK, 50 balls in one box and 1 in the other will certainly give you the desired subset--i.e., a subset of boxes that have exactly 50 balls.. But, it is not clear to me that you will still have a desired subset--i.e., a subset of boxes that have exactly 50 balls--after move balls from the box with 50, to the other boxes.

I don't know if you gave the downvote but I will edit my answer. I can also prove we can obtain every partition of $50$ with Pigeonhole Principle. I also think that my last comment was not convincing enough.

@Mike I edited my answer. I hope it is clear now.

Hi @ArsenBerk I do see the edits and I did go through it. But I am still not satisfied with the answer. I had--and still have--issues with the last two paragraphs. The counting arguments seem ambiguous. Are the boxes labelled or unlabelled? If they are labelled then I am not seeing how you can exclude 2 boxes. If they are unlabelled then I am not seeing the connection between the number of partitions of 50 versus the number of ways of putting 100 balls into 51 boxes

Can we continue from here?

Hi Arsen, sure

okay, thank you very much

so the boxes are identical

and I think it is clear that at least 2 boxes have 1 ball in it

OK, but then why does the fact that p(50>p(49) imply that every arrangement of 100 balls into 51 boxes have a partitioning of 50?

I am still not able to see that

Yes it is clear that at least 2 boxes have 1 ball

In my edit, I showed that we can have every partition of $50$ for some placement

was that part clear?

Indeed yes, any partition of 50 (into k boxes) can be extended to an arrangement of 100 balls into the remaining 51-k boxes. That I got.

yes that's right

and 50 has p(50) partitions

Sure

I think you also got why we have p(49) partitions in total

or was the question about excluding 2 boxes for this part

OK but you take two distinct partitions of 50 and extend them, you may not get distinct arrangements of 100 balls into 51 (identical boxes)

I extended all the partitions actually

If the boxes are identical you can exclude two boxes

because edit was for a general k

so if we have a partition with 1 to 50 summands, it holds

Here is the issue I have: You have p(50) partitions. You extend each of these p(50) partitions into an arrangement of 100 balls into 51 boxes. You may not get p(50) distinct arrangements of 100 balls into 51 boxes though.

I am not convinced you would get p(49) distinct arrangements

*I am not even convinced

okay, first I will explain the total number of placements p(49)

No need that I get

You may not get p(50) distinct arrangements of 100 balls into 51 boxes though.

this is a good point actually and indeed they can't all be distinct since we already know that we have p(49) placements in total

So, why given an arrangement of 100 balls into 51 boxes, is there guarantted to be a partition of 50 in there somewhere?

hmm now I see

I think the problem here is I am overcounting

because when there is an extended placement for a partition of 50, there should be another

like if we have k = 10 for instance, we also have a partition of k = 41

Yes I think so

So I should compare the number without overcounting with p(49) in this case

OR something. But there is more work that needs to be donw

and finding amount of overcounting is pretty hard..

because there are more than one partitioning for 2 <= k <= 49

and I need to know how many partitionings are there for some k

okay then I will delete the answer for now

If I can come up with something to fix it, I can edit and undelete it

That is good. We all have had to pull back answers every now and then. There are plenty of other interesting questions on the forum to work on in the meanwhile...

yes, that's right :)

so I thank you for your time and your feedback. It was a nice one and others seem not to notice my mistake (I couldn't either until you said so).