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I'm not a native speaker of English. I usually pride myself of my proficiency, but I think I may be stumped here. My problem arises out of this question, which among other things asked for a combinatorial proof of the identity

$$\sum_{k = 0}^{n} \binom{x+k}{k} = \binom{x+n+1}{n}$$

(Equation shown only to provide context. I know how to prove this. I've written an answer. That's not the problem.) Another answer, however, provided a hint that reads in its entirety:

Hint. If I have a group of J objects, and I want to see how many ways I can count K of them, I can either do this by taking $J \choose K$ or by counting how many ways I can choose 1, or 2, or 3, or 4, ..., or K of them.

This hint received upvotes, which I consider empirical proof that it is not simply nonsense. My problem is that the hint makes no sense to me. I find this particularly distressing because its author claims that the solution it was a hint for is the same solution I gave in my answer -- but still I can't figure out what the hint even means.

What confuses me is the phrasing "how many ways I can count K of them". This conveys no meaning to me -- what does it mean to "count K of them"? What does a way to count K of them constitute, such that we can speak of how many such ways there are?

I'm fairly certain that "count _ of _" is not a standard mathematical concept with a technical meaning, so I'm assuming that it is an everyday English phrasing whose ordinary meaning I'm simply unfamiliar with. I'm asking because I suspect it may hint at a combinatorial shortcut that I don't know, which I'd like to add to my toolbox.

Yes, I did ask directly as in a comment to the answerer, but apparently I was unable to convince the answerer that I was asking for mere linguistic clarification rather than demanding additional technical details in the hint, and the discussion got a bit heated, and perhaps some fresh eyes will be better able to find a way to express whichever obvious thing I'm missing in a way that will penetrate my skull.

I think I have eliminated the possibility that the hint merely says that $\binom JK$ is one way to count K of them and there's another way involving "1, or 2, or 3, or 4, ... or K" plus possibly other different but irrelevant ways. I have also eliminated a theory that "count K of them" was a typo for "choose K of them". Beyond that I'm drawing blanks.

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hmakholm left over Monica
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    "I have a group of J objects, and I want to see how many ways I can count K of them" — I guess it's just a typo for how many ways I can choose K of them. – ShreevatsaR Sep 24 '11 at 04:45
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    'I have also eliminated a theory that "count K of them" was a typo for "choose K of them". — how have you eliminated this theory? – ShreevatsaR Sep 24 '11 at 04:47
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    On looking again, I must say I do find the linked answer incomprehensible too. – ShreevatsaR Sep 24 '11 at 04:50
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    @ShreevatsaR: One of the few explicit clarifications I could get out of the answerer was that he did mean count not choose. It's a high-rep user who I don't think would normally have a problem with saying "oops, that's a typo" or "yeah, that was confusingly worded" if that were actually the case. So I'm assuming there's actually something to it and the problem is at my end. – hmakholm left over Monica Sep 24 '11 at 05:02
  • @Henning: I am very sorry to have caused you to go so far, but I am absolutely stunned that this is a question here. It was a non-explicit hint referring to the idea of counting a number of objects in one go or multiple goes. Again, I use this ambiguous word 'count' - and it is ambiguous. There are many ways to count things. I wrote that answer that way because I do not ever answer homework questions explicitly. And this, in particular, was a hint given to me when I first encountered this question as a freshmen undergrad... – davidlowryduda Sep 24 '11 at 05:16
  • ... But I truly do not intend to make this a big deal, and I think this is not the intention of MSE. So I have removed the offending answer. For what it's worth, Ross's interpretation is exactly right. – davidlowryduda Sep 24 '11 at 05:16
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    @mixedmath I had a question about the meaning of "count X of Y". I failed to elicit any answer from you, so I decided to ask someone else if they could help me rather than continue to beat a dead horse. How can that stun you? Is it any different from the run-of-the-mill "here is a line from a textbook that I don't understand (and the author does not answer my emails / is dead / speaks only French)" type questions here? – hmakholm left over Monica Sep 24 '11 at 05:28
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    Would the people who have voted to close speak up, please? I vote against closing. – J. M. ain't a mathematician Sep 24 '11 at 05:28
  • @Henning: Ok, I have a couple of things to say. Firstly, we sent over 30 comments back and forth to each other, so I have a hard time believing that one could say that I was unresponsive. Secondly, I repeatedly said that I meant no particular method of counting. In particular, I told you that I did not think that my answer, as written, would indicate a particular method of counting. And I defended this as it is an answer to a homework question. Thirdly, the period of time from your last comment to the creation of this was 45 minutes, during which I went to eat... – davidlowryduda Sep 24 '11 at 05:42
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    Under no circumstance would I ever expect an author to answer an email within 45 minutes. So when I read this question, I read it as a premature extension of a conversation between us. And let us suppose there is an answer: I would also suppose that I, as the person who wrote the offending passage, would be the one to give it? So ultimately, I find it very surprising that although I gave no indication that I would stop responding, you would post this here. I would also like to note that I voted to close, as I find this abrupt and a bit offensive. – davidlowryduda Sep 24 '11 at 05:43
  • @mixedmath: Actually, can you please undelete your answer, so that it's clear what is being talked of? Maybe someone can interpret it more clearly (or you could even post an answer here). As I remember it, your answer said (something like) the number of ways to choose K out of J objects could be counted either as $\binom{J}{K}$, or by counting separately the number of ways to choose 1 or 2 or … K objects (what one might call $\binom{J}{\le K}$). This obviously doesn't make sense, but I'm probably misremembering. Having the original context would help. – ShreevatsaR Sep 24 '11 at 06:01
  • @ShreevatsaR: No, the original wording of the question is included in this question (in the shaded box). And when you say someone can interpret it more clearly, I suggest you read Ross's answer. It is exactly what I intended. – davidlowryduda Sep 24 '11 at 06:08
  • @mixedmath: Ah ok. The part "if I want to see how many ways I can count K of them, I can […] by counting how many ways I can choose 1, or 2, or 3, or 4, ..., or K of them" is what's confusing. You say that "how many ways I can count K of them" includes "how many ways I can choose K of them" plus many others (1, 2, … K-1). So does "how many ways I can count K of them" actually mean something other than "how many ways I can choose K of them", that you had in mind? This is HM's question. Ross's answer is clear as a hint to the original question, but it's not clear how your wording relates to it. – ShreevatsaR Sep 24 '11 at 06:29
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    @mixedmath, I suppose there is some criticism implicit in taking the question here. But I really did want to know what the hint was supposed to mean, and the 30+ comments we had already exchanged made it abundantly clear that I was not going to get you to tell me. Even now, it appears that you still think I was being rhetorical (i.e. that I understood the hint but was complaining about how you chose to present it), when the naked honest truth is that I did not understand what you were saying. What's one supposed to do then? – hmakholm left over Monica Sep 24 '11 at 14:33
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    @Henning: I hope that this question has gotten you the elucidation that you wanted. But I absolutely do not condone this because of your guise of wanting to understand. To answer your question, I would expect you to be well-mannered and patient. To be explicit, I expect you to wait and discuss for more than a day, rather than 45 minutes. I do not think you were rhetorical, or I would not have had an extended conversation with you. But this has left a very poor taste in my mouth. Perhaps you would agree that this question is undeniable local, and will consider deleting it on your own. Good day. – davidlowryduda Sep 24 '11 at 16:16
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    @mixedmath, if you still think that wanting to understand is a "guise", then I can do nothing more for you. Sorry. – hmakholm left over Monica Sep 24 '11 at 16:22
  • @Henning: It is not the desire to understand that is my problem. It is that doing so impatiently is. And in a manner which I find offensive. – davidlowryduda Sep 24 '11 at 16:23

3 Answers3

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The hint in question makes no sense as written. If one knows the combinatorial argument suggested here by Ross and André, which is essentially the one offered by Henning for the other question, one might suspect that the author of the hint had something like it in mind, but I don’t see how one could be sure. It doesn’t even make sense if select or choose is substituted for count, since one cannot select $K$ objects by selecting at most $K$ of them

Brian M. Scott
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The idea of the hint was that if you want to select $n$ objects out of $x+n+1$, you could have selected $k$ out of the first $x+k,$ then skip the next, then take all the rest. To get the total you have to sum over the allowable $k$'s. I think select (or choose) is a better verb in this case. To me, count would be to count the number of ways to select a set of objects subject to the given condition.

Ross Millikan
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Something connected with the hint can be made to work.

We have the $x$ negative numbers $-x$ to $-1$, together with the $n+1$ numbers $0$ to $n$. We want to choose $n$ numbers from these $x+n+1$ numbers. There are $\binom{x+n+1}{n}$ such choices.

Another way to do the choosing is to first pick a $k$ from $0$ to $n$. Let $\mathbb{C}_k$ be the collection of all sets $S$ of $n$ of our numbers such that:

(i) $S$ contains all numbers from $k+1$ to $n$;

(ii) $S$ does not contain the number $k$.

Then the collection of all choices of $n$ numbers is the disjoint union of the $\mathbb{C}_k$.

If $S\in \mathbb{C}_k$, then $S$ consists of the $n-k$ numbers from $k+1$ to $n$, together with any $k$ numbers chosen from the $x+k$ numbers $-x$ to $k-1$. Thus $\mathbb{C}_k$ has $\binom{x+k}{k}$ elements. Add up, $k=0$ to $n$.

André Nicolas
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