If I have $n$ posts in the ground, arranged in a horizontal line, how many distinguishable placements of rings are there over the posts, where:

  1. The $i$th ring encloses $k_i$ posts.
  2. There are exactly $\kappa$ rings.
  3. No two rings enclose the same post (a "non-overlapping" partition).
  4. Each ring encloses only adjacent posts (partition into adjacent elements).
  5. Every un-ringed post, by the end, is then assigned a null ring.

These would be "non-crossing" partitions of $\{1,2,\dots,n\}$ into equivalence classes, with specific counts $k_1,\dots,k_\kappa$ in each class.

So, in the language of set partitions:

If I want to partition an $n$-set into classes such that the first class has $k_1$ elements, the second $k_2$, and so on, with potentially some elements not in a class, and write this number


then reject partitions where two classes overlap, such that if e.g. $\{1,2,3,4,5\}$ is partitioned into $\{\{1,2\},\{3\},\{4,5\}\}$, where $k_1=2,k_2=1,k_3=2$, then this does not overlap, but, $\{\{1,3\},\{5\},\{2,4\}\}$, where similarly $k_1=2,k_2=1,k_3=2$, does overlap, since there exists at least one class which contains non-adjacent integers. In fact, in the example there are two classes with this property: $\{1,3\}$ and $\{2,4\}$.

If I write this number


what is this number? It appears to be related to the Stirling numbers of the second kind, but this does not include the adjacency idea.

  • $\begingroup$ Your list of $4$ criteria doesn't match what you describe afterwards. To get that, you'd have to include a fifth criterion that every post is enclosed by a ring. Please clarify which of these two versions of the question you intended. $\endgroup$ – joriki Feb 3 '20 at 11:29
  • $\begingroup$ How do I describe a partition which doesn't necessarily include every element of the set being partitioned? So not every element of the set needs to be in the partition. $\endgroup$ – apkg Feb 3 '20 at 12:38
  • $\begingroup$ I don't understand how the fifth criterion you added resolves the discrepancy. If I understand correctly, the "null rings" don't count towards the $\kappa$ rings? But then the "null rings" don't make a difference, and the five criteria are still not in correspondence with the description afterwards, which assigns all rings to $\kappa$ equivalence classes. $\endgroup$ – joriki Feb 3 '20 at 12:49
  • $\begingroup$ How would I correct the second explanation, such that not all the posts need to be in a ring? Basically the union of the parts of the partition does not need to be the whole set. $\endgroup$ – apkg Feb 3 '20 at 12:54
  • $\begingroup$ Maybe don't use the equivalence class idea? And just use a set partition? I edited the question. $\endgroup$ – apkg Feb 3 '20 at 12:55

I hope I’ve now correctly understood what you want to do. As I understand it, you want to count the compositions of $n$ with $\kappa$ parts coloured black and no two parts coloured white adjacent.

There are $\binom{n-1}{j-1}$ compositions of $n$ into $j$ parts. This needs to be multiplied by the number of ways to select $w=j-\kappa$ out of $j$ balls to be white without any white balls being adjacent. To count these, glue a black ball to the right of every white ball to obtain $\binom{j-w}w$ selections. But that misses the selections where the rightmost ball is white, so we have to add another $\binom{j-w}{w-1}$ selections, for a total of $\binom{j-w}w+\binom{j-w-1}{w-1}=\binom{j-w+1}w$. Then summing over $j$ yields a count of

$$ \sum_{j=\kappa}^n\binom{n-1}{j-1}\binom{\kappa+1}{j-\kappa}\;. $$

According to Wolfram|Alpha, this is $\binom{n+\kappa}{2\kappa}$. This simple form suggests that there might be a more elegant way to derive it without the detour through a summation.

  • 1
    $\begingroup$ Hmm -- I just realized that you wanted the counts for specific $k_i$, not just the total count. I'll edit the answer later... $\endgroup$ – joriki Feb 3 '20 at 13:46
  • $\begingroup$ Yes the reason I ask here is I can’t see a way of using a multi set argument which conditions on the specific way in which the set is partitioned into subsets with multiplicities $k_i$. But this answer I can use as a foundation. $\endgroup$ – apkg Feb 3 '20 at 14:11
  • $\begingroup$ I have discovered that I simply need the count of, as you say, compositions (this is what I was looking for) with specific counts of all multiplicities $k_i$. But in fact, there is only one of each which is unique up to permutation of the parts. $\endgroup$ – apkg Feb 6 '20 at 5:03

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