Answers.Body: <p>We denote the number as <span class="math-container">$S(N,k)$</span> and recall obvious special cases as follows.</p>
  <p><span class="math-container">$S(N,1)=1$</span>: All points belong to one cluster.</p>
  <p><span class="math-container">$S(N,N)=1$</span>: Each point is in its own cluster.</p>
  <p><span class="math-container">$S(N,k)=0$</span>, if <span class="math-container">$k&gt;N$</span>: It is impossible to have more clusters than points.</p>
  <p>For the more general case, refer <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling number of the second kind</a>.</p>
  <blockquote>
  <p>In mathematics, particularly in combinatorics, a Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of n objects into k non-empty subsets and is denoted by <span class="math-container">${\displaystyle S(n,k)}$</span> or <span class="math-container">${\displaystyle \textstyle \left\{{n \atop k}\right\}}$</span>.</p>
  </blockquote>
  <blockquote>
  <p><span class="math-container">${\displaystyle \left\{{n \atop k}\right\}={\frac {1}{k!}}\sum _{i=0}^{k}(-1)^{k-i}{\binom {k}{i}}i^{n}=\sum _{i=0}^{k}{\frac {(-1)^{k-i}i^{n}}{(k-i)!i!}}.}$</span></p>
  </blockquote>
  <blockquote>
  <p>Stirling numbers of the second kind obey the recurrence relation <span class="math-container">${\displaystyle \left\{{n+1 \atop k}\right\}=k\left\{{n \atop k}\right\}+\left\{{n \atop k-1}\right\}\quad {\mbox{for}}\;0&lt;k&lt;n}$</span>
  with initial conditions <span class="math-container">${\displaystyle \left\{{n \atop n}\right\}=1\quad {\mbox{ for}}\;n\geq 0\quad {\text{ and }}\quad \left\{{n \atop 0}\right\}=\left\{{0 \atop n}\right\}=0\quad {\text{ for }}n&gt;0{\text{.}}}$</span></p>
  </blockquote>
  
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Answers.UserDisplayName: cinch
Answers.Body: <p>cinch handles the general case, but since you say that <span class="math-container">$k$</span> is fixed, there's a simpler approach:</p>
  <p>Assuming each cluster must contain at least one element, we need to distribute the remaining <span class="math-container">$N-k$</span> elements. Each is assigned to one of <span class="math-container">$k$</span> possibilities, and since each such assignment is independent of the others, we can just multiply those together. So there are <span class="math-container">$k^\left(N-k\right)$</span> possible assignments.</p>
  <p>(In addition to assuming fixed <span class="math-container">$k$</span>, this also assumes that it matters <em>which</em> point is assigned to which cluster, not just what the cluster sizes are.)</p>
  
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Answers.UserDisplayName: Ray
Body: <p>Assume that we have N data points and we divide into k clusters.
  What is the number of all possible k clusters for N data points?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-12-12T16:38:44.107Z
Comments.Text: Is this a homework question? If yes, please, edit your post to add the tag [tag:homework]. For these questions, we shouldn't reveal the complete answer. That's why. Thanks for creating a post with a clear title, but please also include some context in your posts, like "what are your findings or ideas?"
Comments.UserDisplayName: nbro
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-12-13T20:45:15.110Z
Comments.Text: No, it is not a HW question.
Comments.UserDisplayName: DSPinfinity
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-12-13T21:36:52.460Z
Comments.Text: Is the value of $k$ fixed?
Comments.UserDisplayName: Ray
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-12-14T12:10:16.710Z
Comments.Text: yes, it is  fixed.
Comments.UserDisplayName: DSPinfinity
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2024-12-18T11:52:29.773Z
Comments.Text: Please, use **also** the [tag:math] tag then when your question is purely mathematical, and if it requires a proof, use the tag [tag:proofs]. If this is not a homework question, then where did you find it? Provide more context in your posts. If you just came up with this question, which I doubt, then you should say that. Also, edit your post to clarify that k is fixed because comments are temporary and can be deleted.
Comments.UserDisplayName: nbro
Tags: <machine-learning><math><clustering>
Title: What is the number of all possible k clusters for N data points?
UserDisplayName: DSPinfinity
ThreadText: # What is the number of all possible k clusters for N data points?
  
  (Asked by: username_0 on 2024-12-12)
  
  Assume that we have N data points and we divide into k clusters.
  What is the number of all possible k clusters for N data points?
  
  **Tags:** machine-learning, math, clustering
  
  > **Comments:**
  > **username_1**: Is this a homework question? If yes, please, edit your post to add the tag [tag:homework]. For these questions, we shouldn't reveal the complete answer. That's why. Thanks for creating a post with a clear title, but please also include some context in your posts, like "what are your findings or ideas?"  
  > **username_0**: No, it is not a HW question.  
  > **username_2**: Is the value of $k$ fixed?  
  > **username_0**: yes, it is  fixed.  
  > **username_1**: Please, use **also** the [tag:math] tag then when your question is purely mathematical, and if it requires a proof, use the tag [tag:proofs]. If this is not a homework question, then where did you find it? Provide more context in your posts. If you just came up with this question, which I doubt, then you should say that. Also, edit your post to clarify that k is fixed because comments are temporary and can be deleted.  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_3 on 2024-12-12. Score: 1 | Accepted answer
  
  We denote the number as $S(N,k)$ and recall obvious special cases as follows.
  
  $S(N,1)=1$: All points belong to one cluster.
  
  $S(N,N)=1$: Each point is in its own cluster.
  
  $S(N,k)=0$, if $k>N$: It is impossible to have more clusters than points.
  
  For the more general case, refer [Stirling number of the second kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind).
  
  > In mathematics, particularly in combinatorics, a Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of n objects into k non-empty subsets and is denoted by ${\displaystyle S(n,k)}$ or ${\displaystyle \textstyle \left\{{n \atop k}\right\}}$.
  
  > ${\displaystyle \left\{{n \atop k}\right\}={\frac {1}{k!}}\sum \_{i=0}^{k}(-1)^{k-i}{\binom {k}{i}}i^{n}=\sum \_{i=0}^{k}{\frac {(-1)^{k-i}i^{n}}{(k-i)!i!}}.}$
  
  > Stirling numbers of the second kind obey the recurrence relation ${\displaystyle \left\{{n+1 \atop k}\right\}=k\left\{{n \atop k}\right\}+\left\{{n \atop k-1}\right\}\quad {\mbox{for}}\;0<k<n}$
  > with initial conditions ${\displaystyle \left\{{n \atop n}\right\}=1\quad {\mbox{ for}}\;n\geq 0\quad {\text{ and }}\quad \left\{{n \atop 0}\right\}=\left\{{0 \atop n}\right\}=0\quad {\text{ for }}n>0{\text{.}}}$
  
  
  ---
  
  ### Answer by username_2 on 2024-12-16. Score: 0
  
  username_3 handles the general case, but since you say that $k$ is fixed, there's a simpler approach:
  
  Assuming each cluster must contain at least one element, we need to distribute the remaining $N-k$ elements. Each is assigned to one of $k$ possibilities, and since each such assignment is independent of the others, we can just multiply those together. So there are $k^\left(N-k\right)$ possible assignments.
  
  (In addition to assuming fixed $k$, this also assumes that it matters *which* point is assigned to which cluster, not just what the cluster sizes are.)
  
  
  ---
  
  