<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>

CC BY-SA 4.0
2024-12-12T07:56:41.713Z
2024-12-12T07:56:41.713Z
cinch
<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>

CC BY-SA 4.0
2024-12-16T16:21:17.953Z
2024-12-16T16:21:17.953Z
Ray
<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>

CC BY-SA 4.0
2024-12-12T16:38:44.107Z
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?"
nbro
CC BY-SA 4.0
2024-12-13T20:45:15.110Z
No, it is not a HW question.
DSPinfinity
CC BY-SA 4.0
2024-12-13T21:36:52.460Z
Is the value of $k$ fixed?
Ray
CC BY-SA 4.0
2024-12-14T12:10:16.710Z
yes, it is  fixed.
DSPinfinity
CC BY-SA 4.0
2024-12-18T11:52:29.773Z
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.
nbro
<machine-learning><math><clustering>
What is the number of all possible k clusters for N data points?
DSPinfinity
# 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.  

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## 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
> with initial conditions ${\displaystyle \left\{{n \atop n}\right\}=1\quad {\mbox{ for}}\;n\geq 0\quad {\text{ and }}\quad \left\{{n \atop 0}\right\}=\left\{{0 \atop n}\right\}=0\quad {\text{ for }}n>0{\text{.}}}$


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### 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.)


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