Number of Set Partitions by Number of Components
Theorem
Let $S$ be a (finite) set whose cardinality is $n$.
Let $\map f {n, k}$ denote the number of different ways $S$ can be partitioned into $k$ components.
Then:
- $\ds \map f {n, k} = {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
Proof
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \map f {n, k} = {n \brace k}$
$\map P 0$ is the degenerate case:
- $\ds \map f {0, k} = \delta_{0 k} = {0 \brace k}$
That is: the empty set can be partitioned one and only one way: into $0$ subsets.
Thus $\map P 0$ is seen to hold.
The remainder of the proof considers $n \in \Z_{> 0}$.
First we note that when $k < 1$ or $k > n$:
- $\ds \map f {n, k} = 0 = {n \brace k}$
Hence, throughout, we consider only such $k$ as $1 \le k \le n$.
We define the representative set of cardinality $n$ to be:
- $S_n := \set {1, 2, \ldots, n}$
Basis for the Induction
$\map P 1$ is the case $\map f {1, 1}$.
There is exactly one way to partition $\set 1$, and that is:
- $\set {\set 1}$
From Stirling Number of the Second Kind of Number with Self:
- $\ds {1 \brace 1} = 1$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
- $\ds \map f {m, k} = {m \brace k}$
from which it is to be shown that:
- $\ds \map f {m + 1, k} = {m + 1 \brace k}$
Induction Step
This is the induction step:
By definition, the number of partitions of $S_m$ into $k$ subsets is $\map f {m, k}$.
A partition of $S_{m + 1}$ can be generated by adding element $m + 1$ into one of the existing partitions of $S_m$.
There are two ways this can be done:
- $(1): \quad$ The subset $\set {m + 1}$ may be added, in one way, to one of the partitions of $S_m$ into $k - 1$ subsets.
- $(2): \quad$ The element $m + 1$ may be added to any one of the $k$ subsets in one of the partitions of $S_m$ into $k$ subsets.
Option $(1)$ gives $1$ partition of $S_{m + 1}$ for each partition of $S_m$ into $k - 1$ subsets, that is: $\map f {m, k - 1}$.
Option $(2)$ gives $k$ partitions of $S_{m + 1}$ for each partition of $S_m$ into $k$ subsets, that is: $k \map f {m, k}$.
Thus:
- $\map f {m + 1, k} = k \map f {m, k} + \map f {m, k - 1}$
By the induction hypothesis:
- $\ds \map f {m + 1, k} = k {m \brace k} + {m \brace k - 1}$
So by definition of Stirling numbers of the second kind:
- $\ds \map f {m + 1, k} = {m + 1 \brace k}$
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n, k \in \Z_{\ge 0}: \map f {n, k} = {n \brace k}$
$\blacksquare$
Examples
$4$-Sets into $2$ Subsets
A set with $4$ elements $\set {1, 2, 3, 4}$ can be partitioned into $2$ subsets in $\ds {4 \brace 2} = 7$ ways:
Let $T$ be the set with $3$ elements $\set {1, 2, 3}$.
The partition of $T$ into $1$ set is:
- $\set {1, 2, 3}$
to which $\set 4$ can be added in one way to achieve:
- $\set {1, 2, 3 \mid 4}$
The partitions of $T$ into $2$ sets are:
- $\set {1, 2 \mid 3}$
- $\set {1, 3 \mid 2}$
- $\set {2, 3 \mid 1}$
to which the element $4$ can be added in $2$ ways each:
- $\set {1, 2, 4 \mid 3}$
- $\set {1, 2 \mid 3, 4}$
- $\set {1, 3, 4 \mid 2}$
- $\set {1, 3 \mid 2, 4}$
- $\set {2, 3, 4 \mid 1}$
- $\set {2, 3 \mid 1, 4}$
Thus we have a total of $7$ ways of partitioning a $4$-element set into $2$ components.
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $64$