Combinatorics (Fall 2010)/Basic enumeration

Sets and Multisets

Subsets

We want to count the number of subsets of a set.

Let[math]\displaystyle{ S=\{x_1,x_2,\ldots,x_n\} }[/math] be an [math]\displaystyle{ n }[/math]-element set, or [math]\displaystyle{ n }[/math]-set for short. Let [math]\displaystyle{ 2^S=\{T\mid T\subset S\} }[/math] denote the set of all subset of [math]\displaystyle{ S }[/math]. [math]\displaystyle{ 2^S }[/math] is called the power set of [math]\displaystyle{ S }[/math].

We give a combinatorial proof that [math]\displaystyle{ |2^S|=2^n }[/math]. We observe that every subset [math]\displaystyle{ T\in 2^S }[/math] corresponds to a unique bit-vector [math]\displaystyle{ v\in\{0,1\}^S }[/math], such that each bit [math]\displaystyle{ v_i }[/math] indicates whether [math]\displaystyle{ x_i\in S }[/math]. Formally, define a map [math]\displaystyle{ \phi:2^S\rightarrow\{0,1\}^n }[/math] by [math]\displaystyle{ \phi(T)=(v_1,v_2,\ldots,v_n) }[/math], where

[math]\displaystyle{ v_i=\begin{cases} 1 & \mbox{if }x_i\in T\\ 0 & \mbox{if }x_i\not\in T. \end{cases} }[/math]

The map [math]\displaystyle{ \phi }[/math] is a bijection (a 1-1 correspondence). The proof that [math]\displaystyle{ \phi }[/math] is a bijection is left as an exercise.

We know that [math]\displaystyle{ |\{0,1\}^n|=2^n\, }[/math]. Since there is a bijection from [math]\displaystyle{ 2^S }[/math] to [math]\displaystyle{ \{0,1\}^n }[/math], it holds that [math]\displaystyle{ |2^S|=|\{0,1\}^n|=2^n\, }[/math]. Here we use the rule of bijection: if there exists a bijection from [math]\displaystyle{ \mathcal{S} }[/math] to [math]\displaystyle{ \mathcal{T} }[/math], then [math]\displaystyle{ |\mathcal{S}|=|\mathcal{T}| }[/math].

There are two elements of this proof:

• Find a 1-1 correspondence between subsets of an [math]\displaystyle{ n }[/math]-set and [math]\displaystyle{ n }[/math]-bit vectors.

An application of this in Computer Science is that we can use bit-array as a data structure for sets: any set defined over a universe [math]\displaystyle{ U }[/math] can be represented by an array of [math]\displaystyle{ |U| }[/math] bits.

• The rule of bijection: if there is a 1-1 correspondence between two sets, then their cardinalities are the same.

Many counting problems are solved by establishing a bijection between the set to be counted and some easy-to-count set. This kind of proofs are usually called (non-rigorously) combinatorial proofs.

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We give an alternative proof that [math]\displaystyle{ |2^S|=2^n }[/math].

Subsets of fixed size

We then count the number of subsets of fixed size of a set.

The twelvfold way

[math]\displaystyle{ f:N\rightarrow M }[/math]

Elements of [math]\displaystyle{ N }[/math]	Elements of [math]\displaystyle{ M }[/math]	Any [math]\displaystyle{ f }[/math]	Injective (1-1) [math]\displaystyle{ f }[/math]	Surjective (on-to) [math]\displaystyle{ f }[/math]
distinguishable	distinguishable	[math]\displaystyle{ m^n\, }[/math]	[math]\displaystyle{ \left(m\right)_n }[/math]	[math]\displaystyle{ m!S(n, m)\, }[/math]
indistinguishable	distinguishable	[math]\displaystyle{ \left({m\choose n}\right) }[/math]	[math]\displaystyle{ {m\choose n} }[/math]	[math]\displaystyle{ \left({m\choose n-m}\right) }[/math]
distinguishable	indistinguishable	[math]\displaystyle{ \sum_{k=1}^m S(n,k) }[/math]	[math]\displaystyle{ \begin{cases}1 & \mbox{if }n\le m\\ 0& \mbox{if }n\gt m\end{cases} }[/math]	[math]\displaystyle{ S(n,m)\, }[/math]
indistinguishable	indistinguishable	[math]\displaystyle{ \sum_{k=1}^m p_k(n) }[/math]	[math]\displaystyle{ \begin{cases}1 & \mbox{if }n\le m\\ 0& \mbox{if }n\gt m\end{cases} }[/math]	[math]\displaystyle{ p_m(n)\, }[/math]