Method of distinguished element

In the mathematical field of enumerative combinatorics, identities are sometimes established by arguments that rely on singling out one "distinguished element" of a set.

Let ${\displaystyle {\mathcal {A}}}${\displaystyle {\mathcal {A}}} be a family of subsets of the set ${\displaystyle A}${\displaystyle A} and let ${\displaystyle x\in A}${\displaystyle x\in A} be a distinguished element of set ${\displaystyle A}${\displaystyle A}. Then suppose there is a predicate ${\displaystyle P(X,x)}${\displaystyle P(X,x)} that relates a subset ${\displaystyle X\subseteq A}${\displaystyle X\subseteq A} to ${\displaystyle x}${\displaystyle x}. Denote ${\displaystyle {\mathcal {A}}(x)}${\displaystyle {\mathcal {A}}(x)} to be the set of subsets ${\displaystyle X}${\displaystyle X} from ${\displaystyle {\mathcal {A}}}${\displaystyle {\mathcal {A}}} for which ${\displaystyle P(X,x)}${\displaystyle P(X,x)} is true and ${\displaystyle {\mathcal {A}}-x}${\displaystyle {\mathcal {A}}-x} to be the set of subsets ${\displaystyle X}${\displaystyle X} from ${\displaystyle {\mathcal {A}}}${\displaystyle {\mathcal {A}}} for which ${\displaystyle P(X,x)}${\displaystyle P(X,x)} is false, Then ${\displaystyle {\mathcal {A}}(x)}${\displaystyle {\mathcal {A}}(x)} and ${\displaystyle {\mathcal {A}}-x}${\displaystyle {\mathcal {A}}-x} are disjoint sets, so by the method of summation, the cardinalities are additive^[1]

${\displaystyle |{\mathcal {A}}|=|{\mathcal {A}}(x)|+|{\mathcal {A}}-x|}${\displaystyle |{\mathcal {A}}|=|{\mathcal {A}}(x)|+|{\mathcal {A}}-x|}

Thus the distinguished element allows for a decomposition according to a predicate that is a simple form of a divide and conquer algorithm. In combinatorics, this allows for the construction of recurrence relations. Examples are in the next section.

• The binomial coefficient ${\displaystyle {n \choose k}}${\displaystyle {n \choose k}} is the number of size-k subsets of a size-n set. A basic identity—one of whose consequences is that the binomial coefficients are precisely the numbers appearing in Pascal's triangle—states that:

${\displaystyle {n \choose k-1}+{n \choose k}={n+1 \choose k}.}${\displaystyle {n \choose k-1}+{n \choose k}={n+1 \choose k}.}

Proof: In a size-(n + 1) set, choose one distinguished element. The set of all size-k subsets contains: (1) all size-k subsets that do contain the distinguished element, and (2) all size-k subsets that do not contain the distinguished element. If a size-k subset of a size-(n + 1) set does contain the distinguished element, then its other k − 1 elements are chosen from among the other n elements of our size-(n + 1) set. The number of ways to choose those is therefore ${\displaystyle {n \choose k-1}}${\displaystyle {n \choose k-1}}. If a size-k subset does not contain the distinguished element, then all of its k members are chosen from among the other n "non-distinguished" elements. The number of ways to choose those is therefore ${\displaystyle {n \choose k}}${\displaystyle {n \choose k}}.

• The number of subsets of any size-n set is 2ⁿ.

Proof: We use mathematical induction. The basis for induction is the truth of this proposition in case n = 0. The empty set has 0 members and 1 subset, and 2⁰ = 1. The induction hypothesis is the proposition in case n; we use it to prove case n + 1. In a size-(n + 1) set, choose a distinguished element. Each subset either contains the distinguished element or does not. If a subset contains the distinguished element, then its remaining elements are chosen from among the other n elements. By the induction hypothesis, the number of ways to do that is 2ⁿ. If a subset does not contain the distinguished element, then it is a subset of the set of all non-distinguished elements. By the induction hypothesis, the number of such subsets is 2ⁿ. Finally, the whole list of subsets of our size-(n + 1) set contains 2ⁿ + 2ⁿ = 2ⁿ⁺¹ elements.

• Let B_n be the nth Bell number, i.e., the number of partitions of a set of n members. Let C_n be the total number of "parts" (or "blocks", as combinatorialists often call them) among all partitions of that set. For example, the partitions of the size-3 set {a, b, c} may be written thus:

${\displaystyle {\begin{matrix}abc\\a/bc\\b/ac\\c/ab\\a/b/c\end{matrix}}}${\displaystyle {\begin{matrix}abc\\a/bc\\b/ac\\c/ab\\a/b/c\end{matrix}}}

We see 5 partitions, containing 10 blocks, so B₃ = 5 and C₃ = 10. An identity states:

${\displaystyle B_{n}+C_{n}=B_{n+1}.}${\displaystyle B_{n}+C_{n}=B_{n+1}.}

Proof: In a size-(n + 1) set, choose a distinguished element. In each partition of our size-(n + 1) set, either the distinguished element is a "singleton", i.e., the set containing only the distinguished element is one of the blocks, or the distinguished element belongs to a larger block. If the distinguished element is a singleton, then deletion of the distinguished element leaves a partition of the set containing the n non-distinguished elements. There are B_n ways to do that. If the distinguished element belongs to a larger block, then its deletion leaves a block in a partition of the set containing the n non-distinguished elements. There are C_n such blocks.

• Combinatorial principles
• Combinatorial proof

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