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| {{Redirect|Distinguished element|sets with pre-defined distinguished elements|Pointed set}}
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| In enumerative [[combinatorics|combinatorial]] [[mathematics]], [[identity (mathematics)|identities]] are sometimes established by arguments that rely on singling out one '''"distinguished element"''' of a set.
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| ==Definition==
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| Let <math>\mathcal{A}</math> be a family of subsets of the set <math>A</math> and let <math>x \in A</math> be a distinguished element of set <math>A</math>. Then suppose there is a predicate <math>P(X,x)</math> that relates a subset <math>X\subseteq A</math> to <math>x</math>. Denote <math>\mathcal{A}(x)</math> to be the set of subsets <math>X</math> from <math>\mathcal{A}</math> for which <math>P(X,x)</math> is true and <math>\mathcal{A}-x</math> to be the set of subsets <math>X</math> from <math>\mathcal{A}</math> for which <math>P(X,x)</math> is false, Then <math>\mathcal{A}(x)</math> and <math>\mathcal{A}-x</math> are disjoint sets, so by the method of summation, the cardinalities are additive<ref name=Petkovsek2002>{{cite journal|last=Petkovšek|first=Marko|coauthors=Tomaž Pisanski|title=Combinatorial Interpretation of Unsigned Stirling and Lah Numbers|journal=University of Ljubljana preprint series|date=November 2002|volume=40|issue=837|pages=1-6|url=http://www.imfm.si/preprinti/PDF/00837.pdf|accessdate=12 July 2013}}</ref>
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| :<math>|\mathcal{A}| = |\mathcal{A}(x)| + |\mathcal{A}-x|</math> | |
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| 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 relation]]s. Examples are in the next section.
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| ==Examples==
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| * The [[binomial coefficient]] <math>{n \choose k}</math> is the number of size-''k'' subsets of a size-''n'' set. A basic identity, one of whose consequences is that these are precisely the numbers appearing in [[Pascal's triangle]], states that:
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| ::<math>{n \choose k-1}+{n \choose k}={n+1 \choose k}.</math>
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| :'''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 <math>{n \choose k-1}</math>. 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 <math>{n \choose k}</math>.
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| *The number of subsets of any size-''n'' set is 2<sup>''n''</sup>.
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| :'''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 2<sup>0</sup> = 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 2<sup>''n''</sup>. 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 2<sup>''n''</sup>. Finally, the whole list of subsets of our size-(''n'' + 1) set contains 2<sup>''n''</sup> + 2<sup>''n''</sup> = 2<sup>''n''+1</sup> elements.
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| * Let ''B''<sub>''n''</sub> be the ''n''th [[Bell number]], i.e., the number of [[partition of a set|partitions of a set]] of ''n'' members. Let ''C''<sub>''n''</sub> 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:
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| ::<math>\begin{matrix}abc \\ a/bc \\ b/ac \\ c/ab \\ a/b/c \end{matrix}</math>
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| :We see 5 partitions, containing 10 blocks, so ''B''<sub>3</sub> = 5 and ''C''<sub>3</sub> = 10. An identity states:
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| ::<math>B_n+C_n=B_{n+1}.\,</math>
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| :'''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''<sub>''n''</sub> 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''<sub>''n''</sub> such blocks.
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| ==See also==
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| * [[Combinatorial principles]]
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| * [[Combinatorial proof]]
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| ==References==
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| {{reflist}}
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| {{DEFAULTSORT:Method Of Distinguished Element}}
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| [[Category:Combinatorics]]
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