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

## Definition

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

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

## Examples

• The binomial coefficient ${n \choose k}$ 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:
${n \choose k-1}+{n \choose k}={n+1 \choose k}.$ 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 ${n \choose k-1}$ . 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 ${n \choose k}$ .
• 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 Bn be the nth Bell number, i.e., the number of partitions of a set of n members. Let Cn 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 {abc} may be written thus:
${\begin{matrix}abc\\a/bc\\b/ac\\c/ab\\a/b/c\end{matrix}}$ We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 10. An identity states:
$B_{n}+C_{n}=B_{n+1}.$ 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 Bn 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 Cn such blocks.