# Method of distinguished element

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In enumerative combinatorial mathematics, identities are sometimes established by arguments that rely on singling out one **"distinguished element"** of a set.

## Definition

Let be a family of subsets of the set and let be a distinguished element of set . Then suppose there is a predicate that relates a subset to . Denote to be the set of subsets from for which is true and to be the set of subsets from for which is false, Then and are disjoint sets, so by the method of summation, the cardinalities are additive^{[1]}

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 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:

**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 . 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 .

- The number of subsets of any size-
*n*set is 2^{n}.

**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^{0}= 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^{n}. 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^{n}. Finally, the whole list of subsets of our size-(*n*+ 1) set contains 2^{n}+ 2^{n}= 2^{n+1}elements.

- Let
*B*_{n}be the*n*th 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:

- We see 5 partitions, containing 10 blocks, so
*B*_{3}= 5 and*C*_{3}= 10. An identity states:

**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.

## See also

## References

- ↑ {{#invoke:Citation/CS1|citation |CitationClass=journal }}