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.

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 &minus; 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 {a, b, c} 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.