Введение
Метод в перечислительной комбинаторике
Method in enumerative combinatorics
В математической области перечислительной комбинаторики тождества иногда доказываются аргументами, основанными на выделении одного "выделенного элемента" множества.
In the mathematical field of enumerative combinatorics, identities are sometimes established by arguments that rely on singling out one "distinguished element" of a set.
Определение
Пусть будет семейство подмножеств множества , а — выделенный элемент множества . Предположим, что существует предикат , связывающий подмножество с . Обозначим через множество подмножеств из , для которых предикат истинен, а через — множество подмножеств из , для которых предикат ложен. Тогда множества и не пересекаются, поэтому, согласно методу суммирования, их мощности аддитивны.
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
Таким образом, выделенный элемент позволяет осуществить разложение в соответствии с предикатом, который представляет собой простую форму алгоритма "разделяй и властвуй". В комбинаторике это позволяет строить рекуррентные соотношения. Примеры приведены в следующем разделе.
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.
Примеры
Биномиальный коэффициент — это количество подмножеств размера k множества размера n. Основное тождество, одним из следствий которого является то, что биномиальные коэффициенты являются именно числами, появляющимися в треугольнике Паскаля, утверждает, что:
The binomial coefficient 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:
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
Доказательство: В множестве размером (n + 1) выберите один выделенный элемент. Множество всех подмножеств размера k содержит: (1) все подмножества размера k, которые содержат выделенный элемент, и (2) все подмножества размера k, которые не содержат выделенный элемент. Если подмножество размера k множества размера (n + 1) содержит выделенный элемент, то его остальные k − 1 элементов выбираются из остальных n элементов нашего множества размера (n + 1). Таким образом, число способов выбора таких элементов равно . Если подмножество размера k не содержит выделенный элемент, то все его k членов выбираются из числа остальных n "невыделенных" элементов. Таким образом, число способов выбрать их равно .
The binomial coefficient 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:
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
Число подмножеств любого множества размера n равно 2n. Доказательство: Мы используем математическую индукцию. Основанием для индукции является истинность этого утверждения в случае n = 0. Пустое множество имеет 0 членов и 1 подмножество, а 20 = 1. Гипотеза индукции — это утверждение в случае n; мы используем её для доказательства случая n + 1. В множестве размером (n + 1) выберите выделенный элемент. Каждое подмножество либо содержит выделенный элемент, либо не содержит. Если подмножество содержит выделенный элемент, то его остальные элементы выбираются из числа остальных n элементов. По гипотезе индукции, число способов сделать это равно 2n. Если подмножество не содержит выделенный элемент, то оно является подмножеством множества всех невыделенных элементов. По гипотезе индукции, число таких подмножеств равно 2n. Наконец, весь список подмножеств нашего множества размером (n + 1) содержит 2n + 2n = 2n+1 элементов.
The binomial coefficient 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:
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
Пусть Bn — n-е число Белла, т. е. число разбиений множества из n членов. Пусть Cn — общее число "частей" (или "блоков", как их часто называют комбинатористы) среди всех разбиений этого множества. Например, разбиения множества размера 3 {a, b, c} могут быть записаны следующим образом:
The binomial coefficient 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:
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
Мы видим 5 разбиений, содержащих 10 блоков, поэтому B3 = 5 и C3 = 10. Тождество гласит:
The binomial coefficient 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:
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
Доказательство: В множестве размером (n + 1) выберите выделенный элемент. В каждом разбиении нашего множества размером (n + 1) либо выделенный элемент является "синглтоном", т. е. множество, содержащее только выделенный элемент, является одним из блоков, либо выделенный элемент принадлежит к большему блоку. Если выделенный элемент является синглтоном, то удаление выделенного элемента оставляет разбиение множества, содержащего n невыделенных элементов. Таких разбиений Bn. Если выделенный элемент принадлежит к большему блоку, то его удаление оставляет блок в разбиении множества, содержащем n невыделенных элементов. Таких блоков Cn.
The binomial coefficient 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:
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.
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 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:
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 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 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.