{"title":"包容与排斥原则","authors":"James Joseph Sylvester","doi":"10.1142/9789814401920_0013","DOIUrl":null,"url":null,"abstract":"The Principle of Inclusion and Exclusion, hereafter called PIE, gives a formula for the size of the union of n finite sets. Usually the universe is finite too. It is a generalisation of the familiar formulas |A ∪ B| = |A| + |B| − |A ∩ B| and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. That is, the cardinality of the union P 1 ∪ P 2 ∪. .. ∪ P k can be calculated by including (adding) the sizes of all of the sets together, then excluding (subtracting) the sizes of the intersections of all pairs of sets, then including the sizes of the intersections of all triples, excluding the sizes of the intersections of all quadruples, and so on until, finally, the size of the intersection of all of the sets has been included or excluded, as appropriate. If n is odd it is included, and if n is even it is excluded. The formula can be expressed more compactly as |P 1 ∪ P 2 ∪ · · · ∪ P n | = k i=1 (−1) k 1≤i 1 <i 2 <···<i k ≤n |P i 1 ∩ P i 2 ∩ · · · ∩ P i k |. It is important to remember that all sets involved must be finite. To prove PIE, both sides count only elements that belong to some positive number of the sets. Each of these is counted once on the LHS. To determine the number of times it is counted on the RHS, suppose it belongs to t ≥ 1 of the sets, and calculate the contribution it makes to each intersection. You'll end up making use of the Binomial Theorem expansion of (1 + (−1)) t. When to use PIE. Vaguely speaking, you should try PIE when you are trying to count something described by a bunch of conditions, any number of which might hold at the same time, and you can't see how to organise the counting by cases. Often PIE is used in conjunction with counting the complement. That is, you use it to count the number of objects in the universe that you don't want, and subtract this from the size of the universe (which needs to be finite!). In applying PIE, the …","PeriodicalId":193527,"journal":{"name":"Enumerative Combinatorics","volume":"21 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"2018-10-08","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"2","resultStr":"{\"title\":\"THE PRINCIPLE OF INCLUSION AND EXCLUSION\",\"authors\":\"James Joseph Sylvester\",\"doi\":\"10.1142/9789814401920_0013\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"The Principle of Inclusion and Exclusion, hereafter called PIE, gives a formula for the size of the union of n finite sets. Usually the universe is finite too. It is a generalisation of the familiar formulas |A ∪ B| = |A| + |B| − |A ∩ B| and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. That is, the cardinality of the union P 1 ∪ P 2 ∪. .. ∪ P k can be calculated by including (adding) the sizes of all of the sets together, then excluding (subtracting) the sizes of the intersections of all pairs of sets, then including the sizes of the intersections of all triples, excluding the sizes of the intersections of all quadruples, and so on until, finally, the size of the intersection of all of the sets has been included or excluded, as appropriate. If n is odd it is included, and if n is even it is excluded. The formula can be expressed more compactly as |P 1 ∪ P 2 ∪ · · · ∪ P n | = k i=1 (−1) k 1≤i 1 <i 2 <···<i k ≤n |P i 1 ∩ P i 2 ∩ · · · ∩ P i k |. It is important to remember that all sets involved must be finite. To prove PIE, both sides count only elements that belong to some positive number of the sets. Each of these is counted once on the LHS. To determine the number of times it is counted on the RHS, suppose it belongs to t ≥ 1 of the sets, and calculate the contribution it makes to each intersection. You'll end up making use of the Binomial Theorem expansion of (1 + (−1)) t. When to use PIE. Vaguely speaking, you should try PIE when you are trying to count something described by a bunch of conditions, any number of which might hold at the same time, and you can't see how to organise the counting by cases. Often PIE is used in conjunction with counting the complement. That is, you use it to count the number of objects in the universe that you don't want, and subtract this from the size of the universe (which needs to be finite!). 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引用次数: 2
摘要
包含与排除原理(以下简称PIE)给出了n个有限集的并集大小的公式。通常宇宙也是有限的。它是我们熟悉的公式| a∪B| = | a | + |B| - | a∩B| = | a | + |B| + |C| - | a∩B| - |B∩C| + | a∩B∩C|的推广。也就是说，P 1∪P 2∪的基数…可通过将所有集合的大小相加，然后剔除所有集合对的交点大小，再剔除所有三元组的交点大小，再剔除所有四三元组的交点大小，以此类推，直到最后，所有集合的交点大小都被剔除或剔除(视情况而定)。如果n是奇数，则包含它，如果n是偶数，则排除它。该公式可以更简洁地表示为| p1∪p2∪···∪P n | = k i=1 (- 1) k 1≤i 1 本文章由计算机程序翻译，如有差异，请以英文原文为准。
The Principle of Inclusion and Exclusion, hereafter called PIE, gives a formula for the size of the union of n finite sets. Usually the universe is finite too. It is a generalisation of the familiar formulas |A ∪ B| = |A| + |B| − |A ∩ B| and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. That is, the cardinality of the union P 1 ∪ P 2 ∪. .. ∪ P k can be calculated by including (adding) the sizes of all of the sets together, then excluding (subtracting) the sizes of the intersections of all pairs of sets, then including the sizes of the intersections of all triples, excluding the sizes of the intersections of all quadruples, and so on until, finally, the size of the intersection of all of the sets has been included or excluded, as appropriate. If n is odd it is included, and if n is even it is excluded. The formula can be expressed more compactly as |P 1 ∪ P 2 ∪ · · · ∪ P n | = k i=1 (−1) k 1≤i 1
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