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Bijections, Double Counting The Art of Counting Bijections, Double Counting Peng Shi Department of Mathematics Duke University September 16, 2009 Straightforward, careful counting Bijection Counting in multiple ways Today, we will consider some commonly-used paradigms of counting: What we focus on in this talk? Enumerative combinatorics is a huge branch of mathematics, involving many theorems and techniques, which we cannot hope to cover in one class Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 2/13 Bijection Counting in multiple ways What we focus on in this talk? Enumerative combinatorics is a huge branch of mathematics, involving many theorems and techniques, which we cannot hope to cover in one class Today, we will consider some commonly-used paradigms of counting: Straightforward, careful counting Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 2/13 Counting in multiple ways What we focus on in this talk? Enumerative combinatorics is a huge branch of mathematics, involving many theorems and techniques, which we cannot hope to cover in one class Today, we will consider some commonly-used paradigms of counting: Straightforward, careful counting Bijection Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 2/13 What we focus on in this talk? Enumerative combinatorics is a huge branch of mathematics, involving many theorems and techniques, which we cannot hope to cover in one class Today, we will consider some commonly-used paradigms of counting: Straightforward, careful counting Bijection Counting in multiple ways Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 2/13 Sum Rule: S S S T If A = A1 A2 ··· An, Ai Aj = ;, then jAj = jA1j + jA2j + ··· + jAnj Product Rule: If W = W1 × W2 × · · · × Wn (Cartesian set product), then jW j = jW1jjW2j · · · jWnj Paradigm 1: Careful Straightforward Counting Comprehensive enumeration/case work Make sure to count every case Don't double count Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 3/13 Product Rule: If W = W1 × W2 × · · · × Wn (Cartesian set product), then jW j = jW1jjW2j · · · jWnj Paradigm 1: Careful Straightforward Counting Comprehensive enumeration/case work Make sure to count every case Don't double count Sum Rule: S S S T If A = A1 A2 ··· An, Ai Aj = ;, then jAj = jA1j + jA2j + ··· + jAnj Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 3/13 Paradigm 1: Careful Straightforward Counting Comprehensive enumeration/case work Make sure to count every case Don't double count Sum Rule: S S S T If A = A1 A2 ··· An, Ai Aj = ;, then jAj = jA1j + jA2j + ··· + jAnj Product Rule: If W = W1 × W2 × · · · × Wn (Cartesian set product), then jW j = jW1jjW2j · · · jWnj Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 3/13 n ways to choose 1st element, n − 1 ways to choose 2nd, . So n(n − 1) ··· (n − m + 1)? But each subset is counted m! times n n(n−1)···(n−m+1) k = m(m−1)···1 n! = (n−m)!m! Basic Tool: Binomial Coefficients How many subsets of f1; 2; ··· ; ng are there with exactly m elements? Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 4/13 But each subset is counted m! times n n(n−1)···(n−m+1) k = m(m−1)···1 n! = (n−m)!m! Basic Tool: Binomial Coefficients How many subsets of f1; 2; ··· ; ng are there with exactly m elements? n ways to choose 1st element, n − 1 ways to choose 2nd, . So n(n − 1) ··· (n − m + 1)? Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 4/13 n n(n−1)···(n−m+1) k = m(m−1)···1 n! = (n−m)!m! Basic Tool: Binomial Coefficients How many subsets of f1; 2; ··· ; ng are there with exactly m elements? n ways to choose 1st element, n − 1 ways to choose 2nd, . So n(n − 1) ··· (n − m + 1)? But each subset is counted m! times Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 4/13 Basic Tool: Binomial Coefficients How many subsets of f1; 2; ··· ; ng are there with exactly m elements? n ways to choose 1st element, n − 1 ways to choose 2nd, . So n(n − 1) ··· (n − m + 1)? But each subset is counted m! times n n(n−1)···(n−m+1) k = m(m−1)···1 n! = (n−m)!m! Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 4/13 Example: suppose jAj = 9, jBj = 6, jCj = 9, jA T Bj = 3, jA T Cj = 4, jB T Cj = 2, jA T B T Cj = 1, what is jA S B S Cj? jA S B S Cj = jAj + jBj + jCj − jA T Bj − jA T Cj − jB T Cj + jA T B T Cj = 9 + 6 + 9 − 3 − 4 − 2 + 1 = 16 In general, [ [ X X \ X \ \ jA1 ··· Anj = jAi j − jAi Aj j + jAi Aj Ak j · · · i i<j i<j<k Basic Tool: Inclusion-Exclusion Principle S S If we know A = A1 ··· An but the Ai 's are not necessarily disjoint, how do we count jAj? Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 5/13 jA S B S Cj = jAj + jBj + jCj − jA T Bj − jA T Cj − jB T Cj + jA T B T Cj = 9 + 6 + 9 − 3 − 4 − 2 + 1 = 16 In general, [ [ X X \ X \ \ jA1 ··· Anj = jAi j − jAi Aj j + jAi Aj Ak j · · · i i<j i<j<k Basic Tool: Inclusion-Exclusion Principle S S If we know A = A1 ··· An but the Ai 's are not necessarily disjoint, how do we count jAj? Example: suppose jAj = 9, jBj = 6, jCj = 9, jA T Bj = 3, jA T Cj = 4, jB T Cj = 2, jA T B T Cj = 1, what is jA S B S Cj? Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 5/13 In general, [ [ X X \ X \ \ jA1 ··· Anj = jAi j − jAi Aj j + jAi Aj Ak j · · · i i<j i<j<k Basic Tool: Inclusion-Exclusion Principle S S If we know A = A1 ··· An but the Ai 's are not necessarily disjoint, how do we count jAj? Example: suppose jAj = 9, jBj = 6, jCj = 9, jA T Bj = 3, jA T Cj = 4, jB T Cj = 2, jA T B T Cj = 1, what is jA S B S Cj? jA S B S Cj = jAj + jBj + jCj − jA T Bj − jA T Cj − jB T Cj + jA T B T Cj = 9 + 6 + 9 − 3 − 4 − 2 + 1 = 16 Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 5/13 Basic Tool: Inclusion-Exclusion Principle S S If we know A = A1 ··· An but the Ai 's are not necessarily disjoint, how do we count jAj? Example: suppose jAj = 9, jBj = 6, jCj = 9, jA T Bj = 3, jA T Cj = 4, jB T Cj = 2, jA T B T Cj = 1, what is jA S B S Cj? jA S B S Cj = jAj + jBj + jCj − jA T Bj − jA T Cj − jB T Cj + jA T B T Cj = 9 + 6 + 9 − 3 − 4 − 2 + 1 = 16 In general, [ [ X X \ X \ \ jA1 ··· Anj = jAi j − jAi Aj j + jAi Aj Ak j · · · i i<j i<j<k Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 5/13 Solution. Let the set of all permultations of f1; 2; ··· ; ng be U. Let D ⊂ U be the jDj set of permutations without fixed points. We seek jUj . Let S S S Ai = f permutationπjπ(i) = ig. Then D = Un(A1 A2 ··· An). S S P P T jA1 ··· Anj = i jAi j − i<j jAi Aj j · · · n n = (n − 1)! − 2 (n − 2)! + 3 (n − 3)! Pn (−1)i+1 = n! i=1 i! The probability is n jDj X (−1)i 1 = ! jUj i! e i=0 Putting the theory into practice Example 1 [Derangements] Aat a Secret Santa party, there are n guests, who each brings a present. Once all presents are collected, they are permuted randomly, and redistributed to the guests. What is the probability no guest receives his/her own gift? What does this converge to as n ! 1? Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 6/13 Let S S S Ai = f permutationπjπ(i) = ig. Then D = Un(A1 A2 ··· An). S S P P T jA1 ··· Anj = i jAi j − i<j jAi Aj j · · · n n = (n − 1)! − 2 (n − 2)! + 3 (n − 3)! Pn (−1)i+1 = n! i=1 i! The probability is n jDj X (−1)i 1 = ! jUj i! e i=0 Putting the theory into practice Example 1 [Derangements] Aat a Secret Santa party, there are n guests, who each brings a present. Once all presents are collected, they are permuted randomly, and redistributed to the guests. What is the probability no guest receives his/her own gift? What does this converge to as n ! 1? Solution. Let the set of all permultations of f1; 2; ··· ; ng be U. Let D ⊂ U be the jDj set of permutations without fixed points. We seek jUj . Peng Shi, Duke University The Art of Counting, Bijections, Double Counting 6/13 S S P P T jA1 ··· Anj = i jAi j − i<j jAi Aj j · · · n n = (n − 1)! − 2 (n − 2)! + 3 (n − 3)! Pn (−1)i+1 = n! i=1 i! The probability is n jDj X (−1)i 1 = ! jUj i! e i=0 Putting the theory into practice Example 1 [Derangements] Aat a Secret Santa party, there are n guests, who each brings a present.
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