The principle of inclusion and exclusion; inversion formulae Speaker: 藍國元 Date: 2005.11.28

Outline An overview of principle of inclusion-exclusion Applications of PIE Möbius Inversion formula Applications of MIF

Overview of PIE

Overview of PIE 1 1

Overview of PIE 1 1 2 1

Overview of PIE 1 1 1

Overview of PIE

Overview of PIE 1 2 1 2 1 3 2 1 1 1 2 1

Overview of PIE 1 2 1 2 1 3 2 1 1 1 2 1

Overview of PIE 1 1 1 1 1 1 1

Principle of Inclusion-Exclusion Notations: S: a set of N-elements E1,E2,…,Er: not necessarily distinct subsets of S N(M): # of elements of S in for Nj

Principle of Inclusion-Exclusion Theorem 1 (PIE) Proof: Case1. and is in none of contributes 1 to left-hand side and right-hand side. Case2. and is in exactly k of the sets contributes 0 to left-hand. The contribution of to right-hand side equals

Principle of Inclusion-Exclusion Theorem (Purest form of PIE) Let be an n-set. Let be the 2n-dimensional vector space (over some field K) of all functions . Let be the linear transformation defined by Then exists and is given by

Principle of Inclusion-Exclusion Combinatorial situation involving previous Theorem : given set : properties set : subset of : # of objects in that have exactly the properties in : # of objects in that have at least the properties in

Principle of Inclusion-Exclusion Combinatorial situation involving previous Theorem Then we have By previous Theorem In particular,

Principle of Inclusion-Exclusion Dual form We can reformulate the PIE by interchanging with , with , and so on. : # of objects in that have at most the properties in Then we have

Applications of PIE Application 1: Derangements How many permutations have no fixed points? i.e. Sol: Let We have By PIE,

Applications of PIE Application 1: Derangements Since , is a good approximation of From it is not difficult to derive

Applications of PIE Application 2: Surjections Give X be an n-set and Y be a k-set. How many surjection of X to Y? Sol: Let We have By PIE,

Applications of PIE Application 2: Surjections Note that if k > n then if k = n then Note that where S(n,k) is the Stirling number of second kind.

Applications of PIE Application 3: Show that Combinatorial proof: Let be an n-set of blue balls be an m-set of red balls. How many k-subsets consist of red balls only? The answer is trivially the right-hand side. The left-hand side is letting S be all the k-subset of X∪Y and Ei those k-subset that contain bi and then PIE gives the left-hand side.

Applications of PIE Application 3: Show that Algebra proof: We use the expansion: The coefficient of in is The coefficient of in is The coefficient of in is

Applications of PIE Application 4: Euler function Let be a positive integer and be the number of integers k with such that Show that Proof: Let We have By PIE,

Theorem 2 [Gauss] Let with . Then Proof: Let Since Since every integer from 1 to n belongs only one , we have Since d runs all positive divisor of n, we have

Example We show that We have

Möbius Inversion formula Definition Let with . The Möbius function, denoted by is 1, if n =1 0, if , if Note that satisfies is called squarefree.

Möbius Inversion formula Theorem 3 Let with . 1, if n =1 0, otherwise Proof: There is nothing to show when n=1. If , then

Möbius Inversion formula Remarks Using the Möbius function, we can reformulate as

Möbius Inversion formula Definition An arithmetic function is a function whose domain is the set of positive integers. Example

Möbius Inversion formula Theorem 4 (Möbius Inversion Formula) Let be arithmetic functions. Then if and only if

Möbius Inversion formula Theorem 4 (Möbius Inversion Formula) the inner sum is 0 unless Proof:

Möbius Inversion formula Theorem 4 (Möbius Inversion Formula) Proof: the inner sum is 0 unless

Applications of MIF The Möbius Inversion Formula can be used to obtain nontrivial identities among arithmetic functions from trivial identities among arithmetic functions.

Applications of MIF Example 1 Let with and let for all such n. We have By MIF, we obtain the nontrivial identity or , equivalently

Applications of MIF Example 2 Let with and let . We have By MIF, we obtain the nontrivial identity

Applications of MIF Example 3 Let with and let . We have By MIF, we obtain the nontrivial identity

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