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

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

3. Overview of PIE

4. Overview of PIE 1 1

5. Overview of PIE 1 1 2 1

6. Overview of PIE 1 1 1

7. Overview of PIE

8. Overview of PIE 1 2 1 2 1 3 2 1 1 1 2 1

9. Overview of PIE 1 2 1 2 1 3 2 1 1 1 2 1

10. Overview of PIE 1 1 1 1 1 1 1

11. 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

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

13. 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

14. 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

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

16. 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

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

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

19. 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,

20. 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.

21. 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.

22. 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

23. 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,

24. 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

25. Example We show that
We have

26. 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.

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

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

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

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

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

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

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

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

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

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

37. -END-