Counting III: Pigeonhole principal, Pascal’s Triangle and more

1. X1 • X2 • + • + • X3 Counting III: Pigeonhole principal, Pascal’s Triangle and more

2. Pigeonhole Principle If there are more pigeons than pigeonholes, then some pigeonholes must contain two or more pigeons

3. Pigeonhole Principle

4. Pigeonhole Principle If there are more pigeons than pigeonholes, then some pigeonholes must contain two or more pigeons

5. Pigeonhole Principle If there are more pigeons than pigeonholes, then some pigeonholes must contain two or more pigeons Application: two people in Pitt must have the same number of hairs on their heads

6. Pigeonhole Principle If there are more pigeons than pigeonholes, then some pigeonholes must contain two or more pigeons Problem: among any n integer numbers, there are some whose sum is divisible by n.

7. Pigeonhole Principle Problem: among any n integer numbers, there are some whose sum is divisible by n. Proof: Consider si=x1+…+xi modulo n. Remainders are either zero or {1, 2, …, n-1}. Exist si = sk (mod n). Take si - sk

8. Pigeonhole Principle If there are more pigeons than pigeonholes, then some pigeonholes must contain two or more pigeons Problem: Prove that among any subset of n+1 integers selected from the set {1,2,3,…,2n} there are two elements a and b such that a is a multiple of b.

9. Pigeonhole Principle Problem: Prove that among any subset of n+1 integers selected from the set {1,2,3,…,2n} there are two elements a and b such that a is a multiple of b. Proof. Let S = {x1,x2,..., xn+1} be a set of n+1 integers xk=2r(k) *yk where yk is odd.

10. Pigeonhole Principle Problem: Prove that among any subset of n+1 integers selected from the set {1,2,3…,2n} there are two elements a and b such that a is a multiple of b. Proof. xk=2r(k) *yk The set {1,2,3…,2n} has n odd integers. Therefore, exist yk = yi.

11. - + To Exclude Or Not To Exclude?

12. A school has 100 students. 50 take French, 40 take Latin, and 20 take both. How many students take neither language? • How many positive integers less than 70 are relatively prime to 70? (70=257)

13. 100 – 70 = 30 F  French students L  Latin students |F|=50|L|=40 French Latin French AND Latin students: |FL|=20 French OR Latin students: |F|+|L|-|F L| = 50 + 40 – 20 = 70 |FL|= Neither language:

14. U  universe of elements Lesson: |AB| = |A| + |B| - |AB| U B A

15. U A2 A1 A3 • How many positive integers less than 70 are relatively prime to 70? 70 = 257 U = [1..70] A1 integers in U divisible by 2 A2  integersin U divisible by 5 A3  integersin U divisible by 7 |A1| = 35 |A2| = 14 |A3|=10

17. U A2 A1 A3 Lesson: Let Sk be the sum of the sizes of All k-tuple intersections of the Ai’s.

18. The Principle of Inclusion and Exclusion Let A1,A2,…,An be sets in a universe U. Let Sk denote the sum of size of all k-tuple intersections of Ai’s.

19. x gets counted times by S1 “ “ “ times by S2 “ “ “ times by S3 “ “ “ times by Sn The formula counts x Let x A1  An be an element appearing in m of the Ai’s.

20. The formula counts x 1 time.

21. The Binomial Formula

22. One polynomial, two representations “Closed form” or “Generating form” “Power series” or “Expanded form”

23. Let x=1. The number of subsets of an n-element set By playing these two representations against each other we obtain a new representation of a previous insight:

24. By varying x, we can discover new identities Let x= -1. Equivalently,

25. Proofs that work by manipulating algebraic forms are called “algebraic” arguments.Proofs that build a 1-1 onto correspondence are called “combinatorial” arguments.

26. Let On be the set of binary strings of length n with an odd number of ones. Let En be the set of binary strings of length n with an even number of ones. We gave an algebraic proof that On =  En

27. A Combinatorial Proof Let On be the set of binary strings of length n with an odd number of ones. Let En be the set of binary strings of length n with an even number of ones. A combinatorial proof must construct a one-to-one correspondence betweenOn and En

28. An attempt at a correspondence Let fn be the function that takes an n-bit string and flips all its bits. ...but do even n work? In f6 we have 110011  001100 101010  010101 Uh oh… fn is clearly a one-to-one and onto function for odd n. E.g. in f7 we have 0010011  1101100 1001101  0110010

29. A correspondence that works for all n Let fn be the function that takes an n-bit string and flips only the first bit. For example, 0010011  1010011 1001101  0001101 110011  010011 101010  001010

30. Problem. Consider all words of length n made from {a,b,c,d}. How many such words have an even number of a’s?

31. Pascal’s Triangle

32. The binomial coefficients have so many representations that many fundamental mathematical identities emerge…

33. Pascal’s Triangle:kth row are the coefficients of (1+X)k 1 (1+X)0 = 1 + 1X (1+X)1 = 1 + 2X + 1X2 (1+X)2 = 1 + 3X + 3X2 + 1X3 (1+X)3 = 1 + 4X + 6X2 + 4X3 + 1X4 (1+X)4 =

34. kth Row Of Pascal’s Triangle: 1 (1+X)0 = 1 + 1X (1+X)1 = 1 + 2X + 1X2 (1+X)2 = 1 + 3X + 3X2 + 1X3 (1+X)3 = 1 + 4X + 6X2 + 4X3 + 1X4 (1+X)4 =

35. Inductive definition of kth entry of nth row:Pascal(n,0) = Pacal (n,n) = 1; Pascal(n,k) = Pascal(n-1,k-1) + Pascal(n-1,k) 1 (1+X)0 = 1 + 1X (1+X)1 = 1 + 2X + 1X2 (1+X)2 = 1 + 3X + 3X2 + 1X3 (1+X)3 = 1 + 4X + 6X2 + 4X3 + 1X4 (1+X)4 =

36. “Pascal’s Triangle” Blaise Pascal 1654

37. Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 “It is extraordinary how fertile in properties the triangle is. Everyone can try his hand.”

38. Summing The Rows 1 1 + 1 1 + 2 + 1 1 + 3 + 3 + 1 1 + 4 + 6 + 4 + 1 1 + 5 + 10 + 10 + 5 + 1 1 + 6 + 15 + 20 + 15 + 6 + 1 =1 =2 =4 =8 =16 =32 =64

39. 6+20+6 = 1+15+15+1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

40. Summing on 1st Avenue 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

41. Summing on kth Avenue 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

42. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 =2 =3 =5 =8 =13

43. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 2 2 2 2 2 2 2

44. Al-Karaji Squares 1 1 1 1 2 +2*1 1 3 +2*3 1 1 4 +2*6 4 1 1 5 +2*10 10 5 1 1 6 +2*15 20 15 6 1 =0 =1 =4 =9 =16 =25 =36

45. All these properties can be proved inductively and algebraically. We will give combinatorial proofs using the Manhattan block walking representation of binomial coefficients.

46. How to count pathways Meandering in a nameless modern metropolis Scenario: You’re in a city where all the streets, numbered 0 through x, run north-south, and all the avenues, numbered 0 through y, run east-west. How many [sensible] ways are there to walk from the corner of 0th St. and 0th avenue to the opposite corner of the city? 0 x y 0

47. Meandering in a nameless modern metropolis • All paths require exactly x+y steps: • x steps east, y steps north • Counting paths is the same as counting which of the x+y steps are northward steps: 0 x y 1 0

48. Meandering in a nameless modern metropolis Now, what if we add the constraint that the path must go through a certain intersection, call it (i,j)? (i,j) 0 x y 1 0

49. 0 0 j’th Street k’th Avenue 1 1 2 2 3 3 4 4 Manhattan

50. Level n 0 2 4 3 1 0 k’th Avenue ……… 1 …………. 2 …………………… 3 ………………………… 4 ………………………………… Manhattan