1. COMPSCI 102 Discrete Mathematics for Computer Science

2. X1 X1 X2 X2 + + X3 X3 Lecture 8 Counting III

3. How many ways to rearrange the letters in the word ? How many ways to rearrange the letters in the word “SYSTEMS”?

4. SYSTEMS 7 places to put the Y, 6 places to put the T, 5 places to put the E, 4 places to put the M, and the S’s are forced 7 X 6 X 5 X 4 = 840

5. SYSTEMS Let’s pretend that the S’s are distinct: S 1 YS 2 TEMS 3 There are 7! permutations of S 1 YS 2 TEMS 3 But when we stop pretending we see that we have counted each arrangement of SYSTEMS 3! times, once for each of 3! rearrangements of S 1 S 2 S 3 7! 3! = 840

6. Arrange n symbols: r 1 of type 1, r 2 of type 2, …, r k of type k n r1r1 n-r 1 r2r2 … n - r 1 - r 2 - … - r k-1 rkrk (n-r 1 )! (n-r 1 -r 2 )!r 2 ! n! (n-r 1 )!r 1 ! = … = n! r 1 !r 2 ! … r k !

7. 14! 2!3!2! = 3,632,428,800 CARNEGIEMELLON

8. Remember: The number of ways to arrange n symbols with r 1 of type 1, r 2 of type 2, …, r k of type k is: n! r 1 !r 2 ! … r k !

9. 5 distinct pirates want to divide 20 identical, indivisible bars of gold. How many different ways can they divide up the loot?

10. Sequences with 20 G’s and 4 /’s GG/G//GGGGGGGGGGGGGGGGG/ represents the following division among the pirates: 2, 1, 0, 17, 0 In general, the ith pirate gets the number of G’s after the (i-1) st / and before the i th / This gives a correspondence (bijection) between divisions of the gold and sequences with 20 G’s and 4 /’s

11. Sequences with 20 G’s and 4 /’s How many different ways to divide up the loot? 24 4

12. How many different ways can n distinct pirates divide k identical, indivisible bars of gold? n + k - 1 n - 1 n + k - 1 k =

13. How many integer solutions to the following equations? x 1 + x 2 + x 3 + x 4 + x 5 = 20 x 1, x 2, x 3, x 4, x 5 ≥ 0 Think of x k are being the number of gold bars that are allotted to pirate k 24 4

14. How many integer solutions to the following equations? x 1 + x 2 + x 3 + … + x n = k x 1, x 2, x 3, …, x n ≥ 0 n + k - 1 n - 1 n + k - 1 k =

15. Identical/Distinct Dice Suppose that we roll seven dice How many different outcomes are there, if order matters? 6767 What if order doesn’t matter? (E.g., Yahtzee) 12 7 (Corresponds to 6 pirates and 7 bars of gold)

16. Multisets A multiset is a set of elements, each of which has a multiplicity The size of the multiset is the sum of the multiplicities of all the elements Example: {X, Y, Z} with m(X)=0 m(Y)=3, m(Z)=2 Unary visualization: {Y, Y, Y, Z, Z}

17. Counting Multisets = n + k - 1 n - 1 n + k - 1 k There number of ways to choose a multiset of size k from n types of elements is:

18. ++ () + () = +++++ Polynomials Express Choices and Outcomes Products of Sum = Sums of Products

19. b2b2 b3b3 b1b1 t1t1 t2t2 t1t1 t2t2 t1t1 t2t2 b1t1b1t1 b1t2b1t2 b2t1b2t1 b2t2b2t2 b3t1b3t1 b3t2b3t2 (b 1 +b 2 +b 3 )(t 1 +t 2 ) = b1t1b1t1 b1t2b1t2 b2t1b2t1 b2t2b2t2 b3t1b3t1 b3t2b3t2 +++++

20. There is a correspondence between paths in a choice tree and the cross terms of the product of polynomials!

21. 1X1X1X1X 1X 1X 1X 1 X X X2X2 X X2X2 X2X2 X3X3 Choice Tree for Terms of (1+X) 3 Combine like terms to get 1 + 3X + 3X 2 + X 3

22. What is a Closed Form Expression For c k ? (1+X) n = c 0 + c 1 X + c 2 X 2 + … + c n X n (1+X)(1+X)(1+X)(1+X)…(1+X) After multiplying things out, but before combining like terms, we get 2 n cross terms, each corresponding to a path in the choice tree c k, the coefficient of X k, is the number of paths with exactly k X’s n k c k =

23. binomial expression Binomial Coefficients The Binomial Formula n 1 (1+X) n = n 0 X0X0 +X1X1 +…+ n n XnXn

24. The Binomial Formula (1+X) 0 = (1+X) 1 = (1+X) 2 = (1+X) 3 = (1+X) 4 = 1 1 + 1X 1 + 2X + 1X 2 1 + 3X + 3X 2 + 1X 3 1 + 4X + 6X 2 + 4X 3 + 1X 4

25. 5! What is the coefficient of EMSTY in the expansion of (E + M + S + T + Y) 5 ?

26. What is the coefficient of EMS 3 TY in the expansion of (E + M + S + T + Y) 7 ? The number of ways to rearrange the letters in the word SYSTEMS

27. What is the coefficient of BA 3 N 2 in the expansion of (B + A + N) 6 ? The number of ways to rearrange the letters in the word BANANA

28. What is the coefficient of (X 1 r 1 X 2 r 2 …X k r k ) in the expansion of (X 1 +X 2 +X 3 +…+X k ) n ? n! r 1 !r 2 !...r k !

29. There is much, much more to be said about how polynomials encode counting questions!

30. Power Series Representation (1+X) n = n k XkXk  k = 0 n n k XkXk   = “Product form” or “Generating form” “Power Series” or “Taylor Series” Expansion For k>n, n k = 0

31. By playing these two representations against each other we obtain a new representation of a previous insight: (1+X) n = n k XkXk  k = 0 n Let x = 1, n k  k = 0 n 2 n = The number of subsets of an n-element set

32. By varying x, we can discover new identities: (1+X) n = n k XkXk  k = 0 n Let x = -1, n k  k = 0 n 0 = (-1) k Equivalently, n k  k even n n k  k odd n =

33. The number of subsets with even size is the same as the number of subsets with odd size

34. Proofs that work by manipulating algebraic forms are called “algebraic” arguments. Proofs that build a bijection are called “combinatorial” arguments (1+X) n = n k XkXk  k = 0 n

35. Let O n be the set of binary strings of length n with an odd number of ones. Let E n be the set of binary strings of length n with an even number of ones. We gave an algebraic proof that  O n  =  E n  n k  k even n n k  k odd n =

36. A Combinatorial Proof Let O n be the set of binary strings of length n with an odd number of ones Let E n be the set of binary strings of length n with an even number of ones A combinatorial proof must construct a bijection between O n and E n

37. An Attempt at a Bijection Let f n be the function that takes an n-bit string and flips all its bits f n is clearly a one-to- one and onto function...but do even n work? In f 6 we have for odd n. E.g. in f 7 we have: 110011  001100 101010  010101 0010011  1101100 1001101  0110010 Uh oh. Complementing maps evens to evens!

38. A Correspondence That Works for all n Let f n 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

39. The binomial coefficients have so many representations that many fundamental mathematical identities emerge… (1+X) n = n k XkXk  k = 0 n

40. The Binomial Formula (1+X) 0 = (1+X) 1 = (1+X) 2 = (1+X) 3 = (1+X) 4 = 1 1 + 1X 1 + 2X + 1X 2 1 + 3X + 3X 2 + 1X 3 1 + 4X + 6X 2 + 4X 3 + 1X 4 Pascal’s Triangle: k th row are coefficients of (1+X) k Inductive definition of k th entry of n th row: Pascal(n,0) = Pascal (n,n) = 1; Pascal(n,k) = Pascal(n-1,k-1) + Pascal(n-1,k)

41. “Pascal’s Triangle” 0 0 = 1 1 0 1 1 2 0 2 1 = 2 2 2 = 1 Al-Karaji, Baghdad 953-1029 Chu Shin-Chieh 1303 Blaise Pascal 1654 3 0 = 1 3 1 = 3 3 2 3 3 = 1

42. Pascal’s Triangle “It is extraordinary how fertile in properties the triangle is. Everyone can try his hand” 11 121 1331 14641 15101051 1615201561

43. 11 121 1331 14641 15101051 1615201561 Summing the Rows +++++++++++++++++++++++++++++++++++++++++++ n k  k = 0 n 2 n = = 1 = 2 = 4 = 8 = 16 = 32 = 64

44. 11 121 1331 14641 15101051 1615201561 1 + 15 + 15 + 16 + 20 + 6= Odds and Evens

45. 11 121 1331 14641 15101051 1615201561 Summing on 1 st Avenue  i = 1 n i 1 =  n i n+1 2 =

46. 11 121 1331 14641 15101051 1615201561 Summing on k th Avenue  i = k n i k n+1 k+1 =

47. 11 121 1331 14641 15101051 1615201561 Fibonacci Numbers = 2 = 3 = 5 = 8 = 13

48. 11 121 1331 14641 15101051 1615201561 Sums of Squares 222 2222

49. 11 121 1331 14641 15101051 1615201561 Al-Karaji Squares +2  = 1 = 4 = 9 = 16 = 25 = 36

50. Pascal Mod 2

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

52. How many shortest routes from A to B? B A 10 5

53. Manhattan j th streetk th avenue 1 0 2 4 3 0 1 2 3 4 There areshortest routes from (0,0) to (j,k) j+k k

54. Manhattan Level nk th avenue 1 0 2 4 3 0 1 2 3 4 There areshortest routes from (0,0) to (n-k,k) n k

55. Manhattan Level nk th avenue 1 0 2 4 3 0 1 2 3 4 There are shortest routes from (0,0) to n k level n and k th avenue

56. Level nk th avenue 1 0 2 4 3 0 1 2 3 4 1 1 1 1 1 1 1 1 1 2 3 3 4 4 6 1 1 5 5 10 6 6 15 20

57. Level nk th avenue 1 0 2 4 3 1 1 1 1 1 1 1 1 1 2 3 3 4 4 6 1 1 5 5 10 6 6 15 20 n k n-1 k-1 n-1 k = + +

58. Level nk th avenue 1 0 2 4 3 0 1 2 3 4 2n n n k  k = 0 n 2 =

59. Level nk th avenue 1 0 2 4 3 0 1 2 3 4 n+1 k+1 i k  i = k n =

60. Vector Programs Let’s define a (parallel) programming language called VECTOR that operates on possibly infinite vectors of numbers. Each variable V ! can be thought of as:

61. Vector Programs Let k stand for a scalar constant will stand for the vector = V ! + T ! means to add the vectors position-wise + =

62. Vector Programs RIGHT(V ! ) means to shift every number in V ! one position to the right and to place a 0 in position 0 RIGHT( ) =

63. Vector Programs Example: V ! := ; V ! := RIGHT(V ! ) + ; V ! = Store: V ! =

64. Vector Programs Example: V ! := ; Loop n times V ! := V ! + RIGHT(V ! ); V ! = n th row of Pascal’s triangle Store: V ! =

65. X1 X1 X2 X2 + + X3 X3 Vector programs can be implemented by polynomials!

66. Programs  Polynomials The vector V ! = will be represented by the polynomial:  i = 0  aiXiaiXi P V =

67. Formal Power Series The vector V ! = will be represented by the formal power series:  i = 0  aiXiaiXi P V =

68.  i = 0  aiXiaiXi P V = V ! = is represented by 0 k V ! + T ! is represented by(P V + P T ) RIGHT(V ! ) is represented by(P V X)

69. Vector Programs Example: V ! := ; Loop n times V ! := V ! + RIGHT(V ! ); V ! = n th row of Pascal’s triangle P V := 1; P V := P V + P V X;

70. Vector Programs Example: V ! := ; Loop n times V ! := V ! + RIGHT(V ! ); V ! = n th row of Pascal’s triangle P V := 1; P V := P V (1+X);

71. Vector Programs Example: V ! := ; Loop n times V ! := V ! + RIGHT(V ! ); V ! = n th row of Pascal’s triangle P V = (1+ X) n

72. Polynomials count Binomial formula Combinatorial proofs of binomial identities Vector programs Here’s What You Need to Know…