1. Introduction to Discrete Mathematics
COMPSCI 102
Introduction to Discrete Mathematics

2. Counting III
Lecture 8 (September 21, 2007) X1 X2 + X3

3. 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: S1YS2TEMS3
There are 7! permutations of S1YS2TEMS3
But when we stop pretending we see that we have counted each arrangement of SYSTEMS 3! times, once for each of 3! rearrangements of S1S2S3 7! 3!
= 840

6. Arrange n symbols: r1 of type 1, r2 of type 2, …, rk of type k
n-r1 r2
n - r1 - r2 - … - rk-1 rk … n!
(n-r1)!r1! =
(n-r1)! …
(n-r1-r2)!r2! = n!
r1!r2! … rk!

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

8. Remember:
The number of ways to arrange n symbols with r1 of type 1, r2 of type 2, …, rk of type k is: n!
r1!r2! … rk!

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 ith /
This gives a correspondence (bijection) between divisions of the gold and sequences with 20 G’s and 4 /’s

11. 24 4 How many different ways to divide up the loot?
Sequences with 20 G’s and 4 /’s 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?
x1 + x2 + x3 + x4 + x5 = 20
x1, x2, x3, x4, x5 ≥ 0 24 4
Think of xk are being the number of gold bars that are allotted to pirate k

14. How many integer solutions to the following equations?
x1 + x2 + x3 + … + xn = k
x1, x2, x3, …, xn ≥ 0
n + k - 1
n - 1
n + k - 1 k =

15. 67 12 7 Identical/Distinct Dice Suppose that we roll seven dice
How many different outcomes are there, if order matters? 67 12 7
What if order doesn’t matter? (E.g., Yahtzee)
(Corresponds to 6 pirates and 7 bars of gold)

16. Unary visualization: {Y, Y, Y, Z, Z}
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. size k from n types of elements is:
Counting Multisets
There number of ways
to choose a multiset of
size k from n types of elements is:
n + k - 1
n - 1
n + k - 1 k =

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

19. b1 b3 b2 t1 t2 t1 t2 t1 t2
b1t1
b1t2
b2t1
b2t2
b3t1
b3t2
(b1+b2+b3)(t1+t2) =
b1t1 +
b1t2 +
b2t1 +
b2t2 +
b3t1 +
b3t2

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

21. Choice Tree for Terms of (1+X)3
Combine like terms to get 1 + 3X + 3X2 + X3

22. What is a Closed Form Expression For ck?
(1+X)n = c0 + c1X + c2X2 + … + cnXn
(1+X)(1+X)(1+X)(1+X)…(1+X)
After multiplying things out, but before combining like terms, we get 2n cross terms, each corresponding to a path in the choice tree
ck, the coefficient of Xk, is the number of paths with exactly k X’s n k
ck =

23. Binomial Coefficients
The Binomial Formula n n 1 n
(1+X)n = X0 + X1
+…+ Xn
Binomial Coefficients
binomial expression

24. The Binomial Formula
(1+X)0 = 1
(1+X)1 =
1 + 1X
(1+X)2 =
1 + 2X + 1X2
(1+X)3 =
1 + 3X + 3X2 + 1X3
(1+X)4 =
1 + 4X + 6X2 + 4X3 + 1X4

25. What is the coefficient of EMSTY in the expansion of5!

26. What is the coefficient of EMS3TY in the expansion of
The number of ways to rearrange the letters in the word SYSTEMS

27. What is the coefficient of BA3N2 in the expansion of
The number of ways to rearrange the letters in the word BANANA

28. What is the coefficient of (X1r1X2r2…Xkrk)
in the expansion of
(X1+X2+X3+…+Xk)n? n!
r1!r2!...rk!

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

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

31. 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: n n k 
(1+X)n = Xk
k = 0 n k 
k = 0
2n =
Let x = 1,
The number of subsets of an n-element set

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

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

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

35. We gave an algebraic proof thatk 
k even
k odd =
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 

36. A combinatorial proof must construct a bijection between On and En
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 bijection between On and En

37. An Attempt at a Bijection
Let fn be the function that takes an
n-bit string and flips all its bits
fn is clearly a one-to-one and onto function
for odd n. E.g. in f7 we have:
...but do even n work? In f6 we have    
Uh oh. Complementing maps evens to evens!

38. 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,    

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

40. Pascal’s Triangle: kth row are coefficients of (1+X)k
The Binomial Formula
(1+X)0 = 1
(1+X)1 =
1 + 1X
(1+X)2 =
1 + 2X + 1X2
(1+X)3 =
1 + 3X + 3X2 + 1X3
(1+X)4 =
1 + 4X + 6X2 + 4X3 + 1X4
Pascal’s Triangle: kth row are coefficients of (1+X)k
Inductive definition of kth entry of nth row: Pascal(n,0) = Pascal (n,n) = 1; Pascal(n,k) = Pascal(n-1,k-1) + Pascal(n-1,k)

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

42. Pascal’s Triangle “It is extraordinary how fertile in 1 properties the
triangle is.
Everyone can
try his
hand” 1
1 1
1 2 1

43.  Summing the Rows n 2n = k k = 0 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
1 2 1
= 1 +
+ +
+ + +
= 2
= 4
= 8
= 16
= 32
= 64

44. Odds and Evens 1
1 1
1 2 1 =

45.   Summing on 1st Avenue i i 1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
1 1
1 2 1
n+1 2 = =

46.  Summing on kth Avenue 1 1 1 n+1 i = 1 2 1 k+1 k 1 3 3 1 1 4 6 4 1
1 1
1 2 1 
i = k n i k
n+1
k+1 =

47. Fibonacci Numbers 1
1 1
1 2 1
= 2
= 3
= 5
= 8
= 13

48. Sums of Squares 1
1 1
1 2 1 2 2 2 2 2 2 2

49. Al-Karaji Squares 1
1 1
1 2 1
= 1
+2
= 4
+2
= 9
+2
= 16
+2
= 25
+2
= 36

50. Pascal Mod 2

51. All these properties can be proved inductively and algebraically
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?10 5 B

53. shortest routes from (0,0) to (j,k)
Manhattan 1 1
jth street
kth avenue 2 2 3 3 4 4
j+k k
There are
shortest routes from (0,0) to (j,k)

54. shortest routes from (0,0) to (n-k,k)
Manhattan 1
Level n
kth avenue 1 2 2 3 3 4 4 n k
There are
shortest routes from (0,0) to (n-k,k)

55. Manhattan 1 Level n kth avenue 1 2 2 3 3 4 4 n k There are1
Level n
kth avenue 1 2 2 3 3 4 4 n k
There are
shortest routes from (0,0) to
level n and kth avenue

56. 1
Level n
kth avenue 1 1 2 2 1 1 3 1 2 1 3 4 1 3 3 1 4 1 4 6 4 1 1 5 1 10 10 5 6 15 20 15 6

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

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

59. 1
Level n
kth avenue 1 2 2 3 3 4 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
<k> will stand for the vector <k,0,0,0,…>
<0> = <0,0,0,0,…>
<1> = <1,0,0,0,…>
V! + T! means to add the vectors position-wise
<4,2,3,…> + <5,1,1,….> = <9,3,4,…>

62. RIGHT( <1,2,3, …> ) = <0,1,2,3,…>
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( <1,2,3, …> ) = <0,1,2,3,…>

63. Vector Programs Example: Store: V! := <6>;
V! := RIGHT(V!) + <42>;
V! := RIGHT(V!) + <2>;
V! := RIGHT(V!) + <13>;
V! = <6,0,0,0,…>
V! = <42,6,0,0,…>
V! = <2,42,6,0,…>
V!= <13,2,42,6,…>
V! = < 13, 2, 42, 6, 0, 0, 0, … >

64. V! = nth row of Pascal’s triangle
Vector Programs
Example:
Store:
V! := <1>;
Loop n times
V! := V! + RIGHT(V!);
V! = <1,0,0,0,…>
V! = <1,1,0,0,…>
V! = <1,2,1,0,…>
V!= <1,3,3,1,…>
V! = nth row of Pascal’s triangle

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

66. Programs  Polynomials
The vector V! = < a0, a1, a2, > will be represented by the polynomial: 
i = 0 
aiXi
PV =

67.  Formal Power Series aiXi PV =
The vector V! = < a0, a1, a2, > will be represented by the formal power series: 
i = 0 
aiXi
PV =

68.  V! = < a0, a1, a2, . . . > aiXi PV = i = 0 
<0> is represented by
<k> is represented by k
V! + T! is represented by
(PV + PT)
RIGHT(V!) is represented by
(PV X)

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

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

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

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