1. Discrete Math For Computing II

2. Discrete Math For Computing II
Main Text:
Topics in enumeration; principle of inclusion and exclusion, Partial orders and lattices. Algorithmic complexity; recurrence relations, Graph theory.
Prerequisite: CS 2305
"Discrete Mathematics and its Applications" Kenneth Rosen, 5th Edition, McGraw Hill.

3. Contact Information
B. Prabhakaran
Department of Computer Science
University of Texas at Dallas
Mail Station EC 31, PO Box
Richardson, TX 75083
Phone: ; Fax:
URL:
Office: ES 3.706
Office Hours: 1-2pm Tuesdays, Thursdays
Other times by appointments through
Announcements: Made in class and on course web page.
TA: TBA.

4. Course Outline Selected topics in chapters 6 through 9.
Chapter 6: Advanced Counting Techniques: recurrence relations, principle of inclusion and exclusion
Chapter 7: Relations: properties of binary relations, representing relations, equivalence relations, partial orders
Chapter 8: Graphs: graph representation, isomorphism, Euler paths, shortest path algorithms, planar graphs, graph coloring
Chapter 9: Trees: tree applications, tree traversal, trees and sorting, spanning trees

5. Course ABET Objectives
Ability to construct and solve recurrence relations
Ability to use the principle of inclusion and exclusion to solve problems
Ability to understand binary relations and their applications
Ability to recognize and use equivalence relations and partial orderings
Ability to use and construct graphs and graph terminology
Ability to apply the graph theory concepts of Euler and Hamilton circuits
Ability to identify and use planar graphs and shortest path problems
Ability to use and construct trees and tree terminology
Ability to use and construct binary search trees

6. Evaluation
1 Mid-terms: in class. 75 minutes. Mix of MCQs (Multiple Choice Questions) & Short Questions.
1 Final Exam: 75 minutes or 2 hours (depending on class room availability). Mix of MCQs and Short Questions.
2 - 3 Quizzes: 5-6 MCQs or very short questions minutes each.
Homeworks/assignments: 3 or 4 spread over the semester.

7. Homeworks Each homework will be for 10 marks. Homeworks Submission:
Submit on paper to TA/Instructor.

8. Grading
Home works: 5%
Quizzes: 15%
Mid-term : 40%
Final: 40%

9. Likely Letter Grades Mostly Relative grading
A-, A, A+: 1.2 – 1.4 times class average
B-, B, B+: 1 – 1.2 class average
C-, C, C+: 0.8 – 1.0 class average
D-, D, D+: 0.7 – 0.8 class average
F: Below 0.6 times class average
Absolute scores will also influence above the ranges.

10. Schedule
Quizzes: Dates announced in class & web, a week ahead. Mostly before midterm and final.
Mid-term: February 23, 2006
Final Exam: Last day of class (April 20th) OR 11am, May 1, 2006 (As per UTD schedule)
Subject to minor changes
Quiz and homework schedules will be announced in class and course web page, giving sufficient time for preparation.

11. Cheating Academic dishonesty will be taken seriously.
Cheating students will be handed over to Head/Dean for further action.
Remember: home works (and exams too!) are to be done individually.
Any kind of cheating in home works/exams will be dealt with as per UTD guidelines.

12. Advanced Counting Techniques
Chapter 6
Advanced Counting Techniques

13. § 6.1: Recurrence Relations
Definition: A recurrence relation for the sequence {an} is an equation expressing an in terms of one or more of the previous terms of the sequence:
a1,a2,a3,…,an-1, with n>=n0, (n0 being a nonnegative integer).
A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.

14. Recurrence Example Consider the recurrence relation
an = 2an−1 − an−2 (n≥2).
Which of the following are solutions? an = 3n Yes -> 2 [3(n-1)] – 3(n-2)
= 3n => an
an = 2n No -> a0 = 1, a1 = 2, a2 = 4;
a2 = 2a1 – a0 = 2.2 – 1 = 3 ≠ a2
an = Yes -> an = 2.5 – 5 = 5 = an

15. Initial conditions The initial conditions (base conditions):
Specify the terms that precede the first term where the recurrence relation takes effect.
i.e., specify a0

16. Example Applications Growth of bank account Initial Amount P0=$10,000
After n Years= Pn
Compound Interest I = 11%
Soln: Pn =Pn-1+(I/100)Pn-1=(1.11)Pn-1
Using Iteration we get, Pn =(1.11)n P0
i.e P30 =(1.11)3010,000=$228,922.97

17. Example Application Rabbits and Fibonacci Numbers
Growth of rabbit population in which each rabbit yields 1 new one every period starting 2 periods after its birth.
Pn = Pn−1 + Pn−2 (Fibonacci relation)

18. Classic Tower of Hanoi Example
Problem: Get all disks from peg 1 to peg 2.
Only move 1 disk at a time.
Never set a larger disk on a smaller one.

19. Hanoi Recurrence Relation
Let Hn = # moves for a stack of n disks.
Optimal strategy:
Move top n−1 disks to spare peg. (Hn−1 moves)
Move bottom disk. (1 move)
Move top n−1 to bottom disk. (Hn−1 moves)
Note: Hn = 2Hn−1 + 1

20. Why is Hn = 2Hn-1 + 1 Only move 1 disk at a time.
Never set a larger disk on a smaller one.

21. Solving Tower of Hanoi RR
Hn = 2 Hn−1 + 1
= 2 (2 Hn−2 + 1) + 1 = 22 Hn−
= 22(2 Hn−3 + 1) = 23 Hn− …
= 2n−1 H1 + 2n−2 + …
= 2n−1 + 2n−2 + … (since H1=1)
= 2n − 1

22. §6.2: Solving Recurrences
Definition: A linear homogeneous recurrence relation of degree k with constant coefficient is a recurrence relation of the form
an = c1an−1 + … + ckan−k, where the ci are all real, and ck ≠ 0.
The solution is uniquely determined if k initial conditions a0…ak−1 are provided

23. §6.2: Solving Recurrences..
Linear?: Right hand side is sum of multiples of previous terms.
Homogenous?: No terms that are NOT multiples of the ajs.
Degree?: k-degree since previous k terms are used.

24. Solving with const. Coefficients
Basic idea: Look for solutions of the form an = rn, where r is a constant.
This requires the characteristic equation: rn = c1rn−1 + … + ckrn−k, i.e., rk − c1rk−1 − … − ck = 0 (Dividing both sides by rn-k and subtracting right hand side from left).
The solutions (characteristic roots) can yield an explicit formula for the sequence.

25. Solving …… Theorem1: Let c1 and c2 be real numbers.
Suppose that r2 − c1r − c2 = 0 has two distinct roots r1and r2. Then the sequence {an} is a solution of the recurrence relation an = c1an−1 + c2an−2 if and only if an = α1r1n + α2r2n for n≥0, where α1, α2 are constants.

26. Theorem 1: Proof
2 things to prove
Case 1: Roots are r1 & r2, i.e., {an = α1r1n + α2r2n}  an is a solution.
r1 & r2 are roots of r2 − c1r − c2 = 0  r12 = c1r1 + c2; r22 = c1r2 + c2.
 c1an−1 + c2an−2 = c1(α1r1n-1 + α2r2n-1) + c2(α1r1n-2 + α2r2n-2)
 α1r1n-2 (c1r1 + c2) + α2r2n-2 (c1r2 + c2)
 α1r1n-2 r12 + α2r2n-2 r22
 an

27. Theorem 1: Proof … an is a solution  a0 = C0 = α1 + α2.
2 things to prove
Case 2: an is a solution  an = α1r1n + α2r2n for some α1 & α2.
an is a solution  a0 = C0 = α1 + α2.
 C1 = α1r1 + α2r2
 α1 = (C1 – C0r2)/(r1- r2)
 α2 = C0 – α1 = (C0r1 - C1)/(r1- r2)
 Works only if r1≠ r2
an = α1r1n + α2r2n works for 2 initial conditions
 Since the initial conditions uniquely determine the sequence, an = α1r1n + α2r2n

28. Example
Solve the recurrence an = an−1 + 2an−2 given the initial conditions a0 = 2, a1 = 7.
An = rn  rn = rn-1 + 2rn-2  r2 = r +2
Solution: Use theorem 1
c1 = 1, c2 = 2
Characteristic equation: r2 − r − 2 = 0
Solutions: r = [−(−1)±((−1)2 − 4·1·(−2))1/2] / 2·1 = (1±91/2)/2 = (1±3)/2, so r = 2 or r = −1.
So an = α1 2n + α2 (−1)n.

29. Example Continued…
To find α1 and α2, solve the equations for the initial conditions a0 and a1: a0 = 2 = α120 + α2 (−1)0
a1 = 7 = α121 + α2 (−1)1
Simplifying, we have the pair of equations: 2 = α1 + α2
7 = 2α1 − α2 which we can solve easily by substitution:
α2 = 2−α1; 7 = 2α1 − (2−α1) = 3α1 − 2;
9 = 3α1; α1 = 3; α2 = -1.
Final answer: an = 3·2n − (−1)n

30. The Case of Degenerate Roots
Theorem2: Let c1 and c2 be real numbers with c2 ≠ 0. Suppose that r2 − c1r − c2 = 0 has only one root r0.
A sequence {an} is a solution of the recurrence relation an=c1an-1 + c2an-2 if and only if an = α1r0n + α2nr0n, for all n≥0, for some constants α1, α2.

31. k-Linear Homogeneous Recurrence Relations with Constant Coefficients
Theorem3: Let c1,c2,….ck be real numbers.
Suppose the C.E.
If this has k distinct roots ri, then the solutions
to the recurrence are of the form:
if and only if
for all n≥0, where the αi are constants.

32. Theorem 3: Example Let an = 6an-1 – 11an-2 +6an-3; a0=2,a1=5, & a2=15.
C.E.: r3 – 6r2 + 11r – 6; Roots = 1,2, & 3.
Solution: an = α1.1n + α2.2n + α3.3n
a0 = 2 = α1 + α2 + α3
a1 = 5 = α1 + α2.2 + α3.3
a2 = 15 = α1 + α2.4 + α3.9
α1 = 1; α2 = 1; α3 = 2.
an = 1 – 2n + 2.3n

33. Degenerate t-roots
Theorem 4: Suppose there are t roots r1,…,rt with multiplicities m1,…,mt. mi >= 1 for i = 1…t. Then:
for all n≥0, where all the α are constants.

34. Theorem 4: Example E.g., Roots of C.E. are 2, 2, 2, 5, 5, & 9.
Solution:
(α1,0 + α1,1n + α1,2n2).2n + (α2,0 + α2,1n).5n + α3,09n

35. Linear NonHomogeneous Recurrence Relations with Constant Coefficients
Linear NonHomogeneous RRs with constant coefficients may (unlike LiHoReCoCos) contain some terms F(n) that depend only on n (and not on any ai’s). General form:
an = c1an−1 + … + ckan−k + F(n)
The associated homogeneous recurrence relation (associated LiHoReCoCo).

36. Solutions of LiNoReCoCos
Theorem 5: If an(p) is a particular solution to the LiNoReCoCo, then
Then all its solutions are of the form: an = an(p) + an(h) ,
where an(h) is a solution to the associated
homogeneous RR

37. Theorem 5: Proof {an(p)} is a particular solution:
an(p) = c1an-1(p) + c2an-2(p) +..+ckan-k(p) + F(n)  (1)
Let bn be a 2nd solution:
bn = c1bn-1 + c2bn ckbn-k + F(n)  (2)
(2) – (1):
bn – an(p) = c1(bn-1 - an-1(p)) + c2(bn-2 - an-2(p)) +… + ck(bn-k - an-k(p))
Hence, {bn – an(p) } is a solution of the associated homogeneous RR, say {an(h) }.  bn = an(p) + an(h) .

38. Example
Find all solutions to an = 3an−1+2n. Which solution has a1 = 3?
To solve this LiNoReCoCo, solve its associated LiHoReCoCo equation: an = 3an−1, and its solutions are an(h) = α3n, where α is a constant.
By Theorem 5, the solutions to the original problem are all of the form an = an(p) + α3n. So, all we need to do is find one an(p) that works.

39. Trial Solutions
Since F(n)=2n, i.e it is linear so a reasonable trial solution is a linear function in n, say pn = cn + d.
Then the equation an = 3an−1+2n becomes,
cn+d = 3(c(n−1)+d) + 2n, (for all n)
Simplifying, we get
(−2c+2)n + (3c−2d) = 0 (collect terms) So c = −1 and d = −3/2.
So a(p)n = −n − 3/2 is a solution.
3 x -5/2 = -15/2. Plus 2x2=4=8/2 is -7/2.

40. Finding a Desired Solution
From Theorem 5, we know that all general solutions to our example are of the form:
an = −n − 3/2 + α3n.
Solve this for α for the given case, a1 = 3:
3 = −1 − 3/2 + α31
α = 11/6
The answer is an = −n − 3/2 + (11/6)3n

41. Theorem 6 Suppose {an} satisfies LiNoRR:
And F(n)= ; bt and s are real nos.
When s is not a root of C.E. of the associated LiHoRR, there is a particular solution of the form

42. Theorem 6 continue…..
When s is a root of this C.E. and its multiplicity is m, there is a particular solution of the form
This factor nm ensures that the proposed particular solution will not already be a solution of the associated LiHoRR.

43. Theorem 6: Example an = 6an-1 – 9an-2 + F(n); F(n) = 3n
C.E.: r2 – 6r + 9 => (r-3)2 = 0; r = 3.
Solution: By Theorem 6, s = 3, multiplicity m = 2. Solution is of the form p0n23n, where p0 is polynomial constant.

44. §6.3: Divide & Conquer R.R.s
Many types of problems are solvable by reducing a problem of size n into some number a of independent subproblems, each of size n/b, where a1 and b>1.
The time complexity to solve such problems is given by a recurrence relation:
f(n) = a·f(n/b) + g(n)
where g(n) is the extra operations required.
This is called a divide-and-conquer recurrence relation

45. Divide+Conquer Examples
Binary search: Break list into 1 sub-problem (smaller list) (so a=1) of size n/2 (so b=2).
So f(n) = 1.f(n/2)+2 (g(n)=2 constant)
Merge sort: Break list of length n into 2 sublists (a=2), each of size  n/2 (so b=2), then merge them, in g(n) = n operations
So M(n) = 2M(n/2) + n

46. Fast Multiplication Example
This algorithm splits each of two 2n-bit integers into two n bits blocks. Thus, from multiplications of two 2n-bit integers, it is reduced to only three multiplications of n bit integers, plus shifts and additions
To find the product ab of two 2n-digit-base 2 numbers, a=(a2n-1a2n-2…a0)2 and b=(b2n-1b2n-2…b0)2,
first, we break them in half: a=2nA1+A0, b=2nB1+B0,

47. Derivation of Fast Multiplication
(Multiply out
polynomials)
Zero
Three multiplications, each with n-digit numbers
(Factor last polynomial)

48. Recurrence Rel. for Fast Mult.
Notice that the time complexity f(n) of the fast multiplication algorithm obeys the recurrence:
f(2n)=3f(n)+C(n) i.e.,
f(n)=3f(n/2)+C(n)
So a=3, b=2.
Time to do the needed adds &
subtracts of n-digit and 2n-digit
numbers

49. Theorem1
Let f(n) = af(n/b) + c whenever n is divisible by b, where a1, b is an integer greater than 1, and c is a positive real number. Then
if a > 1
if a = 1
Furthermore, when n =bk, where k is a positive integer,
f(n) = C C2.
where C1 = f(1) + c/(a-1) and C2 = -c/(a-1) is

50. Theorem1: Proof n = bk; f(n) = akf(1) +
a = 1: f(n) = f(1) + ck; Since n = bk, k = logbn  f(n) = f(1) + clogbn.
n is a power of b or not: f(n) is O(logn), when a = 1.
a > 1: (from Theorem 1, Section 3.2)
f(n) = akf(1) + c(ak – 1)/(a – 1)
= ak[f(1) + c/(a – 1)] – c/(a – 1)
= C1nlogba + C2, since ak = alogbn = nlogba
 f(n) is O(nlogba)

51. MASTER Theorem
Let f(n) = af(n/b) + cnd whenever n=bk, where a1, b>1, and c and d are real number, with c positive and d nonnegative. Then
if a<bd
f(n) is if a=bd
if a>bd

52. Example
We know that number of comparisons used by the merge sort to sort a list of n elements is less then M(n), where M(n)=2M(n/2) + n
From Master Theorem we find that M(n) is O(nlogn)

53. §6.4 Generating Functions
Generating functions: Used to represent sequences efficiently by coding the terms of a sequence as coefficients of powers of a variable x in a formal power series
Definition: Let S = {a0, a1, a2, a3, ...} be an (infinite) sequence of real numbers. Then the generating function G(x), of S is the series
G(x)=a0+a1x+…+akxk+…=

54. Example of GF Let S = {1, 1, 1, 1, ... }. Then
G(x) =1+ x + x xk +... =
if S=1,1,1,1,1,1
By Theorem 1 of Section 3.2,
(x6-1)/(x-1) = 1+ x + x x5
G(x) =(x6-1)/(x-1) is the generating function of sequence 1,1,1,1,1,1

55. Theorems of GF Theorem 1: Let f(x) = and g(x)= . Then,
f(x) + g(x)= and
f(x)g(x)=

56. Example
A expression for the generating function of the sequence S= {1, 1, 1, 1, 1, 1, ...} is 1/(1- x)
1/(1- x) = 1 + x + x2 + x3 + …
Let f(x) = 1/(1-x)2; f(x) =
What are the coefficients of ak?
From theorem 1, we have (1 + x + x2 + ..) x (1 + x + x2 + ..)  x
 1/(1- x)2 = =

57. Permutations & Combinations
Permutation: An ordered arrangement of objects from a set of distinct objects.
P(n,r): n – distinct objects; r – number of objects in the permutation
P(n,r) = n(n-1)(n-2)…(n-r+1)
 P(n,r) = n!/(n-r)!
Example: Selecting first 3 prize winners from 100 different people  100 x 99 x 98 = 970,200.
Combination: Unordered selection of r elements from a set.
C(n,r) = n!/[r!(n-r)!]; n is non-negative, 0 ≤ r ≤ n
Proof: r-permutation is ordered r-combinations.

58. Permutations & Combinations …
Proof: r-permutation is ordered r-combinations.:
 P(n,r) = C(n,r) x P(r,r)
 C(n,r) = P(n,r) / P(r,r)
 C(n,r) = [n!/(n-r)!]/[r!/(r-r)!]
= n!/[r!(n-r)!]
Corollary 1: C(n,r) = C(n,n-r) if n & r are non-negative and r ≤ n.
Proof: C(n,n-r) = n!/[(n-r)!(n- (n-r))!]
= n!/[(n-r)!r!]
Note: C(n,n) = C(n,0) = 1.
Example: Choosing 5 players from 10 member team.
 C(10,5) = 10!/(5!5!) = 252.

59. Binomial Coefficients & Theorem
C(n,r): denoted by n r
C(n,r): Binomial coefficient  these numbers occur as co-efficients in the expansion of powers of binomial expressions such as (a+b)n
Binomial Theorem: x, y are variables; n – non-negative integer.
(x + y)n = ∑j=0 n C(n,j).xn-j.yj
= C(n,0)xn + C(n,1)xn-1y + C(n,2)xn-2y2
+….C(n,n-1)xyn-1 + C(n,n)yn
Example: (x + y)2 = (x+y)(x+y) = xx + xy + xy + yy = x2 + 2xy + y2

60. Binomial Coefficients & Theorem
Proof: Count the number of terms of form xn-j.yj
 Choose xs from the n sums (so that the other terms are ys)
 Coefficient of xn-jyj is C(n,n-j) = C(n,j)
Example 1: Expansion of (x + y)4
= ∑j=0 4 C(4,j)x4-jyj
= C(4,0)x4 + C(4,1)x3y + C(4,2)x2y2 + C(4,3)xy3 + C(4,4)y4
= x4 + 4x3y + 6x2y2 + 4xy3 + y4
Example 2: Coeffiecient of x12y13 in (x+y)25
= C(25,13) = 25!/(13! 12!) = 5,200,300

61. Binomial Theorem: Corollary 1
Corollary 1: n is non-negative  ∑k=0 n C(n,k) = 2n
Proof: 2n = (1 + 1)n = ∑k=0 n C(n,k)1k1n-k
= ∑k=0 n C(n,k)
 Basically, ∑k=0 n C(n,k) is the total number of subsets of a set with n elements
 2n

62. Binomial Theorem: Corollary 2
Corollary 2: n is positive  ∑k=0 n (-1)kC(n,k) = 0
Proof: 0 = 0n = ((-1) + 1)n = ∑k=0 n C(n,k)(-1)k1n-k
= ∑k=0 n C(n,k)(-1)k
 C(n,0) + C(n,2) + C(n,4) + .. = C(n,1) + C(n,3) + C(n,5) + …

63. Binomial Theorem: Corollary 3
Corollary 3: n is positive  ∑k=0 n 2kC(n,k) = 3n
Proof: 3n = (1+2)n = ∑k=0 n C(n,k)(2)k1n-k
= ∑k=0 n C(n,k)2k

64. Permutations With Repetitions
Chapter 4.5
Theorem 1: Number of r-permutations of a set of n objects with repetitions allowed is nr.
Theorem 2: There are C(n+r-1, r) r-combinations from a set with n elements when repetition of elements is allowed.

65. Extended Binomial Coefficients
Definition: Let u be a real number and k a non-negative integer. Then the extended binomial coefficient (EBC) is:
if k>0,
if k=0.
e.g:EBC(-2,3) = (-2)(-3)(-4) / 3! = -4.
EBC(0.5,3)=(0.5)(-0.5)(-1.5) / 3! = 1/16

66. Example
When the top parameter (u) is negative integer, the EBC can be expressed in terms of ordinary binomial coefficients. Using earlier definition:
EBC = (-n)(-n-1)….(-n-r+1)/r!
= (-1)rn(n+1)…(n+r-1)/r!
= (-1)r(n+r-1)!/(r!*(n-1)!). Therefore,

67. The Extended Binomial Theorem
Let x be a real number with |x| < 1
and let u be a real number. Then
(1+x)u =

68. Example Find GF for (1+x)-n and (1-x)-n By EB Theorem, it follows that
We know that
Replacing x with –x we get
(1-x)-n =

69. Counting Problems and Generating Functions
Generating Functions can be used to solve a wide variety of counting problems. In particular they can be used to count the number of combinations of various types.

70. Example
Use GF to find the number of ways to select r objects of n different kinds {ar} if we must select at least one object of each kind?
Chapter 4, theorem.
Each of the n kinds of objects contributes the factor (x+x2+x3…+) to the GF G(x), Hence
G(x)= (x+x2+x3…+)n=xn(1+ x+x2+x3…+) =xn/(1-x)n
Using EB Theorem, we have
G(x) = xn.(1-x)-n = xn = xn
G(x)=

71. Continue… Substituting t = n+r, we get G(x)=
Replacing t by r as the index of summation,
we get
Hence, there are C(r-1,r-n) ways to select r objects of n different kinds if we must select at least one object of each kind.

72. Using GF to Solve Recurrence Relations
We can find the solution to a recurrence relation and its initial conditions by finding an explicit formula for the associated generating function.

73. Example
Solve the R.R., ak=3ak-1 for k=1,2,3… and initial condition a0=2.
Uses identity1/(1-ax) =
Consequently, ak=2 . 3k

74. §6.5 The principle of Inclusion-Exclusion
If A, B and C are finite sets then
|A  B| = |A| + |B| – |A  B|

75. Continue… If A, B and C are finite sets then
|A  B  C| = |A| + |B| + |C| - |A  B| - |B  C| - |A  C| - |A  B  C |
|A  B|
|A|
|B|
|A  C|
|B  C|
|C|

76. General Inclusion-Exclusion
Theorem 1: Let A1,…,An be finite sets.
Then | A1  A2 …… An | =

77. General Theorem: Proof
| A1  A2 …… An |: Show an element in the union is counted only once in the right hand side.
Let a be a member of exactly r sets (A1, …).
a is counted C(r,1) times by 1st summation, C(r,2) by 2nd summation of the intersections, and C(r,m) times by mth summation involving m sets Ai.
a is counted exactly C(r,1) – C(r,2) + C(r,3) - … + (-1)r+1C(r,r).
By Corollary 2 of Section 4.4, C(r,0) - C(r,1) + C(r,2) - C(r,3) - … + (-1)rC(r,r) = 0.
 C(r,0) = C(r,1) – C(r,2) + C(r,3) - … + (-1)r+1C(r,r).
 a is counted only once.

78. §6.6 Applications of Inclusion-Exclusion
Alternative Form: Let Ai be the subset containing the elements that have Property Pi. Writing these quantities in terms of sets, we have
If the number of elements with none of the properties P1,…,Pn is denoted by N(P1’ P2’…Pn’)
then, N(P1’ P2’…Pn’)=N-

79. Continue From the inclusion-exclusion principle, we see that
N(P1’ P2’…Pn’)=N

80. The Number of Onto Functions
Let m and n be positive integers with (m  n). How many ways to assign m different jobs to n different employees if every employee is assigned at least one job?
Assume m = 5, n = 4, following Theorem 1 (next slide):
45 – C(4,1).35 + C(4,2).25 – C(4,3)15 = 240

81. The Number of Onto Functions
Theorem: Let m and n be positive integers with (m  n). Then, there are
nm - C (n,1)·(n -1)m + C (n,2)·(n -2)m+…
+(-1)iC (n,i )·(n-i )m +…+ (-1)n-1C (n,n-1)·1m
onto functions from a set with m elements to a set with n elements

82. Derangements
A derangement is a permutation of objects that leaves no object in its original position.
Theorem: The number of derangements of a set with n elements is

83. Theorem Proof Property Pi: A permutation that fixes element i.
Number of derangements: number of permutations having none of the properties Pi for i = 1,2,…,n.
Dn = N(P1’,P2’,…,Pn’) (from inclusion-exclusion)
= N -
N: number of permutations of n elements. So, N = P(n).
Similarly: N(Pi) = P(n-1); N(Pi,Pj) = P(n-2).
Inserting these values in the expression for Dn, we get the formula.