Climbing an Infinite Ladder

1
Climbing an Infinite Ladder

• Suppose we have an infinite ladder
• We can reach the first rung of the ladder.
• If we can reach a particular rung of the
  ladder,
• then we can reach the next rung.

From (1), we can reach the first rung. Then by
applying (2), we can reach the second rung.
Applying (2) again, the third rung. And so on.
We can apply (2) any number of times to reach any
particular rung, no matter how high up.
This example motivates proof by mathematical
induction.
2
Principle of Mathematical Induction

• Principle of Mathematical Induction To
  prove that P(n) is true for all positive integers
  n, we complete these steps
• Basis Step Show that P(1) is true.
• Inductive Step Show that P(k) ? P(k 1) is
  true for all positive integers k.
• To complete the inductive step, assuming the
  inductive hypothesis that P(k) holds for an
  arbitrary integer k, show that must P(k 1) be
  true.
• Climbing an Infinite Ladder Example
• BASIS STEP By (1), we can reach rung 1.
• INDUCTIVE STEP Assume the inductive hypothesis
  that we can reach rung k. Then by (2), we can
  reach rung k 1.
• Hence, P(k) ? P(k 1) is true for all
  positive integers k. We can reach every rung on
  the ladder.

3
Important Points About Using Mathematical
Induction

• Mathematical induction can be expressed as the
  rule of inference
• where the domain is the set of positive
  integers.
• In a proof by mathematical induction, we dont
  assume that P(k) is true for all positive
  integers! We show that if we assume that P(k) is
  true, then P(k 1) must also be true.
• Proofs by mathematical induction do not always
  start at the integer 1. In such a case, the basis
  step begins at a starting point b where b is an
  integer.

(P(1) ? ?k (P(k) ? P(k 1))) ? ?n P(n),
4
Proving a Summation Formula by Mathematical
Induction

• Example Show that
• Solution
• BASIS STEP P(1) is true since 1(1 1)/2 1.
• INDUCTIVE STEP Assume true for P(k).
• The inductive hypothesis is
• Under this assumption,

Note Once we have this conjecture, mathematical
induction can be used to prove it correct.
5
Proving Inequalities

• Example Use mathematical induction to prove
  that n lt 2n for all positive integers n.
• Solution Let P(n) be the proposition that n lt
  2n.
• BASIS STEP P(1) is true since 1 lt 21 2.
• INDUCTIVE STEP Assume P(k) holds, i.e., k lt 2k,
  for an arbitrary positive integer k.
• Must show that P(k 1) holds. Since by the
  inductive hypothesis, k lt 2k, it follows that
• k 1 lt 2(k 1) 2k 21 2 2k
  2k1
• Therefore n lt 2n holds for all positive
  integers n.

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7
Strong Induction and Well-Ordering

• Section 5.2

8
Strong Induction

• Strong Induction To prove that P(n) is true for
  all positive integers n, where P(n) is a
  propositional function, complete two steps
• Basis Step Verify that the proposition P(1) is
  true.
• Inductive Step Show the conditional statement
  P(1) ? P(2) ? ? P(k) ? P(k 1)
  holds for all positive integers k.

Strong Induction is sometimes called the second
principle of mathematical induction or complete
induction.
9
Strong Induction and the Infinite Ladder

• Strong induction tells us that we can reach all
  rungs if
• We can reach the first rung of the ladder.
• For every integer k, if we can reach the first k
  rungs,
• then we can reach the (k 1)st rung.

• To conclude that we can reach every rung by
  strong induction
• BASIS STEP P(1) holds
• INDUCTIVE STEP Assume P(1) ? P(2) ? ? P(k)
• holds for an arbitrary integer k, and show
  that
• P(k 1) must also hold.
• We will have then shown by strong induction that
  for every positive integer n, P(n) holds, i.e.,
  we can
• reach the nth rung of the ladder.

10
Which Form of Induction Should Be Used?

• We can always use strong induction instead of
  mathematical induction. But there is no reason to
  use it if it is simpler to use mathematical
  induction.
• In fact, the principles of mathematical
  induction, strong induction, and the
  well-ordering property are all equivalent.
• Sometimes it is clear how to proceed using one of
  the three methods, but not the other two.

11
Proof using Strong Induction

• Example
• Prove that every amount of postage of 12 cents
  or more can be formed using just 4-cent and
  5-cent stamps.

12
Proof using Strong Induction

• Solution
• Let P(n) be the proposition that postage of n
  cents can be formed using 4-cent and 5-cent
  stamps.
• BASIS STEP P(12), P(13), P(14), and P(15) hold.
• P(12) uses three 4-cent stamps.
• P(13) uses two 4-cent stamps and one 5-cent
  stamp.
• P(14) uses one 4-cent stamp and two 5-cent
  stamps.
• P(15) uses three 5-cent stamps.
• INDUCTIVE STEP
• The inductive hypothesis states that P(j) holds
  for 12 j k, where k 15. Assuming the
  inductive hypothesis, it can be shown that P(k
  1) holds.
• Using the inductive hypothesis, P(k - 3) holds
  since k - 3 12. To form postage of k 1
  cents, add a 4-cent stamp to the postage for k -
  3 cents.
• Hence, P(n) holds for all n 12.

13
Well-Ordering Property

• Well-ordering property Every nonempty set of
  nonnegative integers has a least element.
• The well-ordering property can be used directly
  in proofs, as the next example illustrates.
• Definition A set is well ordered if every subset
  has a least element.
• N is well ordered under .
• The set of finite strings over an alphabet using
  lexicographic ordering is well ordered.
• We will see a generalization of induction to sets
  other than the integers in the next section.

14
Definitions

• The principle of mathematical induction the
  statement ?n P(n), is true if P(1) is true and ?k
  (P(k) ? P(k 1)) is true.
• Basis step the proof of P(1) in a proof by
  mathematical induction of ?n P(n).
• Inductive step the proof of P(k) ? P(k 1) for
  all positive integers k in a proof by
  mathematical induction of ?n P(n).
• Strong induction the statement ?n P(n) is true
  if P(1) is true and ?k (P(1) ? P(2) ? ? P(k))
  ? P(k 1) .
• Well-ordering property every nonempty set of
  nonnegative integers has a least element.

15
Recursively Defined Functions

• Definition A recursive or inductive
  definition of a function consists of two steps.
• BASIS STEP Specify the value of the function at
  zero.
• RECURSIVE STEP Give a rule for finding its value
  at an integer from its values at smaller
  integers.
• A function f(n) is the same as a sequence a0,
  a1, , where ai, where f(i) ai.

16
Recursively Defined Functions

• Example
• Suppose f is defined by
• f(0) 3,
• f(n 1) 2f(n) 3
• Find f(1), f(2), f(3), f(4)
• Solution
• f(1) 2f(0) 3 23 3 9
• f(2) 2f(1) 3 29 3 21
• f(3) 2f(2) 3 221 3 45
• f(4) 2f(3) 3 245 3 93

17
Recursively Defined Functions

• Example
• Give a recursive definition of the factorial
  function n!
• Solution
• f(0) 1
• f(n 1) (n 1) f(n)

18
Recursively Defined Functions

• Example
• Give a recursive definition of
• Solution
• The first part of the definition is
• The second part is

19
Recursively Defined Sets and Structures

• Recursive definitions of sets have two parts
• The basis step specifies an initial collection of
  elements.
• The recursive step gives the rules for forming
  new elements in the set from those already known
  to be in the set.
• Sometimes the recursive definition has an
  exclusion rule, which specifies that the set
  contains nothing other than those elements
  specified in the basis step and generated by
  applications of the rules in the recursive step.
• We will always assume that the exclusion rule
  holds, even if it is not explicitly mentioned.
• We will later develop a form of induction, called
  structural induction, to prove results about
  recursively defined sets.

20
Structural Induction and Binary Trees

• Theorem
• If T is a full binary tree, then n(T)
  2h(T)1 1.
• Proof
• Use structural induction.
• BASIS STEP The result holds for a full binary
  tree consisting only of a root, n(T) 1 and h(T)
  0. Hence, n(T) 1 201 1 1.
• RECURSIVE STEP Assume n(T1) 2h(T1)1 1 and
  also n(T2) 2h(T2)1 1 whenever T1 and T2 are
  full binary trees.

• n(T) 1 n(T1) n(T2)
  (by recursive formula of n(T))
• 1 (2h(T1)1 1) (2h(T2)1 1)
  (by inductive hypothesis)
• 2max(2h(T1)1, 2h(T2)1 ) 1
• 22max(h(T1),h(T2))1 1
  (max(2x, 2y) 2max(x,y) )
• 22h(t) 1
  (by recursive definition of h(T))
• 2h(t)1 1

-2 .
21
Generalized Induction

• Generalized induction is used to prove results
  about sets other than the integers that have the
  well-ordering property.
• For example, consider an ordering on N ? N,
  ordered pairs of nonnegative integers.
• Specify that (x1, y1) is less than or equal to
  (x2, y2) if either x1 lt x2, or x1 x2 and y1
  lty2 .
• This is called the lexicographic ordering.
• Strings are also commonly ordered by a
  lexicographic ordering.
• The next example uses generalized induction to
  prove a result about ordered pairs from N ? N.

22
Generalized Induction

• Example
• Suppose that am,n is defined for (m,n)?N N
  by a0,0 0 and
• Show that am,n m n(n 1)/2 is defined
  for all (m,n)?N N.
• Solution
• Use generalized induction.
• BASIS STEP a0,0 0 0 (01)/2
• INDUCTIVE STEP Assume that am?,n? m? n?(n?
  1)/2 whenever(m?,n?) is less than (m,n) in the
  lexicographic ordering of N N .
• If n 0, by the inductive hypothesis we can
  conclude
• am,n am-1,n 1 m - 1 n(n 1)/2
  1 m n(n 1)/2 .
• If n gt 0, by the inductive hypothesis we can
  conclude
• am,n am-1,n 1 m n(n - 1)/2
  n m n(n 1)/2 .

-2 .
23
Recursive Algorithms

• Definition
• An algorithm is called recursive if it solves a
  problem by reducing it to an instance of the same
  problem with smaller input.
• For the algorithm to terminate, the instance of
  the problem must eventually be reduced to some
  initial case for which the solution is known.

24
Recursive Factorial Algorithm

• Example
• Give a recursive algorithm for computing n!,
  where n is a nonnegative integer.
• Solution
• Use the recursive definition of the factorial
  function.

• procedure factorial(n nonnegative integer)
• if n 0 then return 1
• else return nfactorial(n - 1)
• output is n!

25
Recursive Binary Search Algorithm

• Example Construct a recursive version of a
  binary search algorithm.
• Solution Assume we have a1,a2,, an, an
  increasing sequence of integers. Initially i is 1
  and j is n. We are searching for x.

• procedure binary search(i, j, x integers, 1 i
  j n)
• m ?(i j)/2?
• if x am then
• return m
• else if (x lt am and i lt m) then
• return binary search(i,m-1,x)
• else if (x gt am and j gtm) then
• return binary search(m1,j,x)
• else return 0
• output is location of x in a1, a2,,an if it
  appears, otherwise 0

26
Proving Recursive Algorithms Correct

• Both mathematical and strong induction are
  useful techniques to show that recursive
  algorithms always produce the correct output.
• Example Prove that the algorithm for computing
  the powers of real numbers is correct.
• Solution Use mathematical induction on the
  exponent n.
• BASIS STEP a0 1 for every nonzero real
  number a, and power(a,0) 1.
• INDUCTIVE STEP The inductive hypothesis is
  that power(a,k) ak, for all a ?0. Assuming the
  inductive hypothesis, the algorithm correctly
  computes ak1,
• Since power(a,k 1) a power (a, k)
• a ak
  ak1 .

procedure power(a nonzero real number, n
nonnegative integer) if n 0 then return 1 else
return a power (a, n - 1) output is an
-2 .
27
Definitions

• Recursive definition of a function a definition
  of a function that specifies an initial set of
  values and a rule for obtaining values of this
  function at integers from its values at smaller
  integers.
• Recursive definition of a set a definition of a
  set that specifies an initial set of elements and
  a rule for obtaining other elements from those in
  the set.
• Structural induction a technique for proving
  results about recursively defined sets.
• Recursive algorithm an algorithm that proceeds
  by reducing a problem to the same problem with
  smaller input.
• Merge sort a sorting algorithm that sorts a list
  by splitting it in two, sorting each of the two
  resulting lists, and merging the results into a
  sorted list.
• Iteration a procedure based on the repeated use
  of operations in a loop.

28
Recursively Defined Functions

• Example
• Give a recursive definition of the factorial
  function n!
• Solution
• f(0) 1
• f(n 1) (n 1) f(n)

29
Recursive Algorithms

• Definition
• An algorithm is called recursive if it solves a
  problem by reducing it to an instance of the same
  problem with smaller input.
• For the algorithm to terminate, the instance of
  the problem must eventually be reduced to some
  initial case for which the solution is known.

30
Recursive Factorial Algorithm

• Example
• Give a recursive algorithm for computing n!,
  where n is a nonnegative integer.
• Solution
• Use the recursive definition of the factorial
  function.

• procedure factorial(n nonnegative integer)
• if n 0
• then return 1
• else
• return nfactorial(n - 1)
• output is n!

31
Recursive Binary Search Algorithm

• Example Construct a recursive version of a
  binary search algorithm.
• Solution Assume we have a1,a2,, an, an
  increasing sequence of integers. Initially i is 1
  and j is n. We are searching for x.

• procedure binary search(i, j, x integers, 1 i
  j n)
• m ?(i j)/2?
• if x am then
• return m
• else if (x lt am and i lt m) then
• return binary search(i,m-1,x)
• else if (x gt am and j gtm) then
• return binary search(m1,j,x)
• else return 0
• output is location of x in a1, a2,,an if it
  appears, otherwise 0

32
Proving Recursive Algorithms Correct

• Both mathematical and strong induction are
  useful techniques to show that recursive
  algorithms always produce the correct output.
• Example Prove that the algorithm for computing
  the powers of real numbers is correct.
• Solution Use mathematical induction on the
  exponent n.
• BASIS STEP a0 1 for every nonzero real
  number a, and power(a, 0) 1.
• INDUCTIVE STEP The inductive hypothesis is
  that power(a,k) ak, for all a ? 0. Assuming the
  inductive hypothesis, the algorithm correctly
  computes ak1,
• Since power(a, k 1) a power (a, k)
• a ak
  ak1 .

procedure power(a nonzero real number, n
nonnegative integer) if n 0 then return 1 else
return a power (a, n - 1) output is an
-2 .
33
The Relationship between (mod m) and mod m
Notations

• The use of mod in a b (mod m) and a mod m
  b are different.
• a b (mod m) is a relation on the set of
  integers.
• In a mod m b, the notation mod denotes a
  function.
• The relationship between these notations is made
  clear in this theorem.
• Theorem 3
• Let a and b be integers, and let m be a positive
  integer.
• Then a b (mod m) if and only if a mod m b
  mod m.

34
Base b Representations

• We can use positive integer b greater than 1 as a
  base, because of this theorem
• Theorem 1
• Let b be a positive integer greater than 1.
• Then if n is a positive integer, it can be
  expressed uniquely in the form
• n akbk ak-1bk-1 . a1b
  a0
• where k is a nonnegative integer, a0,a1,. ak
  are nonnegative integers less than b, and ak? 0.
• The aj, j 0,,k are called the base-b digits
  of the representation.
• The representation of n given in Theorem 1 is
  called the base b expansion of n and is denoted
  by (akak-1.a1a0)b.
• We usually omit the subscript 10 for base 10
  expansions.

35
Binary Expansions

• Most computers represent integers and do
  arithmetic with binary (base 2) expansions of
  integers. In these expansions, the only digits
  used are 0 and 1.
• Example
• What is the decimal expansion of the integer
  that has (101011111)2 as its binary expansion?
• Solution
• (1 0101 1111)2
• 128 027 126 025 124 123
  122 121 120 351.

36
Binary Expansions

• Example
• What is the decimal expansion of the integer
  that has (11011)2 as its binary expansion?
• Solution
• (11011)2
• 1 24 123 022 121 120 27.

37
Octal Expansions

• The octal expansion (base 8) uses the digits
  0,1, 2, 3, 4, 5, 6, 7.
• Example
• What is the decimal expansion of the number with
  octal expansion (7016)8 ?
• Solution
• 783 082 181 680 3598

38
Octal Expansions

• Example
• What is the decimal expansion of the number
  with octal expansion (111)8 ?
• Solution
• 182 181 180 64 8 1 73

39
Hexadecimal Expansions

• The hexadecimal expansion needs 16 digits, but
  our decimal system provides only 10.
• So letters are used for the additional symbols.
  
• The hexadecimal system uses the digits
  0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F.
• The letters A through F represent the decimal
  numbers 10 through 15.
• Example
• What is the decimal expansion of the number with
  hexadecimal expansion (2AE0B)16 ?
• Solution
• 2164 10163 14162 0161 11160
  175627

40
Hexadecimal Expansions

• Example
• What is the decimal expansion of the number
  with hexadecimal expansion (1E5)16 ?
• Solution
• 1162 14161 5160 256 224 5 485

41
Base Conversion

• To construct the base b expansion of an integer
  n
• Divide n by b to obtain a quotient and remainder.
• n bq0 a0 0 a0 b
• The remainder, a0 , is the rightmost digit in the
  base b expansion of n. Next, divide q0 by b.
• q0 bq1 a1 0 a1 b
• The remainder, a1, is the second digit from the
  right in the base b expansion of n.
• Continue by successively dividing the quotients
  by b, obtaining the additional base b digits as
  the remainder. The process terminates when the
  quotient is 0.

continued ?
42
Base Conversion

• Example
• Find the octal expansion of (12345)10
• Solution
• Successively dividing by 8 gives
• 12345 8 1543 1
• 1543 8 192 7
• 192 8 24 0
• 24 8 3 0
• 3 8 0 3
• The remainders are the digits from right to
  left yielding (30071)8.

43
Comparison of Hexadecimal, Octal, and Binary
Representations
Initial 0s are not shown
Each octal digit corresponds to a block of 3
binary digits. Each hexadecimal digit corresponds
to a block of 4 binary digits. So, conversion
between binary, octal, and hexadecimal is easy.
44
Conversion Between Binary, Octal, and Hexadecimal
Expansions

• Example
• Find the octal and hexadecimal expansions of
  (11 1110 1011 1100)2.
• Solution
• To convert to octal, we group the digits into
  blocks of three (011 111 010 111 100)2, adding
  initial 0s as needed. The blocks from left to
  right correspond to the digits 3, 7, 2, 7, and 4.
  Hence, the solution is (37274)8.
• To convert to hexadecimal, we group the digits
  into blocks of four (0011 1110 1011 1100)2,
  adding initial 0s as needed.
• The blocks from left to right correspond to the
  digits 3, E, B, and C. Hence, the solution is
  (3EBC)16.

45
Primes

• Definition
• A positive integer p greater than 1 is called
  prime if the only positive factors of p are 1 and
  p.
• A positive integer that is greater than 1 and is
  not prime is called composite.
• Example
• The integer 7 is prime because its only positive
  factors are 1 and 7, but 9 is composite because
  it is divisible by 3.

46
The Fundamental Theorem of Arithmetic

• Theorem
• Every positive integer greater than 1 can be
  written uniquely as a prime or as the product of
  two or more primes where the prime factors are
  written in order of nondecreasing size.
• Examples
• 100 2 2 5 5 22 52
• 641 641
• 999 3 3 3 37 33 37
• 1024 2 2 2 2 2 2 2 2 2 2
  210

47
Greatest Common Divisor

• Definition
• Let a and b be integers, not both zero.
• The largest integer d such that d a and also d
  b is called the greatest common divisor of a
  and b.
• The greatest common divisor of a and b is
  denoted by gcd(a, b).
• One can find greatest common divisors of
  small numbers by inspection.
• Example
• What is the greatest common divisor of 24 and
  36?
• Solution
• gcd(24, 26) 12
• Example
• What is the greatest common divisor of 17 and
  22?
• Solution
• gcd(17, 22) 1

48
Greatest Common Divisor

• Definition
• The integers a and b are relatively prime if
  their greatest common divisor is 1.
• Example
• 17 and 22

49
Greatest Common Divisor

• Definition
• The integers a1, a2, , an are pairwise
  relatively prime if gcd(ai, aj) 1 whenever 1
  iltj n.
• Example
• Determine whether the integers 10, 17, and 21
  are pairwise relatively prime.
• Solution
• Because gcd(10, 17) 1, gcd(10, 21) 1, and
  gcd(17,21) 1, 10, 17, and 21 are pairwise
  relatively prime.

50
Greatest Common Divisor

• Example
• Determine whether the integers 10, 19, and 24
  are pairwise relatively prime.
• Solution
• Because gcd(10, 24) 2, 10, 19, and 24 are
  not pairwise relatively prime.

51
Least Common Multiple

• Definition
• The least common multiple of the positive
  integers a and b is the smallest positive
  integer that is divisible by both a and b.
• It is denoted by lcm(a,b).
• The least common multiple can also be computed
  from the prime factorizations.
• This number is divided by both a and b and no
  smaller number is divided by a and b.
• Example
• lcm(233572, 2433) 2max(3,4) 3max(5,3)
  7max(2,0) 24 35 72
• The greatest common divisor and the least
  common multiple of two integers are related by
• Theorem 5
• Let a and b be positive integers. Then ab
  gcd(a,b) lcm(a,b)
• (proof is Exercise 31)

52
Euclidean Algorithm
Euclid (325 B.C.E. 265 B.C.E.)

• The Euclidian algorithm is an efficient method
  for computing the greatest common divisor of two
  integers.
• It is based on the idea that gcd(a,b) is equal
  to gcd(a,c) when a gt b and c is the remainder
  when a is divided by b.
• Example Find gcd(91, 287)
• 287 91 3 14
• 91 14 6 7
• 14 7 2 0
• gcd(287, 91) gcd(91, 14) gcd(14, 7) 7

Divide 287 by 91
Divide 91 by 14
Divide 14 by 7
Stopping condition
continued ?
53
Euclidean Algorithm

• The Euclidean algorithm expressed in pseudocode
  is
• The time complexity of the algorithm is O(log b),
  where a gt b.

• procedure gcd(a, b positive integers)
• x a
• y b
• while y ? 0
• r x mod y
• x y
• y r
• return x gcd(a,b) is x

54
Cryptographic Protocols Key Exchange

• Cryptographic protocols are exchanges of messages
  carried out by two or more parties to achieve a
  particular security goal.
• Key exchange is a protocol by which two parties
  can exchange a secret key over an insecure
  channel without having any past shared secret
  information.

55
Product Rule

• The Product Rule
• A procedure can be broken down into a sequence
  of i tasks.
• There are n1 ways to do the first task
• and n2 ways to do the second task
• and n ways to do the th task
• and ni ways to do the ith task
• Then there are n1 n2 ni ways to do the
  procedure.

56
The Product Rule

• Example
• How many different license plates can be made
  if each plate contains a sequence of three
  uppercase English letters followed by three
  digits?
• Solution
• By the product rule, there are 26 26 26
  10 10 10
• n1 n2 n3 n4 n5 n6
• 17,576,000 different possible license
  plates.

57
Basic Counting Principles The Sum Rule

• The Sum Rule
• If a task can be done either in one of n1 ways
  or in one of n2 ways to do the task, where none
  of the set of n1 ways is the same as any of the
  n2 ways, then there are n1 n2 ways to do the
  task.
• Example
• The mathematics department must choose either a
  student or a faculty member as a representative
  for a university committee.
• How many choices are there for this
  representative if there are 37 members of the
  mathematics faculty and 83 mathematics majors and
  no one is both a faculty member and a student.
• Solution
• By the sum rule it follows that there are 37
  83 120 possible ways to pick a representative.

58
Counting Functions

• Counting Functions
• How many functions are there from a set with m
  elements to a set with n elements?
• Questions to ask oneself
• What is a function? One item from domain to
  one in codomain.
• How many elements in domain in codomain?
• For each item in the domain, how many choices
  can you make in the codomain?

59
Combining the Sum and Product Rule

• Example
• Suppose statement labels in a programming
  language can be either a single letter or a
  letter followed by a digit.
• Find the number of possible labels.
• Solution
• Use the product rule.
• 26 26 10 286

60
Counting Functions

• Solution
• Since a function represents a choice of one of
  the n elements of the codomain for each of the m
  elements in the domain, the product rule tells us
  that there are n n n nm such
  functions.
• nm codomaindomain (function)

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• Each function is a unique choice from an element
  in m to an element in n.
• But each choice in m may select the any
  element in n.

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Choices

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Other questions

• Other questions to ask
• Is it a combination?
• what is a combination?
• 2 types
• combination of product and sum rules
• combination of r permutations
• Is it a permutation?
• Is repetition allowed?
• What is repetition?
• Are the objects distinguishable?
• Example the S in word SUCCESS. (combination
  of 4 permutations)
• C(7,3)C(4,2)C(2,1)C(1,1) 420

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Is it a combination?

• Example
• Suppose statement labels in a programming
  language can be either a single letter or a
  letter followed by a digit.
• Find the number of possible labels.
• Solution
• Use the product rule.
• 26 26 10 286

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Is it a permutation?

• A permutation of a set of distinct objects is an
  ordered arrangement of these objects.
• Example Let S 1,2,3.
• The ordered arrangement 3,1,2 is a permutation of
  S.
• Example How many ways are there to select a
  first-prize winner, a second prize winner, and a
  third-prize winner from 100 different people who
  have entered a contest?
• P(100,3) 100 99 98 970,200

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r-permuation

• An ordered arrangement of r elements of a set
  is called an r-permuation.
• Example Let S 1,2,3.
• The ordered arrangement 3, 2 is a 2-permutation
  of S.

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A Formula for the Number of Permutations

• The number of r-permuatations of a set with n
  elements is denoted by P(n,r).
• What is n? The number of elements in a set.
• What is r? The number of elements selected.

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Combinations

• An r-combination of elements of a set is an
  unordered selection of r elements from the set.
• Thus, an r-combination is simply a subset of
  the set with r elements.
• The number of r-combinations of a set with n
  distinct elements is denoted by C(n, r). (Also
  (n!/(n-r)!)
• The notation is also used and is called a
  binomial coefficient.

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Combinations

• Example Let S be the set a, b, c, d.
• Then a, c, d is a 3-combination from S.
• It is the same as d, c, a since the order
  listed does not matter.
• C(4,2) 6 because the 2-combinations of a, b,
  c, d are the six subsets a, b, a, c, a, d,
  b, c, b, d, and c, d.

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Combinations

• Example How many poker hands of five cards can
  be dealt from a standard deck of 52 cards?
• Also, how many ways are there to select 47
  cards from a deck of 52 cards?
• Solution Since the order in which the cards
  are dealt does not matter, the number of five
  card hands is
• (n 52, r 5)

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Combinations

• C(n, r) C(n, n - r)

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Permutations with Repetition

• The number of r-permutations of a set of n
  objects with repetition allowed is nr.
• Example How many strings of length r can be
  formed from the uppercase letters of the English
  alphabet?
• Solution The number of such strings is 26r,
  which is the number of r-permutations of a set
  with 26 elements.

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Combinations with Repetition

• Choice
• Example How many ways are there to select five
  bills from a box containing at least five of
  each of the following denominations 1, 2, 5,
  10, 20, 50, and 100? (7 ways)
• n choices 7 (the demominations) are
  indistinguishable!
• r ways 5 (how many bills).
• (must have enough bills to allow choice of any
  5)

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Indistinguishable Object Combinations

• C(n r 1,r) C(n r 1, n 1)
• C(ways choices - 1, ways)
• C(7 5 1, 5) C(7 5 1, 7 1)
• C(11, 5) C(11, 6)
• 11!/(11-5)!5!
• 11!/6!5!

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Indistinguishable Object Combinations

• Example Suppose that a cookie shop has four
  different kinds of cookies. How many different
  ways can six cookies be chosen?
• Solution The number of ways to choose six
  cookies is the number of 6-combinations of a set
  with four elements.
• C(n r 1, r) C(6 4 -1, 4) C(9, 4)
• 9!/4!5!
• is the number of ways to choose six cookies
  from the four kinds.
• Incorrect