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Elementary Counting Techniques & Combinatorics

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1 Elementary Counting Techniques & Combinatorics
Jim Skon

2 Consider: How many license plates are possible with 3 letters followed by 3 digits? How many license plates are possible with 3 letters followed by 3 digits if no letter repeated? How many different ways can we chose from 4 colors and paint 6 rooms?

3 Consider How many different orders may 9 people be arranged in a line?
How many ways can I return your quizzes so that no one gets their own? How many distinct function exist between two given finite sets A and B?

4 Combinatorics the branch of discrete mathematics concerned with determining the size of finite sets without actually enumerating each element.

5 Combinatorics The Sum Rule (task formulation):
Suppose that a task can be completed by performing exactly one task from a collection of disjoint subtasks: subtask1, subtask2, ... , subtaskn; Now suppose each subtask has a choice of ways to perform it, e.g. subtask1 can be performed t1 ways, subtask2 can be performed t2 ways, ... subtaskn can be performed tn ways. Then number the number of ways to perform the task is: t1 + t tn

6 The Sum Rule Example: You have five novels, four magazines, and three devotional books. How many options do you have for taking one for your wait in the bank line? You have 3 subtasks - pick a novel, pick a magazine, or pick a devotional book. The first can be done 5 ways, the second, 4 ways, and the third 3 ways. Thus: = 12 ways to perform.

7 The Sum Rule Example: Suppose either a CS faculty or a CS student must be chosen for a committee, and there are 4 CS faculty and 16 CS students. How many possible choices are there?

8 The Sum Rule Example: Suppose a student can meet the humanities course requirement by taking either a religion, literature, or art course. There are 3 religion, 4 literature, or 4 art courses to chose from. How many possible choices are there?

9 Combinatorics The Product Rule (task formulation):
Suppose a task needs to be done, and the tasks consists of a sequence of n steps or subtasks: task = task1, task2, task3, ..., taskn where each task taskx has a certain number of ways (tx) in which it can be performed after the preceding tasks have been performed, e.g. task1 = t1 ways, task2 = t2 ways after task1 is complete, task3 = t3 ways after task1 and task2 is complete, ... , taskn = tn ways after task1 ... taskn-1 is complete Then the number of ways the task can be performed is: t1 × t2 × t3 × ... × tn

10 Example - Product Rule How many different ways can we chose from 4 colors and paint 3 rooms? Tasks: 1 - paint room ways to perform (4 colors) 2 - paint room ways to perform (4 colors) 3 - paint room ways to perform (4 colors) Thus t1 = 4, t2 = 4, t3 = 4, and 4 × 4 × 4 = 64 ways to paint the rooms

11 Example - Product Rule How many different ways can we chose from 4 colors and paint 3 rooms, if no room is to be the same color? tasks: 1 - paint room ways to perform (4 colors) 2 - paint room ways to perform (3 colors left) 3 - paint room ways to perform (2 colors left) Thus t1 = 4, t2 = 3, t3 = 2, and 4 × 3 × 2 = 24 ways to paint the rooms

12 Example - Product Rule How many different orders may 9 people be arranged in? There are nine tasks - picking the first person, picking the second, … The first task has 9 choices, the second 8, ... and finally the ninth task has 1 choice: 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 =

13 Example - Product Rule How many different 3 people can be selected from a group of 8 people to a president, vice-president, treasure of the group?

14 Example - Product Rule If student ID’s are two capital letters followed by three numeric digits, then how many ID’s are possible? What if the two letters must be distinct? What if the letters and the numbers must all be distinct?

15 Combinatorics The Product Rule (set formulation):
If A and B are finite sets, then: |A ´ B| = |A| × |B| The cardinality of the Cartesian product of two sets is the product of the their cardinalities. Note that ANY ordered list of items is trivially equivalent to a Cartesian product.

16 Example - Product Rule Let A = {a, b, c, d, e}, B = {1, 3, 5, 7}
How many pairs (x, y) exist where x Î A and y Î B? A ´ B has cardinality |A| × |B| = 5 × 4 = 20

17 Example - Product Rule How many license plates are possible with 3 letters followed by 3 digits? 26×26×26×10×10×10 =

18 Combinatorics The Sum Rule (set formulation):
If A and B are disjoint finite sets, then: |A È B| = |A| + |B| The cardinality of the union of two disjoint sets is the sum of their cardinalities.

19 Example - The Sum Rule Let A = {a, b, c, d, e}, B = {1, 3, 5, 7}
How many ways can one element be chosen? |A È B| = |A| + |B| = = 9.

20 Example - The Sum Rule You have five novels, four magazines, and three devotional books. How many options do you have for taking one for your wait in the bank line?

21 Example Suppose that you own 3 pairs of shoes, 6 pairs of socks, 4 pairs of pants, and 6 shirts. How many different outfits can you make out of these articles of clothing? (An outfit consists of one pair of shoes, one pair of socks, one pair of pants, and one shirt.)

22 Example Consider the following road map.
a) How many ways are there to travel from A to B, and back to A, without going through C? b) How many ways are there to go from A to C, stopping once at B? c) How many ways are there to go from A to C, making at most one intermediate stop? A B C

23 Example A certain apartment complex has 26 television antennas. Each pair of apartments shares a common antenna. How many apartments are there in the complex?

24 Example Two cards are drawn from a deck of cards, one at a time. How many outcomes are possible a) if the order in which the cards are draw matters? b) if the order in which the cards are drawn does not matter?

25 Example Suppose that you flip a coin 5 times and record the sequence of heads and tails. How many possibilities are there for this sequence?

26 Example How many integers between 0 and 1,000,000 contain the digit 9?
(Hint, it is easier to count the number of integers that do not contain the digit 9.)

27 The Pigeonhole Principle
If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects.

28 The Pigeonhole Principle
If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects.

29 The Pigeonhole Principle
If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects.

30 The Pigeonhole Principle
If k + 1 or more objects are placed in k boxes, then there is at least one box containing two or more objects. Proof: Suppose that none of the k boxes contains more then one object. Then the maximum number of objects would be k. This is a contradiction, since there is at least k + 1 objects.

31 The Pigeonhole Principle
Among 367 people, there must be at least 2 with the same birthday, since there is only 366 possible birthdays. In a collection of 10 numbers, at least 2 must have the same most significant digit. In a collection of 11 numbers, at least 2 must have the same least significant digit.

32 The Pigeonhole Principle
How many people must we have in the same room to be sure that at least two have the same birthday?

33 The Pigeonhole Principle
Are there two people in Ohio with the same number of hairs? Are there two people at MVNC the same birthday? Are there two people at MVNC with birthdays on July 14?

34 The Pigeonhole Principle
The Generalized Pigeonhole Principle: If N objects are placed into k boxes, then there is at least one box containing at least N/k objects.

35 Generalized Pigeonhole Principle
If N objects are placed into k boxes, then there is at least one box containing at least N/k objects.

36 Generalized Pigeonhole Principle
If N objects are placed into k boxes, then there is at least one box containing at least N/k objects.

37 Generalized Pigeonhole Principle
If N objects are placed into k boxes, then there is at least one box containing at least N/k objects.

38 Generalized Pigeonhole Principle
Proof: Suppose that none of the boxes contains more than N/k - 1 objects. Then the total number of objects is at most: k (N/k - 1). But since N/k < (N/k + 1), we get the following: k (N/k - 1) < k (((N/k + 1) - 1) = N, thus k (N/k - 1) < N which is a contradiction since there is a total of N objects.

39 Generalized Pigeonhole Principle
Among 100 people there are at least 100/12 = 9 people with the same birthday month.

40 Generalized Pigeonhole Principle
In MVNC there are at least 1500/366 = 5 people with the same birthday.

41 Generalized Pigeonhole Principle
In a class of 44 students, how many will receive the same grade on a scale {A, B, C, D, F}.

42 Generalized Pigeonhole Principle
How many people must we survey, to be sure at least 50 have the same political party affiliation, if we use the three affiliations {Democrat, Republican, neither}? (Make N/3 = 50)

43 Generalized Pigeonhole Principle
What is the least number of area codes needed to guarantee that 25,000,000 phones in a state have distinct 10 digit numbers. (Assume that telephone numbers are of the form NXX-NXX-XXXX, where N is a a digit from 2-9, and X represents any digit.)

44 Generalized Pigeonhole Principle
Let n be a positive integer. Show that in any set of n consecutive integers there is exactly one divisible by n.

45 Generalized Pigeonhole Principle
A computer network consists of 6 computers. Each computer is is directly connected with zero or more of the other computers. Show that there is at least two computers with the same numbers of connections.

46 Generalized Pigeonhole Principle
Show that if seven integers are selected from the first eight positive integers, there must be pair of these integers with a sum equal to to 9. Is this still true if four integers are chosen instead of five?

47 Generalized Pigeonhole Principle
What is the minimum number of students MVNC must have to be assured that at least 40 come from the same state?

48 Generalized Pigeonhole Principle
How many students must attend MVNC to assure use that at least two have the same 3 letter initials?

49 Permutations and Combinations
Consider: How many ways can we choose r things from a collection of n things? pick Pick 4 from 9 colored balls

50 Permutations and Combinations
Consider: How many ways can we choose r things from a collection of n things? This statement is ambiguous in several ways: Are the n things distinct or indistinguishable? Do the selected items form a set (unordered collection) or a sequence (ordered)? May the same item be selected from the r items more then once? (Are repetitions permitted?).

51 Permutations and Combinations
Example using balls: Are the balls identical or different colors? Are some different colors, others the same? Are balls tossed in a bucket (unordered) or lined up in a line in the order chosen? Each ball returned to the collection before the next is selected?

52 Permutation An ordered selection of objects.
If there is a collection of n objects to chose from, and we are selecting all n objects, then we call each possible selection a permutation from the collection. In the general case the items are all distinct, and repetitions are not permitted.

53 Permutation Possible permutations of three colored balls:

54 Permutation Consider the set S = {a, b}. What are the permutations?
ab ba

55 Permutation Consider the set S = {a, b, c}. What are the permutations?
abc acb bca bac cab cba

56 Permutation abcd acbd bcad bacd cabd cbad
Consider the set S = {a, b, c, d}. What are the permutations? abcd acbd bcad bacd cabd cbad abdc acdb bcda badc cadb cbda adbc adcb bdca bdac cdab cdba dabc dacb dbca dbac dcab dcba

57 Permutation Theorem: The number of permutations of a set of n objects is the product of the first n positive integers, that is n(n -1) = n

58 Permutation Justification:
Arranging n objects into order requires n tasks. Task 1 Pick first object (n choices) Task 2 Pick second object (n-1 choices) ... Task n Pick last (nth) object (1 choice) Thus, by the product rule, the number of ways to arrange n objects is: n(n -1) = n!

59 Permutation How many ways are there to arrange 9 floats in the Christmas parade? 9  8  7  6  5  4  3  2  1 =

60 r-Permutations Consider ordering a subset of a collection of objects.
If there is a collection of n objects to chose from, and we are selecting r of the objects, where 0<rn, then we call each possible selection a r-permutation from the collection.

61 r-Permutations Consider a 4-permutation of 9 balls pick

62 r-Permutations Consider the set S = {a, b, c}.
What are the 2-permutations of S? ab ba ac ca bc cb What are the 3-permutations of S? abc acb bca bac cab cba

63 ab ac ad ba bc bd ca cb cd da db dc r-Permutations
Consider the set S = {a, b, c, d}. What are the 2-permutations of S? ab ac ad ba bc bd ca cb cd da db dc

64 r-Permutations Consider the set S = {a, b, c, d}.
What are the 3-permutations of S? abc acb bac bca cba cab (from {a, b, c}) abd adb bad bda dba dab (from {a, b, d}) adc acd dac dca cda cad (from {a, c, d}) dbc dcb bdc bcd cbd cdb (from {b, c, d})

65 r-Permutations Theorem: The number of r-permutations of a set of n objects, written P(n, r) is:

66 r-Permutations Justification:
Arranging r of n objects into order requires r tasks. Task 1 Pick first object (n choices) Task 2 Pick second object (n-1 choices) Task r Pick rth object (n - r + 1 choices) Thus, by the product rule, the number of ways to arrange n objects is:

67 r-Permutations Consider a horse race with 8 horses.
If a spectator were select three different horses at random to bet on for first, second and third places, how likely is he to be completely correct? P(8,3) = 8  7  6 = 336 permutations possible Thus he has a 1 in 336 chance.

68 r-Permutations To send secret messages, two ships have three flagpoles and 10 different flags (one flag per pole). If a set of three flags indicates a message, how many messages are possible? If 0, 1, 2, or 3 flags all represent messages?

69 r-Permutations Suppose license plates for some state contain 3 letters followed by 3 numbers, where no number or letter can be repeated? How many possibilities are there? If we consider all possibilities where repetitions are allowed, how many have at least one repeated number or letter?

70 r-Permutations How many ways can we hand out 7 different free books to 15 students (where each gets exactly one book)?

71 Combinations Combination - an unordered selection of objects. Definition: Consider a set S with n objects. Every k sized subset of those objects (0<rn) is a combination of size r, or a r-combination taken from S.

72 Combinations {a, b} {a, c} {b, c} {a, b, c} {a} {b} {c}
Consider the set A = {a, b, c}. What are the 2-combinations of A? {a, b} {a, c} {b, c} What are the 3-combinations of A? {a, b, c} What are the 1-combinations of A? {a} {b} {c}

73 Combinations Consider the set B = {a, b, c, d}.
What are the 2- combinations of B? {a, b} {a, c} {a,d} {b, c} {b, d} {c, d} What are the 3-combinations of B? {a, b, c} {a, c, d} {b, c, d} {a, b, d}

74 Combinations Notice the comparison of 3-combinations of B with 3-permutations: 3-permutations combinations abc acb bac bca cba cab {a, b, c} abd adb bad bda dba dab {a, b, d} adc acd dac dca cda cad {a, c, d} dbc dcb bdc bcd cbd cdb {b, c, d}

75 Combinations This shows that each r-combination has r-permutations possible. Thus the following theorem: Theorem: The number of r -combinations of a set of n distinct objects is:

76 Combinations Justification: We find the number of permutations, then divide by a factor which we have overcounted. Since each r-combination of the n objects can be ordered P(r,r) = r! ways, we divide the number of r-permutations from n objects by the number of r-permutations of r objects.

77 Combinations The number C(n,r) is also written:
And is be read n choose r objects.

78 Combinations How many subsets of size 5 are there of the set {1, 2, 3, ..., 10}?

79 Combinations How many subsets of size 7 are there of the set {1, 2, 3, ..., 10}? How many subsets of size 3? How many subsets of size 2? How many subsets of size 8?

80 Combinations How many different 5-card hands can be made from a deck of 52 cards?

81 Combinations A certain club has 5 male and 7 female members.
How many ways are there to form a 7 member committee consisting of 3 men and 4 women? Two tasks - pick a man, then pick a woman. Thus: C(7,4) · C(5,3) How many ways can any 7 member committee be selected from the membership?

82 Combinations How many ternary words of length 12 have exactly seven 0's, three 1's, and two 2's? Solution: Here we can think in terms of picking slots to put numbers in. We thus have three tasks: Task 1 - pick 7 of 12 places to put 0's Task 2 - pick 3 of 5 places to put 1's Task 1 - pick 2 of 2 places to put 2's Thus: C(12,7)·C(5,3)·C(2,2)

83 Combinations How many subsets of size 2 are there of {1, 2, ..., 20} which do not consist of two consecutive integers? Solution: Count the number of subsets which do contain two consecutive integers, then subtract from total number of subsets of size two. Clearly 19 subset contain two consecutive integers, e.g. {1,2}, {2,3}, {3,4}, ... , {19,20}. Thus: C(20,2) - 19.

84 Combinations How many bytes contain exactly four 1's?
Solution: C(8,4).

85 Combinations A certain club has 5 men and 6 women.
How many ways are there to form a committee of 3 people? How many ways are there to form a committee consisting of 3 men and 4 woman? How many ways are there to form a committee of 6 people if 2 woman refuse to serve together? How many ways are there to form a committee of 4 men and 3 woman if 2 men refuse to serve together?

86 Combinations Solution:
How many ways are there to form a committee of 3 people? C(11,3) How many ways are there to form a committee consisting of 3 men and 4 woman? C(5,3)·C(6,4) How many ways are there to form a committee of 6 people if 2 woman refuse to serve together? C(11,6)-C(9,4) How many ways are there to form a committee of 4 men and 3 woman if 2 men refuse to serve together? (C(5,4)-C(3,2))·C(6,3)


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