Logic and Introduction to Sets Chapter 6 Dr.Hayk Melikyan/ Department of Mathematics and CS/ Basic Counting Principles 6.3 Basic Counting Principles. In this section, we will see how set operations play an important role in counting techniques.

Sets. Operations with sets and their properties.( union, intersection and complement ) 1. Let A be a set with finitely many elements. Then n(A) denotes the number of elements in A. 2. ADDITION PRINCIPLE (for counting) For any two sets A and B, n(A  B) = n(A) + n(B) - n(A  B) For any two sets A and B, n(A  B) = n(A) + n(B) - n(A  B) If A and B are disjoint, i.e., if A  B = , then n(A  B) = n(A) + n(B) If A and B are disjoint, i.e., if A  B = , then n(A  B) = n(A) + n(B)

Opening example To see how sets play a role in counting, consider the following example. To see how sets play a role in counting, consider the following example. In a certain class, there are 23 majors in Psychology, 16 majors in English and 7 students who are majoring in both Psychology and English. If there are 50 students in the class, how many students are majoring in neither of these subjects? In a certain class, there are 23 majors in Psychology, 16 majors in English and 7 students who are majoring in both Psychology and English. If there are 50 students in the class, how many students are majoring in neither of these subjects? B) How many students are majoring in Psychology alone? B) How many students are majoring in Psychology alone?

Solution: We introduce the following principle of counting that can be illustrated using a Venn- Diagram. We introduce the following principle of counting that can be illustrated using a Venn- Diagram. N( A U B) = n(A) + n(B) – n(A B) N( A U B) = n(A) + n(B) – n(A B) This statement says that the number of elements in the union of two sets A and B is the number of elements of A added to the number of elements of B minus the number of elements that are in both A and B. This statement says that the number of elements in the union of two sets A and B is the number of elements of A added to the number of elements of B minus the number of elements that are in both A and B. A B

Venn Diagrams Venn diagrams are useful tools for determining the number of elements in the various sets in survay. Often the results are represented as a table

Do you see how the numbers of each region are obtained from the given information in the problem? We start with the region represented by the intersection of Psych. And English majors (7). Then, because there must be 23 Psych. Majors, there must be 16 Psych majors remaining in the rest of the set. A similar argument will convince you that there are 9 students who are majoring in English alone. Psychology majors English majors Both Psych and English 7 students in this region 16 students here 9 students in this region N(P U E) = n(P)+n(E)-n(P E) – 7 = 32

A second problem A survey of 100 college faculty who exercise regularly found that 45 jog, 30 swim, 20 cycle, 6 jog and swim, 1 jogs and cycles, 5 swim and cycle, and 1 does all three. How many of the faculty members do not do any of these three activities? How many just jog? A survey of 100 college faculty who exercise regularly found that 45 jog, 30 swim, 20 cycle, 6 jog and swim, 1 jogs and cycles, 5 swim and cycle, and 1 does all three. How many of the faculty members do not do any of these three activities? How many just jog? We will solve this problem using a three-circle Venn Diagram in the accompanying slides. We will solve this problem using a three-circle Venn Diagram in the accompanying slides.

We will start with the intersection of all three circles. This region represents the number of faculty who do all three activities (one). Then, we will proceed to determine the number of elements in each intersection of exactly two sets J =joggers C=CyclistsS=swimmers 1 does all 3

Solution: Starting with the intersection of all three circles, we place a 1 in that region (1 does all three). Then we know that since 6 jog and swim so 5 faculty remain in the region representing those who just jog and swim. Five swim and cycle, so 4 faculty just swim and cycle but do not do all three. Since 1 faculty is in the intersection region of joggers and cyclists, and we already have one that does all three activities, there must be no faculty who just jog and cycle. Starting with the intersection of all three circles, we place a 1 in that region (1 does all three). Then we know that since 6 jog and swim so 5 faculty remain in the region representing those who just jog and swim. Five swim and cycle, so 4 faculty just swim and cycle but do not do all three. Since 1 faculty is in the intersection region of joggers and cyclists, and we already have one that does all three activities, there must be no faculty who just jog and cycle.

Multiplication principle The tree diagram illustrates the 24 ways to get dressed. To illustrate this principle, let’s start with an example. Suppose you have 4 pairs of trousers in your closet, 3 different shirts and 2 pairs of shoes. Assuming that you must wear trousers (we hope so!), a shirt and shoes, how many different ways can you get dressed? Let’s assume the colors of your pants are black, grey, rust, olive. You have four choices here. The shirt ]colors are green, marine blue and dark blue. For each pair of pants chosen (4) you have (3) options for shirts. You have 12 = 4*3 options for wearing a pair of trousers and a shirt. Now, each of these twelve options, you have two pair of shoes to choose from (Black or brown). Thus, you have a total of 4*3*2 = 24 options to get dressed.

MULTIPLICATION PRINCIPLE (a) If two operations O 1 and O 2 are performed in order, with N 1 possible outcomes for the first operation and N 2 possible outcomes for the second operation, then there are N 1 ·N 2 possible combined outcomes of the first operation followed by the second. (b) In general, if n operations O 1, O 2,..., O n are performed in order with possible number of outcomes N 1, N 2,..., N n, respectively, then there are N 1, N 2,..., N n, respectively, then there are N 1 ·N 2 ·...·N n N 1 ·N 2 ·...·N n possible combined outcomes of the operations performed in the given order.

More problems… How many different ways can a team consisting of 28 players select a captain and an assistant captain? How many different ways can a team consisting of 28 players select a captain and an assistant captain? Solution: Operation 1: select the captain. If all team members are eligible to be a captain, there are 28 ways this can be done. Solution: Operation 1: select the captain. If all team members are eligible to be a captain, there are 28 ways this can be done. Operation 2. Select the assistant captain. Assuming that a player cannot be both a captain and assistant captain, there are 27 ways this can be done, since there are 27 team members left who are eligible to be the assistant captain. Operation 2. Select the assistant captain. Assuming that a player cannot be both a captain and assistant captain, there are 27 ways this can be done, since there are 27 team members left who are eligible to be the assistant captain. Then, using the multiplication principle there are (28)(27) ways to select both a captain and an assistant captain. This number turns out to be 756. Then, using the multiplication principle there are (28)(27) ways to select both a captain and an assistant captain. This number turns out to be 756.

Next example A sportswriter is asked to rank 8 teams in the NBA from first to last. How many rankings are possible? A sportswriter is asked to rank 8 teams in the NBA from first to last. How many rankings are possible? Solution: We will use 8 slots that need to be filled. In the first slot, we will determine how many ways to choose the first place team, the second slot is the number of ways to choose the second place team and so on until we get to the 8 th place team. There are 8 choices that can be made for the first place team since all teams are eligible. That leaves 7 choices for the second place team. The third place team is determined from the 6 remaining choices and so on. Solution: We will use 8 slots that need to be filled. In the first slot, we will determine how many ways to choose the first place team, the second slot is the number of ways to choose the second place team and so on until we get to the 8 th place team. There are 8 choices that can be made for the first place team since all teams are eligible. That leaves 7 choices for the second place team. The third place team is determined from the 6 remaining choices and so on. Total is the product of 8(7)…1 = Total is the product of 8(7)…1 =

Examples ( Code Words) How many three letter code words are possible using first 8 letters of the alphabet a) No letters can be repeated? b) Letters can be repeated? c) Adjacent letters cannot be same.

Tree Diagram Problem #15. A coin is tossed with possible outcomes head H, or tails, T. Then a single die is tossed with possible outcomes 1, 2, 3, 4, 5, or 6. How many combined outcomes are there?. B) Multiplication Principle O1: Coin N 1 : 2 outcomes O2: Die N 2 : 6 outcomes Thus, there are N 1 ·N 2 = 2·6 = 12 combined outcomes

Tree Diagram for (15) Thus, there are 12 combined outcomes. Thus, there are 12 combined outcomes.

Problem #45. Let T = the people who play tennis, and G = the people who play golf. G = the people who play golf. Then n(T) = 32, n(G) = 37, n(T  G) = 8 and n(U) = 75. Thus, Then n(T) = 32, n(G) = 37, n(T  G) = 8 and n(U) = 75. Thus, n(T  G) = n(T) + n(G) - n(T  G) = n(T  G) = n(T) + n(G) - n(T  G) = = = 61 = = 61 The set of people who play neither tennis nor The set of people who play neither tennis nor golf is represented by T '  G '. golf is represented by T '  G '. Since U = (T  G)  (T '  G ') and Since U = (T  G)  (T '  G ') and (T  G)  (T '  G ')= , (T  G)  (T '  G ')= , it follows that n(T '  G ') =n(U)-n(T  G) = = 14. it follows that n(T '  G ') =n(U)-n(T  G) = = 14. There are 14 people who play neither tennis nor golf. There are 14 people who play neither tennis nor golf.