Dicrete Maths Lectures - Discrete Mathematics Dr. Fred...

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Unformatted text preview: Discrete Mathematics Dr. Fred Phelps Lecture 1 Counting and Set Theory Announcements • • • Please register for discrete math on the site http://dl.iitu.kz. The main textbook is on that site. The syllabus is also on that site. And a copy of the main textbook: “Discrete Mathematics” by Lovasz, Pelikan and Veztergombi. LPV 1.1 Counting Problems from a Birthday Party • • • Alice (♀) has a birthday party. Invites Bob (♂), Carl (♂), Diane (♀), Eve (♀), Frank (♂), George (♂) Total 7 people – four boys, three girls LPV 1.1 Counting Problems from a Birthday Party Problem 1 - Handshakes Everyone shakes everyone else’s hand. How many handshakes does this make? 7 people each shake 6 hands so 7·6=42. But, each handshake was double counted 7 ×6 2 = 21. LPV 1.1 Counting Problems from a Birthday Party Problem 2 – Seating Arrangements How many ways can the guests be seated at the seven chairs at the table if Alice (the birthday girl) stays at the head of the table? Alice’s chair is fixed – only one possibility. Any one of six people (anyone besides Alice) can sit in the next (2nd) chair – six possibilities. Any one of five (anyone besides Alice or the occupant of the first chair) can sit in the third chair – five possibilities. LPV 1.1 Counting Problems from a Birthday Party Problem 2 – Seating Arrangements And so the total number of possible seating arrangements is: 1·6·5·4·3·2·1=6!=720. LPV 1.1 Counting Problems from a Birthday Party Problem 3 – Dance Pairings How many different sets of dancing pairs (boys dance with girls only) can be formed from the 4 boys and 3 girls? The textbook says this is easy. But the textbook’s solution is wrong! Notice that one boy will not have a partner. So lets start with the girls. LPV 1.1 Counting Problems from a Birthday Party Problem 3 – Dance Pairings 1st girl has four choices of partner 2nd girl has three choices 3rd girl has two choices So the number of possible dance pairs = 4x3x2=24. LPV 1.1 Counting Problems from a Birthday Party Problem 3 – Dance Pairings Now let’s try to solve this starting with the boys. First we choose which three of the four boys will dance. One must sit out. There are 4 ways to do this: Bob, Carl, Frank, but not George Bob, Carl, George, but not Frank Bob, Frank, George, but not Carl Carl, Frank, George, but not Bob LPV 1.1 Counting Problems from a Birthday Party Problem 3 – Dance Pairings This is called the number of combinations of 4 things taken 3 at a time. We will see the general formula for combinations later. Next we order the boys (by age or height or alphabetical order of their names). Thus we can uniquely refer to “the first boy” without knowing who he is. 1st boy has three choices of partner 2nd boy has two choices 3rd boy is stuck with the last girl LPV 1.1 Counting Problems from a Birthday Party Problem 3 – Dance Pairings So, for every combination of three boys, there are 3·2·1=6 dancing pairs. Given the four choices for which boys dance, there are 4·6=24 different ways for the friends to “pair off” to dance. Its good that we get the same answer either way! LPV 1.1 Counting Problems from a Birthday Party Problem 4 – Lottery Tickets A lottery ticket consists of the first ninety positive integers {1, 2, 3, …, 88, 89, 90}. To play the lottery, you chose and mark five of the numbers. How many different ways to mark the lottery ticket are there? There are 90 choices for the first number There are 89 choices for the second number And so on 89 87 86 And so the90 × ×88 × × =of 273,912,160 total number 5, ways to mark the ticket is LPV 1.1 Counting Problems from a Birthday Party Problem 4 – Lottery Tickets But winning does not depend on the order you mark the ticket, (i.e. if you choose {2}, {7}, {49}, {32}, {14} it is the same as if you choose {2}, {49}, {14}, {7}, {32}). Each set of five numbers can be chosen in 5·4·3·2·1=5! orders. So there9are × × × 0 × 88 87 86 89 5 ×4 × ×2 × 31 5, 273,912,160 = = 43,949, 268 120 ways to mark the lottery card. In order to win, you need LPV 1.1 Counting Problems from a Birthday Party Problem 4 – Lottery Tickets Note – the number of ways to mark the lottery ticket is the number of ways to choose 5 numbers out of a set of 90 or the number of combinations of 90 things taken 5 at a time. 90! . (90 − 5)!× 5! You can see that the answer is LPV 1.1 Counting Problems from a Birthday Party Problem 5 – Bridge Hands How many different bridge (i. e. the card game) hands are there? (A bridge hand is 13 cards chosen from a deck of 52). The first card is any one of 52 • The second is any one of 51 • etc. • So there are 52 × × × × ways to receive 13 cards 51 50 ... 40 But order does not matter. So we need to divide by 13! LPV 1.1 Counting Problems from a Birthday Party Problem 5 – Bridge Hands So the number of bridge hands is 52 × × 51 ...40 52! = = 635, 013,559, 600. 13! 39!× 13! This is the number of combinations of 52 things taken 13 at a time. LPV 1.1 Counting Problems from a Birthday Party Problem 6 – Chess Pairings How many different pairs can play chess? (Alice will only watch.) Six people can be seated in six chairs (at three tables) in 6! = 720 ways. Each set of three pairs can choose their tables 3x2=6 different ways: White Table Black White Table Black White Table Black 1 2 1 2 3 4 3 4 5 6 1 2 5 6 3 4 5 6 3 4 5 6 5 6 5 6 1 2 3 4 1 2 3 4 1 2 LPV 1.1 Counting Problems from a Birthday Party Problem 6 – Chess Pairings With both the pairs and the tables fixed there are still 23=8 seating arrangements. White Table Black White Table Black White Table Black White Table Black 1 2 1 2 1 2 1 2 3 4 3 4 4 3 4 3 5 6 6 5 5 6 6 5 2 1 2 1 2 1 2 1 3 4 3 4 4 3 4 3 5 6 6 5 5 6 6 5 So each set of three pairs appears 6x8=48 times in the 720 possible seating arrangements and the number of pairs is 6!/ (6 × = 15. 8) LPV 1.1 Counting Problems from a Birthday Party Problem 6 – Chess Pairings An easier way to do the previous problem: Bob is youngest. Let him choose his partner first. He has five choices. Whoever is next (the youngest among the rest) will have three choices. The last pair will be determined because there are only two non-selected people left. This gives 5x3x1=15 pairs. LPV 1.2 Sets A set is a collection of things (elements). If A is a set and b is an element of A, we write b ∈ A. Set Elements deck of cards each card is an element guests at the party real numbers Alice, Bob, Carl, Diane, Eve, Frank, P George ℝ number for every point on the number line ℤ …, -2, -1, 0, 1, 2, 3,… integers non-negative integers empty set Symbo l D 0, 1, 2, 3, … Z+ none ∅ LPV 1.2 Sets The cardinality of a set A is the number of its elements and is denoted by |A|. |P| = 7 |Ø| = 0 ¢= ∞ Ways to define a set: We can define a set by listing elements between braces: P = {Alice, Bob, Carl, Diane, Eve, Frank, George} 2. We can use a verbal description: {the boys at Alice’s party} 3. Use properties inside brackets after a colon: 1. ¢ + = { x ∈ ¢ : x ≥ 0} LPV 1.2 Subsets A set A is called a subset of set B if every element of A is also an element of B. We write this as: A ⊆ B. If two sets A and B have exactly the same elements they are said to be equal. We write this as: A = B. Example of subsets ∅⊆¥ ⊆¢ + ⊆¢ ⊆¤ ⊆¡ • In plain English, the empty set is a subset of the natural numbers (1, 2, 3, …) which is a subset of the nonnegative integers (0, 1, 2, …), which is a subset of the integers (…-2, -1, 0, 1, 2…) which is a subset of the rational numbers (all numbers which can be written as a fraction of integers), which is a subset of the real numbers. intersection of two sets • • The intersection of two sets A and B is the set consisting of those elements which are common to both sets. A We write this as: I B. If A I B = ∅ then A and B are said to be "disjoint". union of two sets • • The union of two sets A and B is defined as the set consisting of elements which are in at least one of A and B. A We write this as: U B. universe • In most cases the sets naturally come from some “universe” of possible sets. For example, let S be the set of all sequences of capital (i.e. big) English letters with length 4 which do not have any A’s. universe complement • • The complement of a set G in universe U is all of the members of U which are not in G. People write this as: C G or G′ or G . difference of A and B A− B B− A Notice that A − B ∩ B − A = ∅. notation Examples of notation: A1 ≠ A2 means that ∃a ∋: a ∈ A1 ∪ A2 and a ∉ A1 I A2 . So, ∃i ∈ {1, 2} ∋: a ∉ Ai . Symmetric Difference: • • • The symmetric difference of sets A and B is the set of all elements of A or B which are not in both A and B. We write this as A Δ B. A Δ B = {x|(x ∈ A ∧ x ∉ B) ∨ (x ∉ A ∧ x ∈ B)} DeMorgan’s Laws Venn Diagram “proof” of DeMorgan Law Proof of 2nd DeMorgan Law Goodbye! ?? ...
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This note was uploaded on 10/15/2011 for the course MATHS 100 taught by Professor Fredphelps during the Spring '11 term at Jordan University of Science & Tech.

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Dicrete Maths Lectures - Discrete Mathematics Dr. Fred...

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