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2000 solved problems in discrete mathematics

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Discrete Mathematics

Discrete mathematics deals with areas of mathematics that are discrete, as opposed to continuous, in nature. Sequences and series, counting problems, graph theory and set theory are some of the many branches of mathematics in this category. Use Wolfram|Alpha to apply and understand these and related concepts.

Use the ideas of permutation and combination to find binomial coefficients or integer partitions or to do other forms of counting.

Compute binomial coefficients (combinations):

Compute a frobenius number:, find the partitions of an integer:.

Compute the properties of these discrete subsets of Euclidean space. Deal with regular or rooted lattices. Compare them.

Compute the properties of a lattice:

Compute the properties of a root lattice:, compare several lattices:.

Work with the famous function known for being a total computable function that is not primitive recursive.

Evaluate the Ackermann function:

Work with known graphs or specify your own via adjacency lists. Check for isomorphism, compute the shortest path and more.

Compute the properties of a named graph:

Compare several graphs:, analyze a graph specified by adjacency rules:.

Deduce patterns in incompletely specified sequences or series. Sum infinite series or perform other operations on them.

Compute a possible formula and continuation for a sequence:

Sum an incompletely specified infinite series:, related examples.

• Applied Mathematics
• Logic & Set Theory
• Number Theory

Solve recurrences, specify initial values or attempt to model sequences with recurrences.

Solve a recurrence:

Specify initial values:, find a recurrence satisfied by a given sequence:.

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Discrete Mathematics

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• Brilliant Mathematics
• Alek Nelson

Discrete mathematics is the study of mathematical structures that are countable or otherwise distinct and separable. Examples of structures that are discrete are combinations , graphs , and logical statements . Discrete structures can be finite or infinite . Discrete mathematics is in contrast to continuous mathematics , which deals with structures which can range in value over the real numbers , or have some non-separable quality.

Since the time of Isaac Newton and until quite recently, almost the entire emphasis of applied mathematics has been on continuously varying processes, modeled by the mathematical continuum and using methods derived from the differential and integral calculus. In contrast, discrete mathematics concerns itself mainly with finite collections of discrete objects. With the growth of digital devices, especially computers, discrete mathematics has become more and more important.

Discrete structures can be counted, arranged, placed into sets , and put into ratios with one another. Although discrete mathematics is a wide and varied field, there are certain rules that carry over into many topics. The concept of independent events and the rules of product , sum , and PIE are shared among combinatorics , set theory , and probability . In addition, De Morgan's laws are applicable in many fields of discrete mathematics.

Often, what makes discrete mathematics problems interesting and challenging are the restrictions that are placed on them. Although the field of discrete mathematics has many elegant formulas to apply, it is rare that a practical problem will fit perfectly to a specific formula. Part of the joy of discovering discrete mathematics is to learn many different approaches to problem-solving, and then be able to creatively apply disparate strategies towards a solution.

Combinatorics

Graph theory, probability.

Main Article: Combinatorics

Combinatorics is the mathematics of counting and arranging. Of course, most people know how to count, but combinatorics applies mathematical operations to count things that are much too large to be counted the conventional way.

Combinatorics is especially useful in computer science . Combinatorics methods can be used to develop estimates about how many operations a computer algorithm will require. Combinatorics is also important for the study of discrete probability . Combinatorics methods can be used to count possible outcomes in a uniform probability experiment.

Combinatorics is often concerned with how things are arranged. In this context, an arrangement is a way objects could be grouped. The most basic rules regarding arrangements are the rule of product and the rule of sum . These rules govern how to count arrangements using the operations of multiplication and addition, respectively.

At a local deli, the following options are given for a sandwich:

• Bread types: White, Rye, Wheat
• Cheese Types: Swiss, Cheddar, Havarti, Provolone
• Meat Types: Roast Beef, Turkey, Ham, Corned Beef, Pulled Pork

If a customer chooses exactly one of each type of item, then how many possible sandwiches can be made?

A more specific type of arrangement is a permutation . A permutation is an arrangement of objects with regard to order.

At the start of a horse race, there are 12 distinct horses in the field. 3 horses can place at the end of the race, and it matters what order the horses placed in. For example, if horses \(\text{A},\) \(\text{B},\) and \(\text{C}\) placed, then it would matter which horse came in \(1^\text{st}, 2^\text{nd},\) and \(3^\text{rd}.\) The order \(\text{ABC}\) would be different than \(\text{ACB}.\)

How many possible fields of placed horses are there?

A combination (not to be confused with combinat orics ) is another type of arrangement that is related to permutations. A combination is an arrangement of objects without regard to order.

There are 12 distinct players about to play a pick-up game of football. The two team captains are Brandon and Meredith (included in the 12). They will each take turns selecting a player to be on their team until all players are selected.

How many ways can the players be divided into teams?

As a field of mathematics, combinatorics is nearly as broad as discrete mathematics. Other topics within combinatorics include

derangements : a permutation such that no object is in its original spot in the order;

rectangular grid walks : determining the number of ways a rectangular lattice can be traversed;

distribution of objects into bins : determining how objects can be grouped into bins.

Main Article: Set Theory See Also: Cardinality Union and Intersection Set Complement Principle of Inclusion and Exclusion De Morgan's Laws

Set theory is the branch of mathematics that is concerned about collections of objects. Sets can be discrete or continuous; discrete mathematics is primarily concerned with the former. At a basic level, set theory is concerned with how sets can be arranged, combined, and counted.

The cardinality of a finite set is the number of elements in that set. For a given set \(A,\) its cardinality is denoted by \(|A|.\)

What is the cardinality of the set of prime numbers less than 25? The set of prime numbers less than 25 is \[\{2,3,5,7,11,13,17,19,23\}.\] There are 9 elements in this set, so the cardinality is 9. \(_\square\)

Cardinality can also be extended to infinite sets. Although this kind of cardinality cannot be counted, each cardinality can be compared with another cardinality.

Let \(A\) and \(B\) be sets. Their cardinalities are compared as follows: If there exists a bijection between \(A\) and \(B,\) then \(|A|=|B|.\) If there exists an injective function from \(A\) to \(B,\) but no bijective function, then \(|A|<|B|.\)

Show that the set of integers and the set of even integers have the same cardinality. It might seem strange that these sets have the same cardinality. After all, the even integers are more "rare." However, these sets are both infinite . Therefore, the "common sense" thinking about finite sets must be discarded. Instead, the goal is to obtain a bijective function from the set of integers to the set of even integers: \[f(n)=2n, \ n \in \mathbb{Z}.\] The function above gives a one-to-one correspondence between each integer \(n\) and each even integer \(2n.\) Since the bijection is established, the set of integers and the set of even integers have the same cardinality. \(_\square\)

A complement of a set \(A\) is the set of elements that are not in \(A.\) The study of set complements gives a number of efficient methods to calculate cardinalities of finite sets. For example, one can efficiently obtain the cardinality of a set that contains "at least one" element of another set.

David is the leader of the David Committee. He wants to appoint 3 people to be on the Head Council. He has to choose from 9 applicants, three of whom are Tommy, Jack, and Michael. In how many ways can he choose the people to be on the Council, so that at least one of Tommy, Jack, and Michael is chosen?

The union and intersection give ways to describe how sets can be combined.

A positive integer less than 1000 is not only a perfect square but also a perfect cube.

How many such numbers are there?

Person "A" says the truth 60% of the time, and person "B" does so 90% of the time.

In what percentage of cases are they likely to contradict each other in stating the same fact?

De Morgan's laws give identities for the complements of unions and intersections.

How many integers between 1 and 1000 (inclusive) are neither multiples of 2 nor multiples of 5?

The principle of inclusion and exclusion , or PIE, gives a method to find the union or intersection of more than two sets.

How many integers from \(1\) to \(10^6\) (inclusive) are neither perfect squares nor perfect cubes nor perfect fourth powers?

Main Article: Graph Theory

Graph theory is the study of graphs , which are a collection of connected nodes.

3-Cycle " /> A basic graph of 3-Cycle

Graphs are useful for representing all kinds of real-world problems.

If there is a greedy algorithm that will traverse a graph, selecting the largest node value at each point until it reaches a leaf of the graph, what path will the greedy algorithm follow in the graph below?

John lives in the Trees of Ten Houses, and it is a most ideal and idyllic place for him and the other dwellers up in the canopy. They have invested a tremendous amount of time in engineering these houses, and to ensure no house felt isolated from the others, they built a fresh, finely crafted bridge between each and every house!

Unfortunately, the Trees of Ten Houses were not immune to thunderstorms, nor were the bridges well engineered. The night was treacherous, howling with wind and freezing with rain, so the odds for the bridges were not good--each bridge seemed just as likely to survive as to be shattered!

Fortunately, as there were so very many bridges in the Trees of Ten Houses, when John did wake the following morning, he found he was able to make his way to each and every house using only the existing bridges, though round-about routes may have been necessary. As they began rebuilding, John became curious... what were the chances that they'd all be so lucky?

More formally, if \(P\) is the probability that, after the storm, John is able to traverse to each and every house, what is \(\big\lfloor 10^{10} P \big\rfloor?\)

Details and Assumptions:

• The Trees of Ten Houses do, in fact, contain precisely 10 houses.
• Before the storm, there exists a single bridge between each and every unique pair of houses.
• The storm destroys each bridge with independent probability \(\frac{1}{2}\).
• John is allowed to traverse through others' houses to try to reach all of them, but he must only use the surviving bridges to get there. No vine swinging allowed.

Discrete Mathematics Tutorial

Mathematical logic.

• Propositional Logic
• Discrete Mathematics - Applications of Propositional Logic
• Propositional Equivalences
• Predicates and Quantifiers
• Mathematics | Some theorems on Nested Quantifiers
• Rules of Inference
• Mathematics | Introduction to Proofs

Sets and Relations

• Types Of Sets
• Irreflexive Relation on a Set
• Reflexive Relation on Set
• Transitive Relation on a Set
• Set Operations
• Types of Functions
• Mathematics | Sequence, Series and Summations
• Representation of Relation in Graphs and Matrices
• What are Relations?
• Closure of Relations
• Mathematical Induction
• Pigeonhole Principle
• Mathematics | Generalized PnC Set 1
• Discrete Maths | Generating Functions-Introduction and Prerequisites
• Inclusion Exclusion principle and programming applications

Boolean Algebra

• Properties of Boolean Algebra
• Number of Boolean functions
• Minimization of Boolean Functions
• Linear Programming

Ordered Sets & Lattices

• Elements of POSET
• Partial Order Relation on a Set
• Axiomatic Approach to Probability
• Properties of Probability

Probability Theory

• Mathematics | Probability Distributions Set 1 (Uniform Distribution)
• Mathematics | Probability Distributions Set 2 (Exponential Distribution)
• Mathematics | Probability Distributions Set 3 (Normal Distribution)
• Mathematics | Probability Distributions Set 5 (Poisson Distribution)

Graph Theory

• Graph and its representations
• Mathematics | Graph Theory Basics - Set 1
• Types of Graphs with Examples
• Mathematics | Walks, Trails, Paths, Cycles and Circuits in Graph
• How to find Shortest Paths from Source to all Vertices using Dijkstra's Algorithm
• Prim’s Algorithm for Minimum Spanning Tree (MST)
• Kruskal’s Minimum Spanning Tree (MST) Algorithm
• Check whether a given graph is Bipartite or not
• Eulerian path and circuit for undirected graph

Special Graph

• Introduction to Graph Coloring
• Edge Coloring of a Graph
• Check if a graph is Strongly, Unilaterally or Weakly connected
• Biconnected Components
• Strongly Connected Components

Group Theory

• Homomorphism & Isomorphism of Group
• Group Isomorphisms and Automorphisms
• Algebraic Structure

GATE PYQs and LMN

• Last Minute Notes – Discrete Mathematics
• Discrete Mathematics - GATE CSE Previous Year Questions

Discrete Mathematics MCQ

• Propositional and First Order Logic.
• Set Theory & Algebra
• Combinatorics
• Top MCQs on Graph Theory in Mathematics
• Numerical Methods and Calculus

Discrete Mathematics is a branch of mathematics that is concerned with “discrete” mathematical structures instead of “continuous”. Discrete mathematical structures include objects with distinct values like graphs, integers, logic-based statements, etc. In this tutorial, we have covered all the topics of Discrete Mathematics for computer science like set theory, recurrence relation, group theory, and graph theory.

Recent Articles on Discrete Mathematics!

• Introduction to Propositional Logic
• Applications of Propositional Logic
• Propositional and Predicate Logic
• Normal and Principle Forms
• Nested Quantifiers Theorem
• Introduction to Proofs
• Types of Sets
• Rough Set Theory
• Sequence and Summations
• Representations of Matrices and Graphs in Relations
• Types of Relation
• Closure of Relation and Equivalence Relations
• Basics of Counting
• Pascal’s Identity
• Permutations and Combinations
• Generalized Permutations and Combinations
• Generating Functions
• Inclusion-Exclusion Principle
• Discrete Probability Theory
• Boolean Functions
• Boolean Algebraic Theorem
• Number of Boolean Functions

Optimization

• Linear Programming 
• Graphical Solution For Linear Programming
• Simplex Algorithm
• PERT  

Ordered Sets & Lattices

• Partially Ordered Sets
• Hasse Diagrams
• Basic Concepts of Probability
• Probability Axioms
• Conditional Probability
• Bayes’ Theorem
• Uniform Distribution
• Exponential Distribution
• Normal Distribution
• Poisson Distribution
• Introduction to Graph
• Basic terminology of a Graph
• Types of a Graph
• Walks, Trails, Paths, and Circuits
• Graph Distance components
• Cut-Vertices and Cut-Edges
• Bridge in Graph
• Independent sets
• Shortest Path Algorithms [Dijkstra’s Algorithm]
• Application of Graph Theory
• Graph Traversals[DFS]
• Graph Traversals[BFS]
• Characterizations of Trees
• Prim’s Minimum Spanning Tree 
• Kruskal’s Minimum Spanning Tree
• Huffman Codes 
• Tree Traversals
• Traveling Salesman Problem
• Bipartite Graphs  
• Independent Sets and Covering
• Eulerian graphs
• Eulerian graphs- Fleury’s algorithm
• Eulerian graphs- Chinese-Postman-Problem Hamilton
• Matching- Basics, Perfect, Bipartite
• Approximation Algorithms

Vertex Colorings

• Chromatic Numbers, Greedy Coloring Algorithm
• Edge Coloring
• Vizing Theorem
• Planar Graph- Basics, Planarity Testing
• Directed Graphs- Degree Centrality
• Directed Graphs- Weak Connectivity
• Directed Graphs- Strong Components 
• Directed Graphs- Eulerian, Hamilton Directed Graphs
• Directed Graphs- Tarjans’ Algorithm To Find Strongly Connected Component
• Handshaking in Graph Theorem
• Groups, Subgroups, Semi Groups
• Isomorphism, Homomorphism
• Automorphism
• Rings, Integral domains, Fields

Quick Links

• Last-Minute Notes (LMNs)
• Quizzes on Discrete Mathematics

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Discrete Mathematics

What is Discrete Mathematics?

Mathematics is a subject that you’ll either love or dread. It is as simple as that. The people who dread Mathematics are the ones who have not witnessed the beauty of numbers and logic. So, remember it’s never too late for absorbing knowledge. This subject not only teaches us how to deal with problems but also instills common sense in us. Mathematics is one of the subjects which can never truly and entirely separate from our lives. The concepts of Mathematics serve as the basis of various other subjects like physics, computer science, architecture etc. Mathematics is divided into 4 branches namely, arithmetic, algebra, geometry, and trigonometry. Did you know that Archimedes is considered as the Father of Mathematics? Today we’ll learn about Discrete Mathematics.

Do you know what Discrete Mathematics is? Do you know about Discrete Mathematics and its applications? We’ll discuss it all here.

Discrete Mathematics is about Mathematical structures. It is about things that can have distinct discrete values. Discrete Mathematical structures are also known as Decision Mathematics or Finite Mathematics. This is very popularly used in computer science for developing programming languages, software development, cryptography, algorithms, etc.

(Image will be uploaded soon)

Discrete Mathematics Problems and Solutions

Now let’s quickly discuss and solve a Discrete Mathematics problem and solution:

Example 1: 

Determine in how many ways can three gifts be shared among 4 boys in the following conditions-

i) No one gets more than one gift.

ii) A boy can get any number of gifts.

i) The first gift can be given in 4 ways as one cannot get more than one gift, the remaining two gifts can be given in 3 and 2 ways respectively.

The total number of ways = 4 x 3 x 2 = 24.

ii) As there is no restriction, each gift can be given in 4 ways.

The total number of ways = 43 = 64.

Discrete Mathematics Topics

Various concepts of Mathematics are covered by Discrete Mathematics like:

Permutation and Combination

Graph Theory

Sequence and series,   set theory.

Set Theory is a branch of Mathematics that deals with collection of objects. It starts with the fundamental binary relation between an object M and set A.

Imagine there are two sets, say, set A and set B. Set A has numbers 1-5 and Set B has numbers 1-10. You can see that all the elements of set A are in set B. This makes set A a subset of set B.

A={1,2,3,4,5}          B={1,2,3,4,5,6,7,8,9,10}

Imagine there are two sets, say, set A and set B. Set A has numbers 1-5 and Set B has numbers 1-10. If we combine the elements of set A and set B, then the set we get is called a union set. So, we get the union of set A and set B. 

(AUB)={1,2,3,4,5,6,7,8,9,10}

Imagine there are two sets, say, set A and set B. Set A has numbers 1-5 and Set B has numbers 1-10. If we take the elements that are present in both sets then we get the intersection.

(A∩B)={1,2,3,4,5}

Imagine there are two sets, say, set A and set B. Set A has numbers 1-5 and Set B has numbers 1-10. When we are trying to find the Cartesian Product of set A and B, we are actually making an ordered pair. (AXB)={(1,1);(1,2).........(5,4);(5,5)}

Remember  (AXB)≠(BXA)

Permutation and Combination 

Permutation and Combination are all about counting and arranging from the given data. The permutation is all about arranging the given elements in a sequence or order. The combination is about selecting elements in any way required and is not related to arrangement.

You can use the formula for permutation – nPr = \[\frac{(n!)}{(n-r)!}\]

Where r objects have to be arranged out of a total of n number of objects

The formula for combination is—  \[nCr=\frac{n!}{r!(n-r)!}\]

 Where r objects have to be chosen out of a total of n number of objects 

Logic 

Logic can be defined as the study of valid reasoning. The logical formulas are discrete structures and so are proofs thus, forming finite trees. The truth values of logical formulas form a finite set. They are restricted to only two values either true or false. 

Graph Theory is about the study of graphs. They are discrete Mathematical structures and are used to model in relation to pairs between the objects. The graph we are discussing here consists of vertices which are joined by edges or lines. Graphs are one of the most important objects of study in Discrete Mathematics. Discrete Mathematics and graph theory are complementary to each other. Graphs are present everywhere. They are models of structures either made by man or nature. They can model various types of relations and process dynamics in physical, biological and social systems. They can also display networks of communication, data organization, the flow of computation, etc. they are also used in geometry and in topology. 

A sequence is a set of numbers which are arranged in a definite order and following some definite rule. A series is a sum of terms which are in a sequence.

Formulas Related to Some Special Series

 The sum of 1st n natural numbers: 

    \[S_{n} = \frac{n(n+1)}{2}\] 

 The sum of the squares of 1st n natural numbers:  

       \[S_{n} = \frac{n(n+1)(2n+1)}{6}\]

 The sum of the cubes of first n natural numbers:  

         \[S_{n} = \text{(Sum of the first n natural numbers)}2\]

    \[= \frac{n(n+1)}{22}\]

On contrary to real numbers that differs "seamlessly", Discrete Mathematics studies objects such as graphs, integers and statements in reasoning

The objects studied in Discrete Mathematics do not differ seamlessly, in fact, have varied, Discrete Mathematics does not include matters in "continuous mathematics" such as algebra and calculus          

Continuous and Discrete Mathematics

Mathematics can be divided into two categories: continuous and discrete.

Continuous Mathematics is based on a continuous number line or real numbers in continuous form. It is defined by the fact that there is virtually always an endless quantity of numbers between any two integers. In continuous Mathematics, for example, a function can be depicted as a smooth curve with no breaks.

Discrete Mathematics involves separate values; that is, there are a countable number of points between any two points in Discrete Mathematics. If we have a finite number of items, for example, the function can be defined as a list of ordered pairs containing those objects and displayed as a complete list of those pairs.

Use of Discrete Mathematics in Real World 

It is commonly stated that Mathematics may be used to solve a wide range of practical problems. Cathy and MathILy-Er focus on Discrete Mathematics, which supports nearly half of pure Mathematics, operations research, and computer science in general. Discrete Mathematics is becoming more prevalent in academia and industry as time goes on. But what exactly are the applications that people are referring to when they claim Discrete Mathematics can be used? What issues are being addressed? This article attempts to answer those questions. There are summaries of Discrete Mathematics applications in our daily lives, as well as in major and interesting research and corporate applications, with links to extended explanations.

Vedantu: A Mentor

In case you're a student who is preparing for an exam, you can refer to the many sorts of courses available on Vedantu's website or app. These courses will help you in many ways like, you will learn how to write both long and short solutions in various sorts of tests. Vedantu's website also provides you with various study materials for exams of all CBSE Classes like 9th, 10 th , 11 th , 12 th , and other sorts of board and state-level examinations. On Vedantu, you will also learn about the pattern of past year question papers as these papers are eventually going to help you study thoroughly for your future examinations.

If as a student, you are interested in learning more about Vedantu and want a friend that would help you to score well in exams, you can visit the Vedantu website. It is a portal through which you can access all of the essential study resources and it can also be dubbed as an educational tool crafted by a large number of dedicated instructors who put their expertise and hard work into making these resources. So, just visit the website and check out the different types of materials available there.

FAQs on Discrete Mathematics

1. What are the different uses of Discrete Mathematics?

Discrete Mathematics can be applied in various fields such as it can be used in computer science where it is used in different programming languages, storing data etc. It can be used in networking, searching the web, finding locations on Google Maps, scheduling different types of tasks and managing the voting systems. There are several other applications of Discrete Mathematics apart from those which we mentioned. 

2. What are the different topics included in Discrete Mathematics?

Discrete Mathematics comprises a lot of topics which are sets, relations and functions, Mathematical logic, probability, counting theory, graph theory, group theory, trees, Mathematical induction and recurrence relations. All these topics include numbers that are not in continuous form and are rather in discrete form and all these topics have a vast range of applications, therefore becoming very important to study. 

3. How will Discrete Mathematics help me in my life?

Discrete Mathematics revolves around the whole quantities or in other words, it comprises the study of quantities that can be counted. This makes it a very useful field of Mathematics, as it has a lot of applications in our day-to-day life. If you master this field of Mathematics, it will help you a lot with your life. For instance, if you know about logic (part of Discrete Mathematics), you can solve a lot of your problems by just applying the concepts of Mathematical logic. 

4. Is Discrete Mathematics easy or difficult and how can I learn the concepts used in it easily?

When it comes to different fields of Mathematics, Discrete Mathematics is by far the easiest one among all fields. It is very simple as it consists of numbers or quantities that are countable. You just need to understand the concepts of Discrete Mathematics and you are good to go. You can learn all the concepts of Discrete Mathematics from the Vedantu website. 

5. Can Discrete Mathematics be Applied in Real-life?

Yes, Discrete Mathematics has its Application in the Real World too. Discrete Mathematics and Application include:-

The research of Mathematical proof is extremely essential when it comes to logic and is applicable in automated theorem showing and everyday verification of software.

Partially ordered sets and sets with other relations are used in various sectors.

Number Theory is applicable in Cryptography and Cryptanalysis.   

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If you don't have a lot of time but want to excel in class, use this book to:

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Compatible with any classroom text, Schaum's Solved Problem Guides let you practice at your own pace and remind you of all the important problem-solving techniques you need to remember--fast! And Schaum's are so complete, they're perfect for preparing for graduate or professional exams.

Inside you will find:

• 2,000 solved problems with complete solutions--the largest selection of solved problems yet published on this subject
• An index to help you quickly locate the types of problems you want to solve
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• ISBN-10 0070380317
• ISBN-13 978-0070380318
• Edition 1st
• Publisher McGraw Hill
• Publication date October 1, 1991
• Language English
• Dimensions 8.1 x 1 x 10.9 inches
• Print length 404 pages
• See all details

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• Publisher ‏ : ‎ McGraw Hill; 1st edition (October 1, 1991)
• Language ‏ : ‎ English
• Paperback ‏ : ‎ 404 pages
• ISBN-10 ‏ : ‎ 0070380317
• ISBN-13 ‏ : ‎ 978-0070380318
• Item Weight ‏ : ‎ 2.06 pounds
• Dimensions ‏ : ‎ 8.1 x 1 x 10.9 inches
• #66 in Discrete Mathematics (Books)
• #307 in Calculus (Books)
• #2,892 in Study Guides (Books)

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Seymour lipschutz.

Seymour Lipschutz is a 87 year old Author who has a Ph.D and a Professor of Mathematics in Temple University

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Discrete Mathematics Questions

The Discrete Mathematics questions and answers will help students quickly grasp almost all of the concepts discussed in “Discrete Mathematics.” Students can use these questions to quickly understand the concept. We have also included some practice questions so they can understand better. Furthermore, you can get thorough explanations for every question and cross-check your answers. Click here to read more about discrete mathematics .

Discrete Mathematics Questions with Solutions

1. If a set B has n elements, then what is the total number of subsets of B. Justify your answer.

If a set B has “n” elements, then the total number of subsets of B is 2 n .

For example, if B contains 5 elements, say B = {1, 2, 3, 4, 5}, then the total number of subsets of B is 2 5 = 32.

2. If X and y are the two finite sets, such that n(X U Y) = 36, n(X) = 20, n(Y) = 28, then find n( X ∩ Y).

Given: n(X) = 20, n (Y) = 28, n (X U Y) = 36.

As we that, n(X U Y) = n(X) + n(Y) – n(X ∩ Y)

On rearranging the above formula, we get;

n(X ∩ Y) = n(X) + n(Y) – n(X U Y)

Now, substitute the given values in the above formula, we get;

n(X ∩ Y) = 20 + 28 – 36

n(X ∩ Y) = 48 – 36

n(X ∩ Y) = 12.

Hence, n(X ∩ Y) is 12.

3. In a group of 100 students, 72 students can speak English and 43 students can speak Hindi. Based on these data, answer the following questions:

• Find the number of students who can speak English only.
• Find the number of students who can speak Hindi only.
• Find the number of students who can speak both English and Hindi.

Let E be the set of students who speak English, and H be the set of students who speak Hindi.

Thus, from the given conditions, we get

n(EUH) = 100

Using the formula n(A U B) = n(A) + n(B) – n(A ∩ B), we can write;

n(E U H) = n(E) + n(H) – n(E ∩ H) …(1)

Finding Number of students who can speak both English and Hindi:

Rearranging the formula (1), we get

n(E ∩ H) = n(E) + n(H) – n(E U H)

Now, substitute the known values in the above equation, we get

n(E ∩ H) = 72 + 43 – 100

n(E ∩ H) = 115 – 100

n(E ∩ H) = 15.

Therefore, the number of students who can speak both English and Hindi is 15.

Finding Number of students who speak English only:

n(E – H) = n(E) – n(E ∩ H)

Thus, n(E – H) = 72 – 15

n(E – H) = 57.

Hence, the number of students who speak English only is 57.

Finding Number of students who speak Hindi only:

n(H – E) = n(H) – n(E ∩ H)

n(H – E) = 43 – 15

n(H – E) = 28.

Therefore, the number of students who speak Hindi only is 28.

Also, read: Sets .

4. Find the missing number in the sequence: 5, 10, 15, ___, 25, 30.

Given sequence: 5, 10, 15, ___, 25, 30.

In the given sequence, each preceding term is added with 5 to get the succeeding term.

I.e., 5 + 5 = 10

10 + 5 = 15

15 + 5 = 20

20 + 5 = 25

25 + 5 = 30.

Hence, the missing term in the given sequence is 20.

Therefore, the complete sequence is 5, 10, 15, 20 , 25, 30.

5. Determine the missing term in the sequence: 4, 12, __, 108, 324, 972.

Given Sequence: 4, 12, __, 108, 324, 972.

In the given sequence, each succeeding term is obtained by multiplying 3 to its previous term.

I.e., 4 × 3 = 12

12 × 3 = 36

36 × 3 = 108

108 × 3 = 324

324 × 3 = 972.

Hence, the missing term in the given sequence is 36.

Thus, the complete sequence is 4, 12, 36 , 108, 324, 972

Also, read: Sequence and Series .

6. Calculate the tenth term of the Arithmetic Progression 13, 8, 3, -2 …?

Given AP: 13, 8, 3, -2…

Here, the first term, a = 13

Common difference, d = -5.

I.e., d = 8 – 13 = -5

3 – 8 = -5

-2 -3 = -5.

And, n = 10.

The formula to find the nth term of Arithmetic Progression is T n = a + (n-1)d

Now, substitute the given values in the formula, we get

T 10 = 13 + (10 – 1)(-5)

T 10 = 13 + (9)(-5)

T 10 = 13 – 45

Hence, the tenth term of given Arithmetic Progression 13, 8, 3, -2… is -32.

7. In a group of students, there are 6 boys and 4 girls. Out of 10 students, 4 students have to be selected. Find out how many different ways the students can be selected such that at least one boy should be selected?

Total number of students = 10

Number of boys = 6

Number of girls = 4.

So, here we have to select 4 students. In that, at least one boy should be selected.

Hence, there are four cases:

Case 1: All four students are boys:

Thus, the number of ways = 6 C 4 = 15

Case 2: 3 Boys and 1 girl are selected:

The number of ways to select 3 boys and 1 girl = 6 C 3 × 4 C 1 = 20 × 4 = 80.

Case 3: 2 boys and 2 girls are selected

The number of ways to select 2 boys and 2 girls = 6 C 2 × 4 C 2 = 15 × 6 = 90.

Case 4: 1 boy and 3 girls are selected

The number of ways to select 1 boy and 3 girls = 6 C 1 × 4 C 3 = 6 × 4 = 24.

Therefore, the number of ways to select 4 students such that at least one boy to be selected = 15 + 80 + 90 + 24

8. Find out the number of ways that the letters of the word “LEADER” can be arranged?

In the word “LEADER”, there are 6 letters.

In that, E is repeated twice.

Hence, the total number of ways that the letters can be arranged = 6!/2!

As we know, 6! = 720 and 2! = 2

Therefore, the number of ways that the letters can be arranged = 720/2 = 360.

9. Find out the number of ways that the letters of the word “DETAIL” can be arranged such that the vowels must occupy odd positions.

In the word “DETAIL”, there are 6 letters.

In that, there are 3 vowels (E, A, I), and 3 consonants (D, T, L).

According to the given question, the vowels must occupy odd positions.

So, first, mark the positions of the letters as 1, 2, 3, 4, 5, and 6.

As per the questions, vowels should occupy positions 1, 3, and 5.

Hence, the number of ways to arrange the vowels = 3 P 3 = 6.

So, the number of ways to arrange the consonants in the remaining positions = 3 P 3 = 6

Therefore, the number of ways to arrange the letters = 6 × 6 = 36.

Also, read: Permutations and Combinations .

10. Simplify the expression (x + y)(x + z) using the laws of boolean algebra.

Let Q = (x + y)(x + z)

Using the distributive law, we can write;

Q = x.x + x.z + y.x + y.z

By applying the idempotent law A.A = A.

Q = x + x.z + y.x + y.z

Q = x(1 + z) + y.x + y.z [Using distributive law]

Applying identity OR law (1 +A = 1), we can write

Q = x. 1 + y.x + y.z

Q = x + y.x + y.z

Again using the distributive law, we get

Q = x. (1 + y) + y.z

Q = x. 1 + y.z (By applying identity OR Law)

Q = x + y.z.

Therefore, the simplification of the expression (x + y)(x + z) is x + y.z.

Explore More Articles

• Boolean Algebra Questions
• Sequence Questions
• Series Questions
• Permutation Questions
• Permutations and Combinations Questions
• Real Number Questions

Practice Questions

Solve the following discrete mathematics questions:

• If U = {2, 4, 6, 8, 10, 12, 14, 16}, A = {2, 6, 10} and B = {4, 8, 10, 12, 14, 16}, then find
• A – B

2. Find out the number of ways that 3 postcards can be posted in 5 post boxes.

3. Simplify the boolean expression: A = XY + X (Y+Z) + Y(Y+Z).

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Discrete Mathematics - Recurrence Relation

In this chapter, we will discuss how recursive techniques can derive sequences and be used for solving counting problems. The procedure for finding the terms of a sequence in a recursive manner is called recurrence relation . We study the theory of linear recurrence relations and their solutions. Finally, we introduce generating functions for solving recurrence relations.

A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms (Expressing $F_n$ as some combination of $F_i$ with $i < n$).

Example − Fibonacci series − $F_n = F_{n-1} + F_{n-2}$, Tower of Hanoi − $F_n = 2F_{n-1} + 1$

Linear Recurrence Relations

A linear recurrence equation of degree k or order k is a recurrence equation which is in the format $x_n= A_1 x_{n-1}+ A_2 x_{n-1}+ A_3 x_{n-1}+ \dots A_k x_{n-k} $($A_n$ is a constant and $A_k \neq 0$) on a sequence of numbers as a first-degree polynomial.

These are some examples of linear recurrence equations −

How to solve linear recurrence relation

Suppose, a two ordered linear recurrence relation is − $F_n = AF_{n-1} +BF_{n-2}$ where A and B are real numbers.

The characteristic equation for the above recurrence relation is −

$$x^2 - Ax - B = 0$$

Three cases may occur while finding the roots −

Case 1 − If this equation factors as $(x- x_1)(x- x_1) = 0$ and it produces two distinct real roots $x_1$ and $x_2$, then $F_n = ax_1^n+ bx_2^n$ is the solution. [Here, a and b are constants]

Case 2 − If this equation factors as $(x- x_1)^2 = 0$ and it produces single real root $x_1$, then $F_n = a x_1^n+ bn x_1^n$ is the solution.

Case 3 − If the equation produces two distinct complex roots, $x_1$ and $x_2$ in polar form $x_1 = r \angle \theta$ and $x_2 = r \angle(- \theta)$, then $F_n = r^n (a cos(n\theta)+ b sin(n\theta))$ is the solution.

Solve the recurrence relation $F_n = 5F_{n-1} - 6F_{n-2}$ where $F_0 = 1$ and $F_1 = 4$

The characteristic equation of the recurrence relation is −

$$x^2 - 5x + 6 = 0,$$

So, $(x - 3) (x - 2) = 0$

Hence, the roots are −

$x_1 = 3$ and $x_2 = 2$

The roots are real and distinct. So, this is in the form of case 1

Hence, the solution is −

$$F_n = ax_1^n + bx_2^n$$

Here, $F_n = a3^n + b2^n\ (As\ x_1 = 3\ and\ x_2 = 2)$

$1 = F_0 = a3^0 + b2^0 = a+b$

$4 = F_1 = a3^1 + b2^1 = 3a+2b$

Solving these two equations, we get $ a = 2$ and $b = -1$

Hence, the final solution is −

$$F_n = 2.3^n + (-1) . 2^n = 2.3^n - 2^n $$

Solve the recurrence relation − $F_n = 10F_{n-1} - 25F_{n-2}$ where $F_0 = 3$ and $F_1 = 17$

$$ x^2 - 10x -25 = 0$$

So $(x - 5)^2 = 0$

Hence, there is single real root $x_1 = 5$

As there is single real valued root, this is in the form of case 2

$F_n = ax_1^n + bnx_1^n$

$3 = F_0 = a.5^0 + (b)(0.5)^0 = a$

$17 = F_1 = a.5^1 + b.1.5^1 = 5a+5b$

Solving these two equations, we get $a = 3$ and $b = 2/5$

Hence, the final solution is − $F_n = 3.5^n +( 2/5) .n.2^n $

Solve the recurrence relation $F_n = 2F_{n-1} - 2F_{n-2}$ where $F_0 = 1$ and $F_1 = 3$

$$x^2 -2x -2 = 0$$

$x_1 = 1 + i$ and $x_2 = 1 - i$

In polar form,

$x_1 = r \angle \theta$ and $x_2 = r \angle(- \theta),$ where $r = \sqrt 2$ and $\theta = \frac{\pi}{4}$

The roots are imaginary. So, this is in the form of case 3.

$F_n = (\sqrt 2 )^n (a cos(n .\sqcap /4) + b sin(n .\sqcap /4))$

$1 = F_0 = (\sqrt 2 )^0 (a cos(0 .\sqcap /4) + b sin(0 .\sqcap /4) ) = a$

$3 = F_1 = (\sqrt 2 )^1 (a cos(1 .\sqcap /4) + b sin(1 . \sqcap /4) ) = \sqrt 2 ( a/ \sqrt 2 + b/ \sqrt 2)$

Solving these two equations we get $a = 1$ and $b = 2$

$F_n = (\sqrt 2 )^n (cos(n .\pi /4 ) + 2 sin(n .\pi /4 ))$

Non-Homogeneous Recurrence Relation and Particular Solutions

A recurrence relation is called non-homogeneous if it is in the form

$F_n = AF_{n-1} + BF_{n-2} + f(n)$ where $f(n) \ne 0$

Its associated homogeneous recurrence relation is $F_n = AF_{n–1} + BF_{n-2}$

The solution $(a_n)$ of a non-homogeneous recurrence relation has two parts.

First part is the solution $(a_h)$ of the associated homogeneous recurrence relation and the second part is the particular solution $(a_t)$.

$$a_n=a_h+a_t$$

Solution to the first part is done using the procedures discussed in the previous section.

To find the particular solution, we find an appropriate trial solution.

Let $f(n) = cx^n$ ; let $x^2 = Ax + B$ be the characteristic equation of the associated homogeneous recurrence relation and let $x_1$ and $x_2$ be its roots.

If $x \ne x_1$ and $x \ne x_2$, then $a_t = Ax^n$

If $x = x_1$, $x \ne x_2$, then $a_t = Anx^n$

If $x = x_1 = x_2$, then $a_t = An^2x^n$

Let a non-homogeneous recurrence relation be $F_n = AF_{n–1} + BF_{n-2} + f(n)$ with characteristic roots $x_1 = 2$ and $x_2 = 5$. Trial solutions for different possible values of $f(n)$ are as follows −

Solve the recurrence relation $F_n = 3F_{n-1} + 10F_{n-2} + 7.5^n$ where $F_0 = 4$ and $F_1 = 3$

This is a linear non-homogeneous relation, where the associated homogeneous equation is $F_n=3F_{n-1}+10F_{n-2}$ and $f(n)=7.5^n$

The characteristic equation of its associated homogeneous relation is −

$$x^2 - 3x -10 = 0$$

Or, $(x - 5)(x + 2) = 0$

Or, $x_1= 5$ and $x_2 = -2$

Hence $a_h = a.5^n + b.(-2)^n$ , where a and b are constants.

Since $f(n) = 7.5^n$, i.e. of the form $c.x^n$, a reasonable trial solution of at will be $Anx^n$

$a_t = Anx^n = An5^n$

After putting the solution in the recurrence relation, we get −

$An5^n = 3A(n – 1)5^{n-1} + 10A(n – 2)5^{n-2} + 7.5^n$

Dividing both sides by $5^{n-2}$, we get

$An5^2 = 3A(n - 1)5 + 10A(n - 2)5^0 + 7.5^2$

Or, $25An = 15An - 15A + 10An - 20A + 175$

Or, $35A = 175$

Or, $A = 5$

So, $F_n = An5^n= 5n5^n=n5^{n+1}$

The solution of the recurrence relation can be written as −

$F_n = a_h + a_t$

$=a.5^n+b.(-2)^n+n5^{n+1}$

Putting values of $F_0 = 4$ and $F_1 = 3$, in the above equation, we get $a = -2$ and $b = 6$

$F_n = n5^{n+1} + 6.(-2)^n -2.5^n$

Generating Functions

Generating Functions represents sequences where each term of a sequence is expressed as a coefficient of a variable x in a formal power series.

Mathematically, for an infinite sequence, say $a_0, a_1, a_2,\dots, a_k,\dots,$ the generating function will be −

$$G_x=a_0+a_1x+a_2x^2+ \dots +a_kx^k+ \dots = \sum_{k=0}^{\infty}a_kx^k$$

Some Areas of Application

Generating functions can be used for the following purposes −

For solving a variety of counting problems. For example, the number of ways to make change for a Rs. 100 note with the notes of denominations Rs.1, Rs.2, Rs.5, Rs.10, Rs.20 and Rs.50

For solving recurrence relations

For proving some of the combinatorial identities

For finding asymptotic formulae for terms of sequences

What are the generating functions for the sequences $\lbrace {a_k} \rbrace$ with $a_k = 2$ and $a_k = 3k$?

When $a_k = 2$, generating function, $G(x) = \sum_{k = 0}^{\infty }2x^{k} = 2 + 2x + 2x^{2} + 2x^{3} + \dots$

When $a_{k} = 3k, G(x) = \sum_{k = 0}^{\infty }3kx^{k} = 0 + 3x + 6x^{2} + 9x^{3} + \dots\dots$

What is the generating function of the infinite series; $1, 1, 1, 1, \dots$?

Here, $a_k = 1$, for $0 \le k \le \infty$

Hence, $G(x) = 1 + x + x^{2} + x^{3}+ \dots \dots= \frac{1}{(1 - x)}$

Some Useful Generating Functions

For $a_k = a^{k}, G(x) = \sum_{k = 0}^{\infty }a^{k}x^{k} = 1 + ax + a^{2}x^{2} +\dots \dots \dots = 1/ (1 - ax)$

For $a_{k} = (k + 1), G(x) = \sum_{k = 0}^{\infty }(k + 1)x^{k} = 1 + 2x + 3x^{2} \dots \dots \dots =\frac{1}{(1 - x)^{2}}$

For $a_{k} = c_{k}^{n}, G(x) = \sum_{k = 0}^{\infty} c_{k}^{n}x^{k} = 1+c_{1}^{n}x + c_{2}^{n}x^{2} + \dots \dots \dots + x^{2} = (1 + x)^{n}$

For $a_{k} = \frac{1}{k!}, G(x) = \sum_{k = 0}^{\infty }\frac{x^{k}}{k!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}\dots \dots \dots = e^{x}$

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