1 Introduction to Counting
Total Page:16
File Type:pdf, Size:1020Kb
Load more
Recommended publications
-
Chapter 1. Combinatorial Theory 1.3: No More Counting Please 1.3.1
Chapter 1. Combinatorial Theory 1.3: No More Counting Please Slides (Google Drive) Alex Tsun Video (YouTube) In this section, we don't really have a nice successive ordering where one topic leads to the next as we did earlier. This section serves as a place to put all the final miscellaneous but useful concepts in counting. 1.3.1 Binomial Theorem n We talked last time about binomial coefficients of the form k . Today, we'll see how they are used to prove the binomial theorem, which we'll use more later on. For now, we'll see how they can be used to expand (possibly large) exponents below. You may have learned this technique of FOIL (first, outer, inner, last) for expanding (x + y)2. We then combine like-terms (xy and yx). (x + y)2 = (x + y)(x + y) = xx + xy + yx + yy [FOIL] = x2 + 2xy + y2 But, let's say that we wanted to do this for a binomial raised to some higher power, say (x + y)4. There would be a lot more terms, but we could use a similar approach. 1 2 Probability & Statistics with Applications to Computing 1.3 (x + y)4 = (x + y)(x + y)(x + y)(x + y) = xxxx + yyyy + xyxy + yxyy + ::: But what are the terms exactly that are included in this expression? And how could we combine the like-terms though? Notice that each term will be a mixture of x's and y's. In fact, each term will be in the form xkyn−k (in this case n = 4). -
Useful Relations in Permutations and Combination 1. Useful Relations
Useful Relations in Permutations and Combination 1. Useful Relations - Factorial n! = n.(n-1)! 2. n퐶푟= n푃푟/r! n n-1 3. Pr = n( Pr-1) 4. Useful Relations - Combinations n n 1. Cr = C(n - r) Example 8 8 C6 = C2 = 8×72×1 = 28 n 2. Cn = 1 n 3. C0 = 1 n n n n n 4. C0 + C1 + C2 + ... + Cn = 2 Example 4 4 4 4 4 4 C0 + C1 + C2 + C3+ C4 = (1 + 4 + 6 + 4 + 1) = 16 = 2 n n (n+1) Cr-1 + Cr = Cr (Pascal's Law) n퐶푟 =n/퐶푟−1=n-r+1/r n n If Cx = Cy then either x = y or (n-x) = y. 5. Selection from identical objects: Some Basic Facts The number of selections of r objects out of n identical objects is 1. Total number of selections of zero or more objects from n identical objects is n+1. 6. Permutations of Objects when All Objects are Not Distinct The number of ways in which n things can be arranged taking them all at a time, when st nd 푃1 of the things are exactly alike of 1 type, 푃2 of them are exactly alike of a 2 type, and th 푃푟of them are exactly alike of r type and the rest of all are distinct is n!/ 푃1! 푃2! ... 푃푟! 1 Example: how many ways can you arrange the letters in the word THESE? 5!/2!=120/2=60 Example: how many ways can you arrange the letters in the word REFERENCE? 9!/2!.4!=362880/2*24=7560 7.Circular Permutations: Case 1: when clockwise and anticlockwise arrangements are different Number of circular permutations (arrangements) of n different things is (n-1)! 1. -
Combinatorics
Combinatorics Problem: How to count without counting. I How do you figure out how many things there are with a certain property without actually enumerating all of them. Sometimes this requires a lot of cleverness and deep mathematical insights. But there are some standard techniques. I That's what we'll be studying. We sometimes use the bijection rule without even realizing it: I count how many people voted are in favor of something by counting the number of hands raised: I I'm hoping that there's a bijection between the people in favor and the hands raised! Bijection Rule The Bijection Rule: If f : A ! B is a bijection, then jAj = jBj. I We used this rule in defining cardinality for infinite sets. I Now we'll focus on finite sets. Bijection Rule The Bijection Rule: If f : A ! B is a bijection, then jAj = jBj. I We used this rule in defining cardinality for infinite sets. I Now we'll focus on finite sets. We sometimes use the bijection rule without even realizing it: I count how many people voted are in favor of something by counting the number of hands raised: I I'm hoping that there's a bijection between the people in favor and the hands raised! Answer: 26 choices for the first letter, 26 for the second, 10 choices for the first number, the second number, and the third number: 262 × 103 = 676; 000 Example 2: A traveling salesman wants to do a tour of all 50 state capitals. How many ways can he do this? Answer: 50 choices for the first place to visit, 49 for the second, . -
Inclusion‒Exclusion Principle
Inclusionexclusion principle 1 Inclusion–exclusion principle In combinatorics, the inclusion–exclusion principle (also known as the sieve principle) is an equation relating the sizes of two sets and their union. It states that if A and B are two (finite) sets, then The meaning of the statement is that the number of elements in the union of the two sets is the sum of the elements in each set, respectively, minus the number of elements that are in both. Similarly, for three sets A, B and C, This can be seen by counting how many times each region in the figure to the right is included in the right hand side. More generally, for finite sets A , ..., A , one has the identity 1 n This can be compactly written as The name comes from the idea that the principle is based on over-generous inclusion, followed by compensating exclusion. When n > 2 the exclusion of the pairwise intersections is (possibly) too severe, and the correct formula is as shown with alternating signs. This formula is attributed to Abraham de Moivre; it is sometimes also named for Daniel da Silva, Joseph Sylvester or Henri Poincaré. Inclusion–exclusion illustrated for three sets For the case of three sets A, B, C the inclusion–exclusion principle is illustrated in the graphic on the right. Proof Let A denote the union of the sets A , ..., A . To prove the 1 n Each term of the inclusion-exclusion formula inclusion–exclusion principle in general, we first have to verify the gradually corrects the count until finally each identity portion of the Venn Diagram is counted exactly once. -
STRUCTURE ENUMERATION and SAMPLING Chemical Structure Enumeration and Sampling Have Been Studied by Mathematicians, Computer
STRUCTURE ENUMERATION AND SAMPLING MARKUS MERINGER To appear in Handbook of Chemoinformatics Algorithms Chemical structure enumeration and sampling have been studied by 5 mathematicians, computer scientists and chemists for quite a long time. Given a molecular formula plus, optionally, a list of structural con- straints, the typical questions are: (1) How many isomers exist? (2) Which are they? And, especially if (2) cannot be answered completely: (3) How to get a sample? 10 In this chapter we describe algorithms for solving these problems. The techniques are based on the representation of chemical compounds as molecular graphs (see Chapter 2), i.e. they are mainly applied to constitutional isomers. The major problem is that in silico molecular graphs have to be represented as labeled structures, while in chemical 15 compounds, the atoms are not labeled. The mathematical concept for approaching this problem is to consider orbits of labeled molecular graphs under the operation of the symmetric group. We have to solve the so–called isomorphism problem. According to our introductory questions, we distinguish several dis- 20 ciplines: counting, enumerating and sampling isomers. While counting only delivers the number of isomers, the remaining disciplines refer to constructive methods. Enumeration typically encompasses exhaustive and non–redundant methods, while sampling typically lacks these char- acteristics. However, sampling methods are sometimes better suited to 25 solve real–world problems. There is a wide range of applications where counting, enumeration and sampling techniques are helpful or even essential. Some of these applications are closely linked to other chapters of this book. Counting techniques deliver pure chemical information, they can help to estimate 30 or even determine sizes of chemical databases or compound libraries obtained from combinatorial chemistry. -
Lecture 6: Entropy
Matthew Schwartz Statistical Mechanics, Spring 2019 Lecture 6: Entropy 1 Introduction In this lecture, we discuss many ways to think about entropy. The most important and most famous property of entropy is that it never decreases Stot > 0 (1) Here, Stot means the change in entropy of a system plus the change in entropy of the surroundings. This is the second law of thermodynamics that we met in the previous lecture. There's a great quote from Sir Arthur Eddington from 1927 summarizing the importance of the second law: If someone points out to you that your pet theory of the universe is in disagreement with Maxwell's equationsthen so much the worse for Maxwell's equations. If it is found to be contradicted by observationwell these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of ther- modynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation. Another possibly relevant quote, from the introduction to the statistical mechanics book by David Goodstein: Ludwig Boltzmann who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics. There are many ways to dene entropy. All of them are equivalent, although it can be hard to see. In this lecture we will compare and contrast dierent denitions, building up intuition for how to think about entropy in dierent contexts. The original denition of entropy, due to Clausius, was thermodynamic. -
Counting: Permutations, Combinations
CS 441 Discrete Mathematics for CS Lecture 17 Counting Milos Hauskrecht [email protected] 5329 Sennott Square CS 441 Discrete mathematics for CS M. Hauskrecht Counting • Assume we have a set of objects with certain properties • Counting is used to determine the number of these objects Examples: • Number of available phone numbers with 7 digits in the local calling area • Number of possible match starters (football, basketball) given the number of team members and their positions CS 441 Discrete mathematics for CS M. Hauskrecht 1 Basic counting rules • Counting problems may be hard, and easy solutions are not obvious • Approach: – simplify the solution by decomposing the problem • Two basic decomposition rules: – Product rule • A count decomposes into a sequence of dependent counts (“each element in the first count is associated with all elements of the second count”) – Sum rule • A count decomposes into a set of independent counts (“elements of counts are alternatives”) CS 441 Discrete mathematics for CS M. Hauskrecht Inclusion-Exclusion principle Used in counts where the decomposition yields two count tasks with overlapping elements • If we used the sum rule some elements would be counted twice Inclusion-exclusion principle: uses a sum rule and then corrects for the overlapping elements. We used the principle for the cardinality of the set union. •|A B| = |A| + |B| - |A B| U B A CS 441 Discrete mathematics for CS M. Hauskrecht 2 Inclusion-exclusion principle Example: How many bitstrings of length 8 start either with a bit 1 or end with 00? • Count strings that start with 1: • How many are there? 27 • Count the strings that end with 00. -
Binomial Coefficients and Subsets
page 1 Finite Mathematics Binomial Coefficients and Subsets In the worksheet “Combinations and Their Sums,” we saw how ibn Mun‘im, living in Morocco around 1200, used combinations C(n,r) to count the number of ways to select colored threads or other things from a pool of items, assuming that repeats aren’t allowed and the order of selection doesn’t matter. Let’s do some practice and warm-up. Exercise 1. a) In how many different ways can 3 employees be selected from an office of 7 employees? C(____, ____) = ______ b) Use the “row-sum pattern” to simplify: C(3,3) + C(4,3) + C(5,3) + C(6,3) = C(____, ____) = ______ c) Use the “column-sum pattern” to simplify: C(9,5) + C(9,6) = C(____, ____) = ______ Ibn Mun‘im wasn’t the first to explore combinations like these, but apparently they’d never been studied by mathematicians in ancient Greece and Mesopotamia, who excelled in many other topics. In the late 900s, an Indian poet named Halayudha presented the arithmetical triangle as a tool in categorizing the different kinds of poetic meter, the sequences of stressed or unstressed syllables in one line of poetry (see Exercise 9 below). A little later, around 1010, the Arab mathematician Abū Bakr al-Karajī in Baghdad used the arithmetical triangle to solve an algebra problem, that of figuring out the coefficients of binomial powers like (x + y) 4 (see Exercises 2-4 below). During the Middle Ages, it was the Arab world that was pre-eminent in science and mathematics. -
Week 3-4: Permutations and Combinations
Week 3-4: Permutations and Combinations March 10, 2020 1 Two Counting Principles Addition Principle. Let S1;S2;:::;Sm be disjoint subsets of a ¯nite set S. If S = S1 [ S2 [ ¢ ¢ ¢ [ Sm, then jSj = jS1j + jS2j + ¢ ¢ ¢ + jSmj: Multiplication Principle. Let S1;S2;:::;Sm be ¯nite sets and S = S1 £ S2 £ ¢ ¢ ¢ £ Sm. Then jSj = jS1j £ jS2j £ ¢ ¢ ¢ £ jSmj: Example 1.1. Determine the number of positive integers which are factors of the number 53 £ 79 £ 13 £ 338. The number 33 can be factored into 3 £ 11. By the unique factorization theorem of positive integers, each factor of the given number is of the form 3i £ 5j £ 7k £ 11l £ 13m, where 0 · i · 8, 0 · j · 3, 0 · k · 9, 0 · l · 8, and 0 · m · 1. Thus the number of factors is 9 £ 4 £ 10 £ 9 £ 2 = 7280: Example 1.2. How many two-digit numbers have distinct and nonzero digits? A two-digit number ab can be regarded as an ordered pair (a; b) where a is the tens digit and b is the units digit. The digits in the problem are required to satisfy a 6= 0; b 6= 0; a 6= b: 1 The digit a has 9 choices, and for each ¯xed a the digit b has 8 choices. So the answer is 9 £ 8 = 72. The answer can be obtained in another way: There are 90 two-digit num- bers, i.e., 10; 11; 12;:::; 99. However, the 9 numbers 10; 20;:::; 90 should be excluded; another 9 numbers 11; 22;:::; 99 should be also excluded. So the answer is 90 ¡ 9 ¡ 9 = 72. -
Combinatorics Combinatorics I Combinatorics II Product Rule Sum Rule I Sum Rule II
Combinatorics I Combinatorics Introduction Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Combinatorics is the study of collections of objects. Specifically, counting objects, arrangement, derangement, etc. of objects along with their mathematical properties. Counting objects is important in order to analyze algorithms and Spring 2006 compute discrete probabilities. Originally, combinatorics was motivated by gambling: counting configurations is essential to elementary probability. Computer Science & Engineering 235 Introduction to Discrete Mathematics Sections 4.1-4.6 & 6.5-6.6 of Rosen [email protected] Combinatorics II Product Rule Introduction If two events are not mutually exclusive (that is, we do them separately), then we apply the product rule. A simple example: How many arrangements are there of a deck of 52 cards? Theorem (Product Rule) In addition, combinatorics can be used as a proof technique. Suppose a procedure can be accomplished with two disjoint A combinatorial proof is a proof method that uses counting subtasks. If there are n1 ways of doing the first task and n2 ways arguments to prove a statement. of doing the second, then there are n1 · n2 ways of doing the overall procedure. Sum Rule I Sum Rule II There is a natural generalization to any sequence of m tasks; If two events are mutually exclusive, that is, they cannot be done namely the number of ways m mutually exclusive events can occur at the same time, then we must apply the sum rule. is n1 + n2 + ··· nm−1 + nm Theorem (Sum Rule) We can give another formulation in terms of sets. Let If an event e1 can be done in n1 ways and an event e2 can be done A1,A2,...,Am be pairwise disjoint sets. -
The Pigeonhole Principle
The Pigeonhole Principle The pigeonhole principle is the following: If m objects are placed into n bins, where m > n, then some bin contains at least two objects. (We proved this in Lecture #02) Why This Matters ● The pigeonhole principle can be used to show a surprising number of results must be true because they are “too big to fail.” ● Given a large enough number of objects with a bounded number of properties, eventually at least two of them will share a property. ● The applications are extremely deep and thought-provoking. Using the Pigeonhole Principle ● To use the pigeonhole principle: ● Find the m objects to distribute. ● Find the n < m buckets into which to distribute them. ● Conclude by the pigeonhole principle that there must be two objects in some bucket. ● The details of how to proceeds from there are specific to the particular proof you're doing. Theorem: For any natural number n, there is a nonzero multiple of n whose digits are all 0s and 1s. Theorem: For any natural number n, there is a nonzero multiple of n whose digits are all 0s and 1s. 1 11 111 1111 11111 There are 10 objects here. 111111 1111111 11111111 111111111 1111111111 Theorem: For any natural number n, there is a nonzero multiple of n whose digits are all 0s and 1s. 0 1 11 1 111 2 1111 11111 3 111111 4 1111111 11111111 5 111111111 6 1111111111 7 8 Theorem: For any natural number n, there is a nonzero multiple of n whose digits are all 0s and 1s. -
Formalization of Some Central Theorems in Combinatorics of Finite
Kalpa Publications in Computing Volume 1, 2017, Pages 43–57 LPAR-21S: IWIL Workshop and LPAR Short Presentations Formalization of some central theorems in combinatorics of finite sets Abhishek Kr Singh School of Technology and Computer Science, Tata Institute of Fundamental Research, Mumbai [email protected] Abstract We present fully formalized proofs of some central theorems from combinatorics. These are Dilworth’s decomposition theorem, Mirsky’s theorem, Hall’s marriage theorem and the Erdős-Szekeres theorem. Dilworth’s decomposition theorem is the key result among these. It states that in any finite partially ordered set (poset), the size of a smallest chain cover and a largest antichain are the same. Mirsky’s theorem is a dual of Dilworth’s decomposition theorem, which states that in any finite poset, the size of a smallest antichain cover and a largest chain are the same. We use Dilworth’s theorem in the proofs of Hall’s Marriage theorem and the Erdős-Szekeres theorem. The combinatorial objects involved in these theorems are sets and sequences. All the proofs are formalized in the Coq proof assistant. We develop a library of definitions and facts that can be used as a framework for formalizing other theorems on finite posets. 1 Introduction Formalization of any mathematical theory is a difficult task because the length of a formal proof blows up significantly. In combinatorics the task becomes even more difficult due to the lack of structure in the theory. Some statements often admit more than one proof using completely different ideas. Thus, exploring dependencies among important results may help in identifying an effective order amongst them.