CIS Spring 2018 (instructor Val Tannen)
|
|
- Godwin Stanley Phillips
- 6 years ago
- Views:
Transcription
1 CIS Sprig 2018 (istructor Val Tae) Lecture 5 Thursday, Jauary 25 COUNTING We cotiue studyig how to use combiatios ad what are their properties. Example 5.1 How may 8-letter strigs ca be costructed by usig the 26 letters of the alphabet if each strig cotais 3,4, or 5 vowels? There is o restrictio o the umber of occurreces of a letter i the strig. Solutio: Let E be the set of 8-letter strigs that cotai at least 3 vowels. Let E i be the set of 8-letter strigs cotaiig exactly i > 0 vowels (so i {1, 2, 3, 4, 5}) A elemet of E i, i.e., a 8-letter strig with exactly i vowels, ca be costructed usig the followig steps. Step 1. Choose i locatios for vowels out of the available 8 locatios. Step 2. Choose the vowels for each of the i locatios. Step 3. Choose the cosoats for each of the remaiig 8 i locatios. Step 1 ca be performed i ( 8 i) ways. Step 2 ca be performed i 5 i ways. Step 3 ca be performed i 21 8 i ways. By the multiplicatio rule, the umber of 8-letter strigs with exactly i vowels is give by ( ) 8 E i 5 i 21 8 i i Sice the sets E 3, E 4, ad E 5 partitio (divide i pairwise disjoit subsets) the set E, by the sum rule we get 5 5 ( ) 8 E E i 5 i 21 8 i i i3 Addedum. The followig questio was raised i class i oe of the previous offerigs of this course. What if we wat to cout all 8-letter strigs with distict letters that have 3, 4, or 5 vowels? I this case, the above procedure still applies. However, the umber of ways of doig each step chages. Step 1 ca be performed i ( 8 i) ways. Step 2 ca be performed i (5)i ways. Step 3 ca be performed i (21) 8 i ways. By the multiplicatio rule, the umber of 8-letter strigs with distict letters that have exactly i vowels is give by ( ) 8 (5) i (21) 8 i i i3 1
2 The total umber of 8-letter strigs with distict letters that have 3, 4, or 5 vowels is 5 i3 ( ) 8 (5) i (21) 8 i i It is worthwhile revisitig Example 2.5 from Lecture 2 ad comparig it with what we did just ow. Example 5.2 Suppose you are i Mahatta at the NY Public Library (5th Ave ad 42d St) ad you wish to walk to Columbus Circle (8th Ave ad 59th St) takig a shortest path (but you caot take Broadway for some reaso :). How may blocks do you eed to walk? How may differet ways are there to walk a shortest path? Solutio: A shortest path cosists of makig oly two kids of decisios: go-west oe block (at most times), or, go-north oe block (at most times), for a total of blocks. How may differet paths of legth 20 blocks are there from the Library to the Circle? Ay such path is a sequece of 20 decisios of oe of two kids: go-west 3 times, or go-north 17 times. Ad ay such sequece is a valid shortest path. Such a sequece has 20 positios ad choosig the 3 positios i which to make go-west decisios ca be doe i ( ) 20 3 ways. If we choose istead the positios i which to make go-north decisios we get ( 20 17) differet ways. But these umbers are equal (why?). More geerally, cosider the grid defied by the poits of iteger coordiates i the plae. Assumig that we ca move oly alog lies of iteger abscissa or iteger ordiate how may differet shortest paths are there from the origi to the poit (m, ) where m, N? The aswer is ( ) ( ) m + m + m By the way, you ca check that these two are equal by usig the expressio with factorial, or, better, wait for a combiatorial proof comig up below. Stars ad bars (sticks ad crosses) coutig certai kid of collectios. Next we explai, via a example, a versatile method for Example 5.3 How may differet ways are there to buy a doze douts whe 5 give glazes are available? Assume (1) that there is a ulimited supply of douts of each glaze, (2) that douts of the same glaze are idistiguishable, ad (3) that the placemet of the douts i, say, a box, does ot distiguishes your purchases. 2
3 Solutio: We apply stars ad bars as follows. Arrage 12 stars i a row. They represet hypothetical douts before glazess are assiged, let s call them proto-douts :) Obviously, all 12 proto-douts are idistiguishable so their orderig is irrelevat. Now place 4 bars betwee some of the stars. Note that this separates the proto-douts ito 5 cotiguous parts. 1 Assig the 1st glaze to the proto-douts i the leftmost part, 2d glaze to the ext part, etc. For example: *** ** * **** ** A B C D E Achocolate, Bmaple, Cdulce-de-leche, Damaretto-cherry, Ehazelut Note that it is possible that you do t buy douts of a specific glaze at all. This meas that we have to allow for bars at the begiig or ed ad for adjacet bars: *** ****** *** ********* *** * * ********** A B C D E AB C D E A BCD E I the first example above there are o maple-glazed douts, ad o hazelut-glazed douts, etc. Note also that because the proto-douts are idistiguishable, the orderig of A,B,C,D,E does ot matter, oly how may proto-douts are i each cotiguous part matters. For example the followig have the same glaze distributio hece they produce the same purchase. *** ** * **** ** * **** ** ** *** A B C D E C D B E A Thus, to avoid overcoutig, we fix a orderig of A,B,C,D,E ad the we cout. So how do we cout the differet ways of costructig a sequece of stars ad bars that correspods to a this fixed orderig of glazes? The 12 stars ad the 4 bars form a sequece with positios. Out of these, we choose 4 positios where we put the bars, or, equivaletly, 12 positios where we put the stars. The aswer is ( ) ( ). Let s review what we have doe so far about coutig differet kids of collectios. We have see that the umber of distict sequeces of legth r made of elemets from a give set of elemets is r This is the same as the umber of differet ways to costruct from elemets of a set with elemets a collectio of size r that is both ordered ad allows repetitios. 1 I am usig part rather tha (sub)sequece because their orderig is irrelevat. 3
4 We have also cosidered coutig the umber of ways to costruct from elemets of a give set with elemets a collectio of size r that is ordered but does ot allow repetitios. We called these permutatios of r out of ad we deoted their umber by () r! ( r)! Fially, we have cosidered coutig the umber of ways to costruct from elemets of a give set with elemets a collectio of size r that is uordered ad does ot allow repetitios. That is the same as coutig the subsets of size r of the set with elemets. We called these combiatios of r out of ad we deoted their umber by ( r ) () r r!! r!( r)! It is ow atural to cosider coutig the umber of ways to costruct from elemets of a give set with elemets a collectio of size r that is uordered but does allow repetitios. Such a collectio is called a bag or a multiset (because it is set-like, but with repetitios). We shall call these combiatios of r out of with repetitio ad deote their umber by (( r)) (read multichoose r). Notatio for bags There is o stadard otatio. Let A {a, b, c, d} be the give set. Some people use the ormal set otatio (braces) for bags (multisets) ad just repeat elemets. For example, the bag {a, b, a, c, a, b} is the same as the bag {a, a, a, b, b, c}. This ca be dagerous for oe who cofuses them with sets i which the copies of the same elemet are somehow distiguishable. To avoid this, some people use double braces: {{a, b, a, c, a, b}} (same as {{a, a, a, b, b, c}}). A iterestig alterative is to use braces, but, for each distict elemet, use a atural umber that gives the umber of copies (repetitios). For istace the bag above would be {3 a, 2 b, 1 c} or eve {3 a, 2 b, 1 c, 0 d}! (we shall see that this ca be thought of a certai kid of fuctio.) Do t use this otatio whe the elemets of A are umbers, say, A {1, 2, 3, 4}, it s cofusig! Coutig bags Let A {a 1,..., a } be the give set. Its elemets are (clearly) distiguishable. It turs out that we ca use exactly the stars ad bars techique that we used to cout dout purchases to cout the umber of bags of size r costructed from elemets of A ad thus obtai a formula for (( )) r. 2 Thik of costructig bag of size r as separatig r stars ito parts correspodig to copies of the same elemet of A. Sice A we use it with 1 bars. This ca be doe by choosig 1 positios for the bars (equivaletly, choosig r positios for the stars) i a sequece of legth + r 1. Therefore (( )) ( ) ( ) + r 1 + r 1 r r 1 2 A otatio used by some textbooks, icludig Scheierma. 4
5 Warig Note that is the umber of bags of size r (so repetitios are couted i the r) made from elemets of a set of size (so o repetitios amog the ). Marbles i bis, cois to childre (( r)) is also the umber of ways of puttig r idistiguishable marbles i distiguishable bis; or, the umber of ways of distributig r idistiguishable cois 3 to distiguishable childre; or, the umber of distict atural umber solutios to the Diophatie equatio x 1 + x x r Example 5.4 I how may ways ca we distribute 11 idistiguishable cois to 3 distiguishable childre? Solutio: (( )) ( ) ( ) 2. Example 5.5 A (somewhat fairer) distributio of 11 idistiguishable cois to 3 distiguishable childre assumes that each child gets at least 3 cois. I how may differet ways ca this be doe? Solutio: We begi by givig 3 cois each to the 3 childre. This leaves us cois to distribute to the 3 childre accordig to the geeral scheme. So the umber of ways is (( 3 ( 2)) ) ( 2 4 ) 2 6. Defiitio 5.6 Let m, be two itegers such that m. The iteger iterval [m..] is defied as the set of itegers m ad. That is [m..] {x Z m x } Note that both m ad are elemets i [m..]. Note also that there are m + 1 elemets i [m..]. That is [m..] m + 1 Do ot cofuse iteger itervals with the usual itervals of real umbers. For istace [0..1] {0, 1} has two elemets while [0, 1] is ifiite. 3 May years ago this was taught with peies, the with ickels, evetually with quarters, keepig up with iflatio, you kow. At some poit we gave up ad started usig cois. I expect that at some poit i the future we ll use bitcois ad beam them directly ito the childre s electroic persoal assistat. 5
6 Example 5.7 What is the umber of o-decreasig sequeces of legth 12 whose elemets are take from [10..30]? What is the umber of icreasig (we ofte say strictly icreasig so there is o cofusio) such sequeces? Solutio: [10..30] has elemets. No-decreasig meas that elemets ca be repeated, as log as copies are adjacet i the sequece. Here is a procedure for costructig a o-decreasig sequece of legth 12 usig itegers i [10..30]. Step 1: choose a uordered collectio of 12 out of the 21 umbers, with repetitio. Step 2: order it. Step 1 ca be doe i (( )) ( ) ways. Step 2 ca be doe i exactly oe way. The aswer is ( 32 12). O the other had, to cout the umber of strictly icreasig sequeces of legth 12 made of umbers from [10..30] we choose a combiatio of 12 out of 21 (without repetitio!) ad the order it. The aswer is ( 21 12). Biomial coefficiets Mathematicias typically use this ame istead of (umber of) combiatios. We are about to see why. A biomial is a sum of two terms, such as a + b. The biomial theorem gives a expressio for (a + b) where a ad b are real umbers ad is a atural umber. Theorem 5.8 (Biomial Theorem) For ay real umbers a ad b ad atural umber ( ) (a + b) a k b k k k0 Proof: The formula is easily checked for 0. For 1, let s thik what happes whe we calculate the expasio (a + b) (a + b) (a + b) (a + b) We obtai a sum of terms (moomials), each of the form a k b k, where k is 0 or 1, etc.,..., or. 4 Most of these terms occur more tha oce (which oes occur exactly oce?). How may times does a k b k occur i the expasio? This is the same umber of times as there are orderigs of k a s ad k b s. This is equal to ( ) k. Thus the coefficiet of like terms of the form a k b k is ( k). This proves the theorem. Pascal s Triagle We kow that (a + b) 0 1, (a + b) 1 a + b, (a + b) 2 a 2 + 2ab + b 2, (a + b) 3 a 3 + 3a 2 b + 3ab 2 + b 3, etc. 4 How may terms? Use the multiplicatio rule to figure out that there are 2 terms! 6
7 Blaise Pascal observed some iterestig relatioships betwee the biomial coefficiets whe arraged i rows as follows: Specifically, observe that each biomial coefficiet is the sum of the two just above it. Formally: Theorem 5.9 (Pascal s Idetity) If ad k are positive itegers with k the ( ) ( ) ( ) k k 1 k Proof: This ca be easily verified usig the factorial formula for combiatios. followig proof is more isightful. However, the Let S {x 1, x 2,..., x } be the set of elemets. We cout the umber of subsets of size k i two ways. Oe way gives the combiatios of k our of, i.e., the LHS of Pascal s Idetity. I the secod way, we observe that k-elemet subsets of S ca be partitioed ito those that cotai x ad those that do t. For the former type of subset the other k 1 elemets come from S \ {x }. There are ( 1 k 1) ways of choosig these subsets. For the latter type of subset all of the k elemets must be chose from S \ {x }. There are ( ) 1 k ways of doig this. Thus, by the sum rule the umber of k-elemet subsets of S is give by the RHS of Pascal s Idetity. Proofs like the oe we just saw are called combiatorial. Combiatorial proofs We ow prove a couple of idetities usig the followig techique. To prove a idetity we will pose a coutig questio. We will the aswer the questio i two ways, oe aswer will correspod to LHS ad the other would correspod to the RHS of the idetity. We have see a example of this techique i the proof of Pascal s Idetity. This techique is ofte called a combiatorial proof. Example 5.10 Prove that ( ) r ( ) r 7
8 Solutio: We give a combiatorial proof. Let X be a set of size. We cout the umber of subsets of size r of X i two ways. We already did oe such cout whe we defied combiatios ad we kow the aswer is ( r). Alteratively we cout the umber of ways we ca choose elemets that are ot i a subset of size r. That is, cout the umber of subsets of size r which is, of course, ( r). The subsets of size r are i oe-to-oe correspodece with their complemets (see Example 3.12 i Lecture 3). These complemets are exactly the subsets of size r. Therefore there must be as may subsets of size r as there are subsets of size r. This proves the idetity. (The argumet that uses, iformally, the oe-to-oe correspodece is a example of somethig called the Bijectio Rule. We will lear about it whe we study fuctios.) Example 5.11 Prove that ( ) 0 ( ) + 1 ( ) ( )... + ( 1) 0 2 Solutio: Oe way to solve this problem is by substitutig a 1 ad b 1 i the Biomial Theorem, yieldig ( ) 0 0 ( 1) k. k k0 However, a combiatorial proof will give us more isight ito what the expressio meas. We wat to prove that ( ) ( ) ( ) ( ) ( ) ( ) Cosider a set X {x 1, x 2, x 3,..., x }. We wat to show that the total umber of subsets of X that have eve size equals the total umber of subsets of X that have odd size. We will ow show that both these quatities equal 2 1 from which the claim follows. A eve-sized subset of X ca be costructed as follows. Step 1. Decide whether x 1 belogs to the subset or ot. Step 2. Decide whether x 2 belogs to the subset or ot.. Step. Decide whether x belogs to the subset or ot. I the first 1 steps oe ca make either oe of the 2 choices, i or out. But i step oly oe choice is possible! This is because if we have chose a eve umber of elemets from X \ {x } to put i the subset the we must leave out x otherwise we must iclude x i the subset. Hece usig the multiplicatio rule the total umber of eve-sized subsets of X equals
9 Aother way of thikig about this is to cout i two steps: i the first step choose a subset of {x 1,..., x 1 }; i the secod step add or ot x to the subset chose i the first step, makig sure the result has eve size (do t forget that 0 is eve!). To compute the umber of odd-sized subsets we could proceed similarly. Or, we could cout complemetarily: sice we kow that the total umber of subsets of X is 2, the total umber of odd-sized subsets of X is (2 1) 2 1. Example 5.12 Prove that k0 ( k) 2. Solutio: We pose the followig coutig questio. Give a set S of distict elemets how may subsets are there of the set S? From earlier lectures, we kow that the aswer is 2. This gives us the RHS. Aother way to compute the aswer to the questio is as follows. The powerset 2 S cotaiig all possible subsets ca be partitioed ito S 0, S 1,..., S, where S i, 0 i, is the set of all subsets of S that have cardiality i. Thus 2 S S 0 + S S ( ) ( ) ( ) ( ) LHS k k0 This proves the claim. (How do you derive the same idetity from the Biomial Theorem?) Example 5.13 Give a combiatorial proof to show that r k0 ( )( ) m k r k Solutio: We pose the followig coutig questio. ( ) + m There are me ad m wome. How may ways are there to form a committee of r people from this group of people? r 9
10 By defiitio, there are ( ) +m r distict committees of r people. This gives us the RHS. The set S of all possible committees of r people ca be partitioed ito subsets S 0, S 1, S 2,..., S r, where S k is the set of committees i which there are exactly k me ad the rest r k are wome. Note that S k ( )( m k r k). Thus we have S r k0 r k0 ( )( ) m k r k which gives us the left had side of the expressio. 10
Lecture Overview. 2 Permutations and Combinations. n(n 1) (n (k 1)) = n(n 1) (n k + 1) =
COMPSCI 230: Discrete Mathematics for Computer Sciece April 8, 2019 Lecturer: Debmalya Paigrahi Lecture 22 Scribe: Kevi Su 1 Overview I this lecture, we begi studyig the fudametals of coutig discrete objects.
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More informationSection 5.1 The Basics of Counting
1 Sectio 5.1 The Basics of Coutig Combiatorics, the study of arragemets of objects, is a importat part of discrete mathematics. I this chapter, we will lear basic techiques of coutig which has a lot of
More informationCombinatorics and Newton s theorem
INTRODUCTION TO MATHEMATICAL REASONING Key Ideas Worksheet 5 Combiatorics ad Newto s theorem This week we are goig to explore Newto s biomial expasio theorem. This is a very useful tool i aalysis, but
More informationLet us consider the following problem to warm up towards a more general statement.
Lecture 4: Sequeces with repetitios, distributig idetical objects amog distict parties, the biomial theorem, ad some properties of biomial coefficiets Refereces: Relevat parts of chapter 15 of the Math
More informationGenerating Functions. 1 Operations on generating functions
Geeratig Fuctios The geeratig fuctio for a sequece a 0, a,..., a,... is defied to be the power series fx a x. 0 We say that a 0, a,... is the sequece geerated by fx ad a is the coefficiet of x. Example
More information( ) GENERATING FUNCTIONS
GENERATING FUNCTIONS Solve a ifiite umber of related problems i oe swoop. *Code the problems, maipulate the code, the decode the aswer! Really a algebraic cocept but ca be eteded to aalytic basis for iterestig
More informationInjections, Surjections, and the Pigeonhole Principle
Ijectios, Surjectios, ad the Pigeohole Priciple 1 (10 poits Here we will come up with a sloppy boud o the umber of parethesisestigs (a (5 poits Describe a ijectio from the set of possible ways to est pairs
More informationCSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo
Coutig Methods CSE 191, Class Note 05: Coutig Methods Computer Sci & Eg Dept SUNY Buffalo c Xi He (Uiversity at Buffalo CSE 191 Discrete Structures 1 / 48 Need for Coutig The problem of coutig the umber
More informationIntermediate Math Circles November 4, 2009 Counting II
Uiversity of Waterloo Faculty of Mathematics Cetre for Educatio i Mathematics ad Computig Itermediate Math Circles November 4, 009 Coutig II Last time, after lookig at the product rule ad sum rule, we
More informationHomework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is
Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of
More informationChapter 7 COMBINATIONS AND PERMUTATIONS. where we have the specific formula for the binomial coefficients:
Chapter 7 COMBINATIONS AND PERMUTATIONS We have see i the previous chapter that (a + b) ca be writte as 0 a % a & b%þ% a & b %þ% b where we have the specific formula for the biomial coefficiets: '!!(&)!
More information1 Counting and Stirling Numbers
1 Coutig ad Stirlig Numbers Natural Numbers: We let N {0, 1, 2,...} deote the set of atural umbers. []: For N we let [] {1, 2,..., }. Sym: For a set X we let Sym(X) deote the set of bijectios from X to
More informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
More informationPermutations, Combinations, and the Binomial Theorem
Permutatios, ombiatios, ad the Biomial Theorem Sectio Permutatios outig methods are used to determie the umber of members of a specific set as well as outcomes of a evet. There are may differet ways to
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More information(ii) Two-permutations of {a, b, c}. Answer. (B) P (3, 3) = 3! (C) 3! = 6, and there are 6 items in (A). ... Answer.
SOLUTIONS Homewor 5 Due /6/19 Exercise. (a Cosider the set {a, b, c}. For each of the followig, (A list the objects described, (B give a formula that tells you how may you should have listed, ad (C verify
More informationLecture 10: Mathematical Preliminaries
Lecture : Mathematical Prelimiaries Obective: Reviewig mathematical cocepts ad tools that are frequetly used i the aalysis of algorithms. Lecture # Slide # I this
More informationThe Random Walk For Dummies
The Radom Walk For Dummies Richard A Mote Abstract We look at the priciples goverig the oe-dimesioal discrete radom walk First we review five basic cocepts of probability theory The we cosider the Beroulli
More informationChapter 0. Review of set theory. 0.1 Sets
Chapter 0 Review of set theory Set theory plays a cetral role i the theory of probability. Thus, we will ope this course with a quick review of those otios of set theory which will be used repeatedly.
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationRandomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018)
Radomized Algorithms I, Sprig 08, Departmet of Computer Sciece, Uiversity of Helsiki Homework : Solutios Discussed Jauary 5, 08). Exercise.: Cosider the followig balls-ad-bi game. We start with oe black
More informationSummer High School 2009 Aaron Bertram
Summer High School 009 Aaro Bertram 3 Iductio ad Related Stuff Let s thik for a bit about the followig two familiar equatios: Triagle Number Equatio Square Number Equatio: + + 3 + + = ( + + 3 + 5 + + (
More informationLecture 2: April 3, 2013
TTIC/CMSC 350 Mathematical Toolkit Sprig 203 Madhur Tulsiai Lecture 2: April 3, 203 Scribe: Shubhedu Trivedi Coi tosses cotiued We retur to the coi tossig example from the last lecture agai: Example. Give,
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationCombinatorics II. Combinatorics. Product Rule. Sum Rule II. Theorem (Product Rule) Theorem (Sum Rule)
Combiatorics Combiatorics I Slides by Christopher M. Bourke Istructor: Berthe Y. Choueiry Fall 27 Computer Sciece & Egieerig 235 to Discrete Mathematics Sectios 5.-5.6 & 7.5-7.6 of Rose cse235@cse.ul.edu
More information1. n! = n. tion. For example, (n+1)! working with factorials. = (n+1) n (n 1) 2 1
Biomial Coefficiets ad Permutatios Mii-lecture The followig pages discuss a few special iteger coutig fuctios You may have see some of these before i a basic probability class or elsewhere, but perhaps
More informationMath 475, Problem Set #12: Answers
Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe
More information3.1 Counting Principles
3.1 Coutig Priciples Goal: Cout the umber of objects i a set. Notatio: Whe S is a set, S deotes the umber of objects i the set. This is also called S s cardiality. Additio Priciple: Whe you wat to cout
More informationMT5821 Advanced Combinatorics
MT5821 Advaced Combiatorics 1 Coutig subsets I this sectio, we cout the subsets of a -elemet set. The coutig umbers are the biomial coefficiets, familiar objects but there are some ew thigs to say about
More informationCombinatorially Thinking
Combiatorially Thiig SIMUW 2008: July 4 25 Jeifer J Qui jjqui@uwashigtoedu Philosophy We wat to costruct our mathematical uderstadig To this ed, our goal is to situate our problems i cocrete coutig cotexts
More informationChapter 1 : Combinatorial Analysis
STAT/MATH 394 A - PROBABILITY I UW Autum Quarter 205 Néhémy Lim Chapter : Combiatorial Aalysis A major brach of combiatorial aalysis called eumerative combiatorics cosists of studyig methods for coutig
More informationDiscrete Mathematics for CS Spring 2007 Luca Trevisan Lecture 22
CS 70 Discrete Mathematics for CS Sprig 2007 Luca Trevisa Lecture 22 Aother Importat Distributio The Geometric Distributio Questio: A biased coi with Heads probability p is tossed repeatedly util the first
More informationRandom Models. Tusheng Zhang. February 14, 2013
Radom Models Tusheg Zhag February 14, 013 1 Radom Walks Let me describe the model. Radom walks are used to describe the motio of a movig particle (object). Suppose that a particle (object) moves alog the
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationDiscrete Mathematics for CS Spring 2005 Clancy/Wagner Notes 21. Some Important Distributions
CS 70 Discrete Mathematics for CS Sprig 2005 Clacy/Wager Notes 21 Some Importat Distributios Questio: A biased coi with Heads probability p is tossed repeatedly util the first Head appears. What is the
More informationProblem Set 2 Solutions
CS271 Radomess & Computatio, Sprig 2018 Problem Set 2 Solutios Poit totals are i the margi; the maximum total umber of poits was 52. 1. Probabilistic method for domiatig sets 6pts Pick a radom subset S
More informationSets. Sets. Operations on Sets Laws of Algebra of Sets Cardinal Number of a Finite and Infinite Set. Representation of Sets Power Set Venn Diagram
Sets MILESTONE Sets Represetatio of Sets Power Set Ve Diagram Operatios o Sets Laws of lgebra of Sets ardial Number of a Fiite ad Ifiite Set I Mathematical laguage all livig ad o-livig thigs i uiverse
More informationWeek 5-6: The Binomial Coefficients
Wee 5-6: The Biomial Coefficiets March 6, 2018 1 Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( 1 1 + 1 The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers
More informationP1 Chapter 8 :: Binomial Expansion
P Chapter 8 :: Biomial Expasio jfrost@tiffi.kigsto.sch.uk www.drfrostmaths.com @DrFrostMaths Last modified: 6 th August 7 Use of DrFrostMaths for practice Register for free at: www.drfrostmaths.com/homework
More informationMath 172 Spring 2010 Haiman Notes on ordinary generating functions
Math 72 Sprig 200 Haima Notes o ordiary geeratig fuctios How do we cout with geeratig fuctios? May eumeratio problems which are ot so easy to hadle by elemetary meas ca be solved usig geeratig fuctios
More informationWhat is Probability?
Quatificatio of ucertaity. What is Probability? Mathematical model for thigs that occur radomly. Radom ot haphazard, do t kow what will happe o ay oe experimet, but has a log ru order. The cocept of probability
More informationIt is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function.
MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study real-valued fuctios defied
More informationPUTNAM TRAINING PROBABILITY
PUTNAM TRAINING PROBABILITY (Last udated: December, 207) Remark. This is a list of exercises o robability. Miguel A. Lerma Exercises. Prove that the umber of subsets of {, 2,..., } with odd cardiality
More informationBasic Counting. Periklis A. Papakonstantinou. York University
Basic Coutig Periklis A. Papakostatiou York Uiversity We survey elemetary coutig priciples ad related combiatorial argumets. This documet serves oly as a remider ad by o ways does it go i depth or is it
More informationSolutions to Final Exam
Solutios to Fial Exam 1. Three married couples are seated together at the couter at Moty s Blue Plate Dier, occupyig six cosecutive seats. How may arragemets are there with o wife sittig ext to her ow
More informationEnd-of-Year Contest. ERHS Math Club. May 5, 2009
Ed-of-Year Cotest ERHS Math Club May 5, 009 Problem 1: There are 9 cois. Oe is fake ad weighs a little less tha the others. Fid the fake coi by weighigs. Solutio: Separate the 9 cois ito 3 groups (A, B,
More informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More informationWorksheet on Generating Functions
Worksheet o Geeratig Fuctios October 26, 205 This worksheet is adapted from otes/exercises by Nat Thiem. Derivatives of Geeratig Fuctios. If the sequece a 0, a, a 2,... has ordiary geeratig fuctio A(x,
More informationPermutations & Combinations. Dr Patrick Chan. Multiplication / Addition Principle Inclusion-Exclusion Principle Permutation / Combination
Discrete Mathematic Chapter 3: C outig 3. The Basics of Coutig 3.3 Permutatios & Combiatios 3.5 Geeralized Permutatios & Combiatios 3.6 Geeratig Permutatios & Combiatios Dr Patrick Cha School of Computer
More informationMA131 - Analysis 1. Workbook 3 Sequences II
MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisher-provided material Overview May mathematical
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationPutnam Training Exercise Counting, Probability, Pigeonhole Principle (Answers)
Putam Traiig Exercise Coutig, Probability, Pigeohole Pricile (Aswers) November 24th, 2015 1. Fid the umber of iteger o-egative solutios to the followig Diohatie equatio: x 1 + x 2 + x 3 + x 4 + x 5 = 17.
More informationLecture 6 Simple alternatives and the Neyman-Pearson lemma
STATS 00: Itroductio to Statistical Iferece Autum 06 Lecture 6 Simple alteratives ad the Neyma-Pearso lemma Last lecture, we discussed a umber of ways to costruct test statistics for testig a simple ull
More informationLecture Notes for CS 313H, Fall 2011
Lecture Notes for CS 313H, Fall 011 August 5. We start by examiig triagular umbers: T () = 1 + + + ( = 0, 1,,...). Triagular umbers ca be also defied recursively: T (0) = 0, T ( + 1) = T () + + 1, or usig
More informationBooks Recommended for Further Reading
Books Recommeded for Further Readig by 8.5..8 o 0//8. For persoal use oly.. K. P. Bogart, Itroductory Combiatorics rd ed., S. I. Harcourt Brace College Publishers, 998.. R. A. Brualdi, Itroductory Combiatorics
More informationSEQUENCES AND SERIES
Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces
More informationMT5821 Advanced Combinatorics
MT5821 Advaced Combiatorics 9 Set partitios ad permutatios It could be said that the mai objects of iterest i combiatorics are subsets, partitios ad permutatios of a fiite set. We have spet some time coutig
More informationMathematical Induction
Mathematical Iductio Itroductio Mathematical iductio, or just iductio, is a proof techique. Suppose that for every atural umber, P() is a statemet. We wish to show that all statemets P() are true. I a
More informationSolutions. tan 2 θ(tan 2 θ + 1) = cot6 θ,
Solutios 99. Let A ad B be two poits o a parabola with vertex V such that V A is perpedicular to V B ad θ is the agle betwee the chord V A ad the axis of the parabola. Prove that V A V B cot3 θ. Commet.
More informationCS 336. of n 1 objects with order unimportant but repetition allowed.
CS 336. The importat issue is the logic you used to arrive at your aswer.. Use extra paper to determie your solutios the eatly trascribe them oto these sheets. 3. Do ot submit the scratch sheets. However,
More information1 What is combinatorics?
1 What is combiatorics Combiatorics is the brach of mathematics dealig with thigs that are discrete, such as the itegers, or words created from a alphabet. This is i cotrast to aalysis, which deals with
More information4 The Sperner property.
4 The Sperer property. I this sectio we cosider a surprisig applicatio of certai adjacecy matrices to some problems i extremal set theory. A importat role will also be played by fiite groups. I geeral,
More information4.3 Growth Rates of Solutions to Recurrences
4.3. GROWTH RATES OF SOLUTIONS TO RECURRENCES 81 4.3 Growth Rates of Solutios to Recurreces 4.3.1 Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer.
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationDiscrete Mathematics and Probability Theory Spring 2012 Alistair Sinclair Note 15
CS 70 Discrete Mathematics ad Probability Theory Sprig 2012 Alistair Siclair Note 15 Some Importat Distributios The first importat distributio we leared about i the last Lecture Note is the biomial distributio
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationCombinatorics I Introduction. Combinatorics. Combinatorics I Motivating Example. Combinations. Product Rule. Permutations. Theorem (Product Rule)
Combiatorics I Itroductio Combiatorics Computer Sciece & Egieerig 235: Discrete Mathematics Christopher M. Bourke cbourke@cse.ul.edu Combiatorics is the study of collectios of objects. Specifically, coutig
More informationLecture 6 April 10. We now give two identities which are useful for simplifying sums of binomial coefficients.
Lecture 6 April 0 As aother applicatio of the biomial theorem we have the followig. For 0. 0 I other words startig with the secod row ad goig dow if we sum alog the rows alteratig sig as we go the result
More informationsubcaptionfont+=small,labelformat=parens,labelsep=space,skip=6pt,list=0,hypcap=0 subcaption ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, 2/16/2016
subcaptiofot+=small,labelformat=pares,labelsep=space,skip=6pt,list=0,hypcap=0 subcaptio ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06. Self-cojugate Partitios Recall that, give a partitio λ, we may
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationThe Binomial Theorem
The Biomial Theorem Robert Marti Itroductio The Biomial Theorem is used to expad biomials, that is, brackets cosistig of two distict terms The formula for the Biomial Theorem is as follows: (a + b ( k
More informationarxiv: v1 [math.nt] 10 Dec 2014
A DIGITAL BINOMIAL THEOREM HIEU D. NGUYEN arxiv:42.38v [math.nt] 0 Dec 204 Abstract. We preset a triagle of coectios betwee the Sierpisi triagle, the sum-of-digits fuctio, ad the Biomial Theorem via a
More informationSome Basic Counting Techniques
Some Basic Coutig Techiques Itroductio If A is a oempty subset of a fiite sample space S, the coceptually simplest way to fid the probability of A would be simply to apply the defiitio P (A) = s A p(s);
More informationDiscrete Mathematics and Probability Theory Summer 2014 James Cook Note 15
CS 70 Discrete Mathematics ad Probability Theory Summer 2014 James Cook Note 15 Some Importat Distributios I this ote we will itroduce three importat probability distributios that are widely used to model
More informationARRANGEMENTS IN A CIRCLE
ARRANGEMENTS IN A CIRCLE Whe objects are arraged i a circle, the total umber of arragemets is reduced. The arragemet of (say) four people i a lie is easy ad o problem (if they liste of course!!). With
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More information6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
More information2 Geometric interpretation of complex numbers
2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that
More informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More informationInternational Contest-Game MATH KANGAROO Canada, Grade 11 and 12
Part A: Each correct aswer is worth 3 poits. Iteratioal Cotest-Game MATH KANGAROO Caada, 007 Grade ad. Mike is buildig a race track. He wats the cars to start the race i the order preseted o the left,
More informationA Combinatorial Proof of a Theorem of Katsuura
Mathematical Assoc. of America College Mathematics Joural 45:1 Jue 2, 2014 2:34 p.m. TSWLatexiaTemp 000017.tex A Combiatorial Proof of a Theorem of Katsuura Bria K. Miceli Bria Miceli (bmiceli@triity.edu)
More informationCounting Well-Formed Parenthesizations Easily
Coutig Well-Formed Parethesizatios Easily Pekka Kilpeläie Uiversity of Easter Filad School of Computig, Kuopio August 20, 2014 Abstract It is well kow that there is a oe-to-oe correspodece betwee ordered
More informationCommutativity in Permutation Groups
Commutativity i Permutatio Groups Richard Wito, PhD Abstract I the group Sym(S) of permutatios o a oempty set S, fixed poits ad trasiet poits are defied Prelimiary results o fixed ad trasiet poits are
More informationMath 105 TOPICS IN MATHEMATICS REVIEW OF LECTURES VII. 7. Binomial formula. Three lectures ago ( in Review of Lectuires IV ), we have covered
Math 5 TOPICS IN MATHEMATICS REVIEW OF LECTURES VII Istructor: Lie #: 59 Yasuyuki Kachi 7 Biomial formula February 4 Wed) 5 Three lectures ago i Review of Lectuires IV ) we have covered / \ / \ / \ / \
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, o-egative fuctio o the closed iterval [a, b] Fid
More information1 Introduction. 1.1 Notation and Terminology
1 Itroductio You have already leared some cocepts of calculus such as limit of a sequece, limit, cotiuity, derivative, ad itegral of a fuctio etc. Real Aalysis studies them more rigorously usig a laguage
More informationarxiv: v1 [math.co] 23 Mar 2016
The umber of direct-sum decompositios of a fiite vector space arxiv:603.0769v [math.co] 23 Mar 206 David Ellerma Uiversity of Califoria at Riverside August 3, 208 Abstract The theory of q-aalogs develops
More informationTopic 5: Basics of Probability
Topic 5: Jue 1, 2011 1 Itroductio Mathematical structures lie Euclidea geometry or algebraic fields are defied by a set of axioms. Mathematical reality is the developed through the itroductio of cocepts
More informationIf a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?
2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a
More informationMath 61CM - Solutions to homework 3
Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig
More informationDiscrete Mathematics Recurrences
Discrete Mathematics Recurreces Saad Meimeh 1 What is a recurrece? It ofte happes that, i studyig a sequece of umbers a, a coectio betwee a ad a 1, or betwee a ad several of the previous a i, i
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More informationLecture 3: Catalan Numbers
CCS Discrete Math I Professor: Padraic Bartlett Lecture 3: Catala Numbers Week 3 UCSB 2014 I this week, we start studyig specific examples of commoly-occurrig sequeces of umbers (as opposed to the more
More informationBeurling Integers: Part 2
Beurlig Itegers: Part 2 Isomorphisms Devi Platt July 11, 2015 1 Prime Factorizatio Sequeces I the last article we itroduced the Beurlig geeralized itegers, which ca be represeted as a sequece of real umbers
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationProduct measures, Tonelli s and Fubini s theorems For use in MAT3400/4400, autumn 2014 Nadia S. Larsen. Version of 13 October 2014.
Product measures, Toelli s ad Fubii s theorems For use i MAT3400/4400, autum 2014 Nadia S. Larse Versio of 13 October 2014. 1. Costructio of the product measure The purpose of these otes is to preset the
More informationCSE 21 Mathematics for
CSE 2 Mathematics for Algorithm ad System Aalysis Summer, 2005 Outlie What a geeratig fuctio is How to create a geeratig fuctio to model a problem Fidig the desired coefficiet Partitios Expoetial geeratig
More informationLecture 2: Monte Carlo Simulation
STAT/Q SCI 43: Itroductio to Resamplig ethods Sprig 27 Istructor: Ye-Chi Che Lecture 2: ote Carlo Simulatio 2 ote Carlo Itegratio Assume we wat to evaluate the followig itegratio: e x3 dx What ca we do?
More information