Let us consider the following problem to warm up towards a more general statement.

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1 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 for CS boo Discrete Structures II Summer 2018 Rutgers Uiversity Istructor: Abhishe Bhrushudi 1 Sequeces with repetitios revisted Let us cosider the followig problem to warm up towards a more geeral statemet Questio How may sequeces are there of the form A, B, C, where A, B ad C are disjoit subsets of S = {1,, 1000}, ad A = 500, B = 325, C = 175? Proof Note that A + B + C = S, ad thus they form a partitio of S To distribute the 1000 elemets of S betwee A, B, ad C, it suffices to specify which elemets of S are chose to go ito A, ad which of them are chose to go to B the rest will the go to C Let s first pic which elemets will go ito A We have to choose 500 elemets from S, so there are ways of doig that Havig piced elemets for A, we would be left with = 500 elemets i S, ad have to choose 325 out of them for B For this, there are ways, ad so the total is: = 1000! !500! 500! 325!175! = 1000! 500!325!175! Let us try to geeralize this to the followig: how may sequeces are there of the form A 1,, A where A 1,, A are disjoit subsets of S = {1,, } ad A 1 = r 1, A 2 = r 2,, A = r with r 1 + r r =? Let us do exactly as we did above: first pic r 1 elemets to put ito A 1, the r 2 to put ito A 2,, ad fially r 1 elemets i A 1 As soo as have filled up the first 1 elemets, we will be left with r r = r elemets, ad thus all of them must go ito A Followig the above calculatios, we see that the umber of ways of picig elemets for A 1,, A 1 is r1 r 1 This ca be writte as r 2 r1 r 2 r 3 r1 r 2 r 1! r 1! r 1 r 2! r 1! r 1! r 2! r 1 r 2! r 3! r 1 r 2 r 3! r 1 r 2! r 1! r 1 r 1! 1

2 If you stare at this expressio for a while you realize that all the umerators except! will get cacelled by some of the deomiators, ad the oly deomiators left will be r 1!r 2! r! where! did the r! eve come from? Thus, the expressio becomes: r 1!r 2!r! Thus, we have Lemma 1 The umber of sequeces of the form A 1,, A where A 1,, A are disjoit subsets! of S = {1,, } ad A 1 = r 1, A 2 = r 2,, A = r with r 1 + r r = is r 1!r 2!r! I other words, the umber of ways of distributig distict objects amog distict parties such! that party i gets exactly r i may objects ad r r = is r 1!r! This also turs out to be the umber of permutatios whe there are some repeated objects as we shall see ow the formula does loo lie the oe we saw i the last lecture, does t it? Suppose we have objects i total, r 1 of type 1, r 2 of type 2,, r of type, such that r 1 + +r = How may distict permutatios are there of these objects? If they were all distict, the the aswer would have bee!, but as we have see before, this will lead to a overcout whe we have repetitios I the last lecture, we suggested that oe ca use the divisio rule to get aroud this problem, ad arrive at the umber of distict permutatios beig! = r 1,, r r 1! r!, where the expressio i the LHS is ow as a multiomial coefficietwe will ow give a alterate ad much cleaer proof of this based o what we just did above with sequece of sets Let s thi of comig up with oe possible arragemet of these objects What eeds to be specified i order to completely specify a arragemet? Well, I eed to first tell you what positios the objects of type 1 go ito, the I eed to tell you what positios the objects of type 2 go ito,, ad fially what positios objects of type 1 go ito As soo as I tell you all that, you ll automatically ow where the objects of type go ito they are forced to go to the remaiig positios Now, how may ways are there to choose the positios for the objects of type 1 There are r 1 objects of type 1, so we eed to choose r 1 out of possible positios, ad thus the umber of ways is r 1 Havig chose the positios for the first type, we the choose positios for the objects of the secod type Now we eed to choose r 2 positios from r 1 available oes, so that s r 1 r 2 waysbut, hold o? Is t this turig out to be exactly what we did above? Turs out you ca do a simple tric thi bijectio method: if you let A i to be all the positios i the permutatio where the r i objects of type i will go ito, the the problem becomes oe of coutig the umber of sequeces of the form A 1,, A, with A 1,, A beig disjoit subsets of {1,, } these are the possible positios i the permutatio of sizes r 1,, r respectively, ad i=1 r i = Now we ca use Lemma 1 to fiish the job, ad coclude that the umber of! distict permutatios is r 1!r! 2 Distributig idetical objects amog distict parties May coutig problems ca be rephrased as the followig problem: Questio Suppose that there are idetical cadies that eed to be distributed amog r childre 2

3 obviously, o two childre are idetical, ad so we have r distict parties here I how may ways ca you do this if you are ot allowed to brea cadies ito smaller pieces? To solve these problems, it s coveiet to covert them ito a liear equatio with costraits thi bijectio method to be able to thi clearly: Let x 1,, x r be r variables such that x i deotes the umber of cadies child i receives Sice all cadies must get distributed, we have that x 1 + x x r = Furthermore, sice you ca t brea the cadies ad must give whole cadies to the childre, we must eforce that the values tae by x 1,, x r be iteger, ie we are iterested i iteger valued solutios to the above equatio Also, it does t mae sese if a variable becomes egative what would that eve mea? istead of givig cadies to the child, as the child for cadies?, so we will say that x 1,, x r 0 So the umber of ways of distributig cadies amog childre become the same as the umber of iteger solutios to the above equatio with the additioal costrait that x 1,, x r 0 While this way of phrasig the problem as fidig solutios to a liear equatio help us thi clearly ad mae sure we got the problem right, it does t really do much i terms of helpig us fid the aswer Istead, we will ot tur to our old frieds: biary strigs! Let s wor with = 10 ad r = 4, ie 10 cadies amog 4 childre, or equivaletly, solutios to x 1 + x 2 + x 3 + x 4 = 10 subject to x 1, x 2, x 3, x 4 0 Let s loo at the biary strig of legth 10 cosistig of all zeros, ad thi of those zeros as cadies We ca ow thi of puttig oes ito the strig as the process of dividig the zeros cadies ito two parts Say we start out with , ad add a oe somewhere i there: This ca be iterpreted as there beig two parts, oe to the left of the oe ad the other to the right of the oe, ad the first part has 3 zeros cadies, ad the secod part has 7 zeros cadies If throw i aother oe, the we get three parts i the same maer: ca be thought of as three parts with part oe ad two havig three zeros each, ad the third part havig 4 zeros Of course there could be parts that are empty, eg, i , the secod part betwee the two oes is empty Now, if we throw i three oes i total, we get a distributio of the 10 zeros ito 4 parts Let s fix the covetio that all the zeros cadies to the left of the first oe go to x 1 or the first child, all the zeros betwee the 1st ad 2d oe go to x 2 the secod child, ad so o Thus, the umber of ways of distributig the 10 cadies betwee 4 childre, or the umber of solutios to the above equatios, is exactly the umber of biary strigs of legth 13 that have 10 zeros ad 3 oes, which we ow usig the subset rule is 13 3 We ca geeralize the above argumet to coclude that the umber of ways of distributig idetical objects amog distict parties is equal to the umber of biary strigs that have exactly zeros ad 1 oes why 1, ad ot? Because a sigle oe created two partitios, two oes creatd three, ad so o, so 1 oes will give us partitios which is Questio I how may ways ca you distribute 20 idivisible gold bars amog 5 pirates so that every pirate gets at least oe bar? 3

4 Proof If we covert this a equatio with costraits we get x 1 + x x 5 = 20 with the costraits that x 1 1, x 2 1,, x 5 1 Ufortuately, the above method oly wors whe we have costraits of the form x 1 0,, x 5 0 To get aroud this, let s first simply satisfy the bare miimum requiremets of all the pirates variables, ie let s give oe bar to each of the pirates variables so that they all have at least oe This meas we are left with 20 5 = 15 bars We ca ow focus o distributig the remaiig 15 bars amog the 5 pirates without havi to worry about ay additioal costraits Notice that ow the equatio would be simply x x 5 = 15 with x 1,, x 5 0, ad we ow that the umber of solutios to this equatio is Ca you geeralize this? Questio I how may ways ca you distribute idivisible gold bars amog pirates so that every pirate gets at least r bars? Sometimes the distributio problems disguise themselves i a very icospicuous maer the oly way to call their bluff is to practice a lot of problems Here is a example: Questio Suppose we roll 10 idetical dice together How may possible outcomes are there? Proof Note that i all the dice are idetical ii there is o order i which we rolled the dice all of them were rolled simultaeously Thus, the oly thig that distiguishes a outcome from aother, or the oly thig that really defies a outcome is the followig set of statistics: how may of the dice tured up 1, how may tured up 2,, how may tured up 6? Let x i be the umber of dice that tured up the umber i here 1 i 6 The we have x x 6 = 10, ad x 1,, x 6 0, ad wat to cout the umber of solutios to this set of equatios We are ow o home turf, ad ca coclude that the aswer must be Note that had the dice bee colored with distict colors, or if they had bee rolled oe at a time, the the problem would be very differet, ad the aswer would simply be 6 10 usig the product rule covice yourself of this! Here is aother oe this oe is from your boo that ca seem totally urelated to distributio problems: Questio S uppose there are 20 boos arraged i a row o a boo shelf I how may ways ca you choose 6 boos so that o two of the chose boos are adjacet to each other o the rac? At first glace, this problem might ot seem to have aythig to do with what we have bee discussig so far A little bit of rephrasig maes this illusio disappear Note that the most obvious way of specifyig a selectio/choice of 6 boos is to literally specify the idex of the boo if oe were to start coutig from 1 startig at the left most boo o the shelf A more o-obvious way would be the followig: specify how may uchose boos there are to the left of the first chose boo, the specify how may uchose boos there are betwee the first ad secod chose boo, the the umber of uchose boos betwee the secod ad third chose boo,, ad fially the umber of uchose boos to the right of the last sixth chose 4

5 boo So, I ca set this up i the followig way: let x 1 deote the umber of boos to the left of the first chose boo, x 2 deote the umber of boos betwee the first ad secod chose boos, ad so o, with x 7 represetig the umber of boos to the right of the sixth chose boo What do coditios do we wat to impose o these variables? Well, first of all there must be 14 uchose boos i all, ad each of those 14 boos must be either betwee two of the chose boos, or the left or right of all the chose boos, ad thus: x 1 + x x 7 = 14 Furthermore, we wat that o two boos be adjacet, so there must be at least oe uchose boo betwee the first ad secod chose boo, so we wat x 2 1, ad similarly we wat x 3,, x 6 1 How about x 1 ad x 7, there is o costrait o them: there could be as low as zero it is possible to choose the first boo from the left o the shelf, or the last boo o the shelf, ad so x 1, x 7 0 Agai, we are bac i familiar territory, so we ca use the machiery we developed earlier i the sectio to fiish the problem 3 Biomial theorem Recall that x + y 2 = x 2 + y 2 + 2xy Actually, there is a lot goig o whe oe tries to derive that: x + y 2 is othig but x + y x + y Sice multiplicatio is distributive over additio, there will be four terms comig from the four pssible multiplicatios, ad the all we be added up Basically, each of the four terms will cosist of exactly oe etry each from the two copies of x + y For the first copy, we have the choice of picig up either x or y, ad we have the same choices for the secod copy This gives 2 2 = 4 terms i total The four terms are xx + xy + yx + yy Now recall that xx = x 2 ad yy = y 2 Also, because of multiplicatio beig commutative we have that xy = yx, ad so we get x 2 + y 2 + 2xy Oe ca ow do the same for higher powers of x + y How about x + y? This is othig but copies of x + y multiplied together Agai, usig distributivity of multiplicatio over additio, we will get 2 terms: there are copies of x + y, ad we have two choices for the first copy, two for the secod copy, ad so o Obviously, as before, may terms will collapse to the same expressio because of commutativity xy = yx, ad the combiig cosecutive xs ad ys ito powers This meas that, i the fial simplified expressio for x + y, differet terms will have differet coefficiets i frot of them depedig o how may of the 2 origial terms collapse to them Let s as the questio, what s the coefficiet of x y? I other words, how may of the 2 iitial terms collapse to x y Clearly, oly those terms that have exactly xs i them will collapse to x y, so how may terms have exactly xs? Recall that we had to pic betwee x ad y i each of the copies of x + y Thus, the terms that give us x y must correspod to outcomes of the picig process where we decided to pic x i exactly of the copies of x + y, ad piced y i the rest How may such outcomes are there? It s exactly the same as decidig i which copies of x + y out of the possible will we pic x, ad there are ways of doig that So the coefficiet of x y must be Sice there was othig special about ad i the above argumet, we ca coclude: 5

6 Theorem 2 Biomial theorem Let 0 be a iteger, the we have that x + y = =0 x y 4 Some properties of biomial coefficiets The biomial theorem lets us derive some iterestig thigs about biomial coefficiets Let us first as the followig questio what is the sum of all biomial coefficiets for a fixed, ie We ca aswer this problem i two ways The first oe is combiatorial: recall that just deoted the umber of size subsets of the set {1,, } So, it seems lie we are addig the umber of subsets of size 0, the umber of subsets of size 1, ad so o up till the umber of subsets of size But if you thi about it this accouts for all possible subsets of {1,, }, ad thus the cout must add up to the total umber of differet possible subsets of {1,, }, which we ow is 2 A more algebraic way of seeig this is tae the statemet of the biomial theorem ad set x = y = 1 What happes if we set y = 1 ad x = 1? The LHS becomes zero, ad we get 0 = is eve ad =0 1 Now otice that 1 becomes 1 wheever is eve, ad becomes 1 wheever is odd, so we ca write: 0 =, ad so is eve ad = is odd ad is odd ad This is aother way of derivig somethig we proved usig the bijectio method before: is eve ad is othig but the umber of eve size subsets of {1,, } ad is odd ad is the umber of odd size subsets of {1,, }, ad the above expressio tells us that there are as may odd size subsets as there are eve size subsets we gave a bijective proof for the case whe was odd, ca you try for eve? We wat to show the followig: < 2 = > 2, ie the umber of subsets of {1,, } of size less tha /2, is the same as the umber of subsets of size more tha /2 I would t give you the whole proof but it follows pretty simply by recallig that = ad so 0 =, 1 = 1, ad so o 6