Lecture 6 April 10. We now give two identities which are useful for simplifying sums of binomial coefficients.

Transcription

1 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 is 0. Give the symmetry it trivially holds whe is odd, it is ot trivial to show that it holds whe is eve. Algebraic proof: I the biomial theorem set x ad y to get Combiatorial proof: First ote that this is equivalet to showig the followig: The left had side couts the umber of subsets of {,,..., } with a eve umber of elemets while the right had side couts the umber of subsets with a odd umber of elemets. To show that we have a equal umber of these two types we will use a ivolutio argumet. A ivolutio o a set X is a mappig φ : X X so that φφx x. Give a ivolutio φ the elemets the aturally split ito fixed poits elemets with φx x or pairs two elemets x y where φx y ad φy x. I our case our ivolutio will act o [] the set of all subsets of {,,..., } ad is defied for a subset A as follows: { A \ {} if is i A, φa A {} if is ot i A. I other words we tae a set ad if is i it we remove it ad if is ot i it we add it. We ow mae some observatios. First, for ay subset A we have φφa A so that it is a ivolutio. Further, there are o fixed poits sice must either be i or ot i a set. So we ca ow brea the collectio of subsets ito pairs {A, φa}. Fially, by the ivolutio A ad φa will differ by exactly oe elemet so oe of them is eve ad oe of them is odd. So we ow have a way to pair every subset with a eve umber of elemets with a subset with a odd umber of elemets. So the umber of such subsets must be equal. We ow give two idetities which are useful for simplifyig sums of biomial coefficiets. 0 ad., To prove the first oe we will cout the umber of wals from 0, 0 to, usig right steps ad up steps i two ways. 0, 0, We must mae steps to the right ad steps up. So the umber of ways to wal from oe corer to the other is the umber of ways to choose whe to mae the up steps which is. We ca also cout the umber of wals by groupig them accordig to which lie segmet is used i the last step to the right i the picture above this correspods to groupig accordig to the lie segmet which itersects the dotted lie. These lie segmets go from, i to, i where i 0,...,. Oce we have crossed the lie segmet there is oly oe way to fiish the wal to the corer straight up the side. O the other had the umber of ways to get to, i is i. So by the rule of additio we have that the i total umber of wals is 0 0. Combiig these two differet ways to cout the paths gives the idetity. A proof of the other result ca be doe similarly usig the followig picture we leave it to the iterested reader to fill i the details. 0, 0,

2 These idetities ca also be proved by usig. For example, we have the followig.. Whe I was taught these idetities they were called the hocey stic idetities. This ame comes from the patter that they form i Pascal s triagle. For istace we have that 6 3 show below i blue is the 5 show below i red The other idetity correspods to looig at the mirror image of Pascal s triagle. We ow give a applicatio of the hocey stic idetity. i i0 6 I the first lecture we saw how to add up the sum of i by coutig dots i two differet ways. Here we wat to sum up i, the problem is that we do t have a easy method to do that. We do have a easy method to add up the biomial coefficiets, so if we ca rewrite i i terms of biomial coefficiets the we ca easily aswer this questio. So cosider the followig. i ii i i i. The ability to rewrite i as a combiatio of biomial coefficiets ca also be used for ay other polyomial expressio of i. The tric is to fid the coefficiets ivolved. For the case of polyomials of the form i l we have that the coefficiet of { i is! l } where { l } are the Stirlig umbers of the secod id, which we will discuss i a later lecture. Now usig this ew way to write i we have i i i i0 i0 i i i0 i The oly difficult step ow is goig from the secod to the third lie, which is doe usig the hocey stic idetity. There is a more geeral form of the biomial theorem ow as the multiomial theorem multiomial referrig to may variables as compared to biomial which refers to two. It states x x x,,..., where 0, 0,..., 0 x x x,!,,...,!!! is the multiomial coefficiet we have see i previous lectures. Returig to the biomial theorem, we have that x x x. 0 I particular for fixed all of the biomial coefficiets are stored as coefficiets iside of the fuctio fx x. I geeral give ay sequece of umbers a 0, a, a,... possibly ifiitely may we ca store these as coefficiets of a fuctio which we call the geeratig fuctio as follows gx a 0 a x a x. There are differet ways that we ca store the sequece as coefficiets leadig to differet classes of geeratig fuctios. This method is what is ow as a ordiary geeratig fuctio, i a later lecture

3 we will see what are ow as expoetial geeratig fuctios. Geerally speaig we will treat the fuctio as a formal series, i.e., as a algebraic rather tha aalytic structure. This maes it coveiet to wor with whe we do ot eed to worry about covergece. Nevertheless we will mae extesive use of aalytic techiques, whe we do that is whe we will start to worry about covergece. Let us ow cosider the problem of fidig the geeratig fuctio for the sequece a where is fixed. I other words we wat to fid a fuctio so that its series expasio ote that there are ifiitely may a i this case is gy y. We ca do this by actually ivestigatig a much more geeral fuctio. Namely we will use a geeratig fuctio with two variables x ad y ad is defied as follows: hx, y x y., Usig the biomial theorem we have hx, y x y, y x y x y x y x y xy. Note that this fuctio stores the etire Pascal s triagle iside of it. I the above derivatio we used the followig importat idetity, z z 0 z z. Aalytically this is true for z <, formally this is always true. Now to fid our desired gy we are looig for the coefficiet of x. So we ow rewrite hx, y as a power series i x ad get hx, y y xy y 0 0 y y y x y y x. y y x Fially, readig off the coefficiet for x i hx, y gives us our desired fuctio, amely gy y y y. Lecture 7 April 3 Today we loo at oe motivatio for studyig geeratig fuctios, amely a coectio betwee polyomials ad distributio problems. Let us start with a simple example. Example: Give a iterpretatio for the coefficiet of x for the polyomial Solutio: gx x x 3 x 5 3. First ote that we ca write gx as x x 3 x 5 x x 3 x 5 x x 3 x 5 ad that a typical term which we will get whe we expad has the form x x x 3 where the term x comes from picig a term from the first polyomial so {0,, 3, 5}, the term x comes from picig a term from the secod polyomial so {0,, 3, 5}, ad the term x 3 comes from picig a term from the third polyomial so 3 {0,, 3, 5}. Sice we are iterested i the coefficiet of x the we eed 3. So this ca be rephrased as a balls ad bis problem where we are tryig to distribute balls amog three bis where i each bi we ca have either 0,, 3 or 5 balls. Example: Give a iterpretatio for the coefficiet of x r for the fuctio gx xx xx xx 3 x 5. Solutio: By the same reasoig as we have before a typical term i the expasio loos lie x 3 with beig 0, or ; 0; ad 3 0 ad eve. So this ca be rephrased as a balls ad bis problem where we are tryig to distribute r balls amog three bis where i the first bi we put 0, or balls; i the secod bi we put ay umber of balls we wat; i the third bi we put i a odd umber of balls. I geeral give a polyomial gx g xg x g x

4 with each g i a sum of some power of xs, i.e., g i x l x qi l. The the coefficiet of x r i gx is the umber of ways to place r balls ito bis so that i the ith bi the umber of balls is ql i for some l. The idea is that the term g i x tells you the restrictios about the id of balls that ca be placed ito that bi. It is importat that the coefficiets of the g i x are all s. We ca also start with a balls ad bis problem ad traslate it ito fidig the coefficiet of some appropriate polyomial. Example: Traslate the followig problem ito fidig a coefficiet of some fuctio: How may ways are there to put balls ito four bis so that the first two bis have betwee ad 5 balls, the third bi has at least 3 balls ad the last bi has a odd umber of balls? Solutio: Sice there will be balls we will be looig for the coefficiet of x. Now let us traslate the coditio o each bi. For the first two bis there are betwee ad 5 balls so g x g x x x 3 x 4 x 5 remember, the powers list the umber of balls that ca be put i the bi. For the third bi we have to have at least 3 balls so g 3 x x 3 x 4. We could also use the fuctio g 3x x 3 x 4 x, the differece beig that we stop ad have a polyomial istead of a ifiite series. The reaso we ca do this is that we are oly iterested i the coefficiet of x, aythig which will ot cotribute to that coefficiet ca be dropped without chagig the aswer. However by eepig the series we ca use the same fuctio to aswer the questio for ay arbitrary umber of balls ad so the first optio is more geeral. Fially for the last bi we have g 4 x x x 3 x 5. x x m x m x 0 0 x x m x x x m x x x for x < x 0 x The first two are simple applicatios of the biomial theorem. The third is easily verifiable by multiplyig both sides by x ad simplifyig. The fourth oe is a well ow sum ad ca be cosidered the limitig case of the third oe. Note that as a fuctio of x, i.e., aalytically, this maes sese oly whe x <. Formally there is o restrictio o x, this is because formally x is actig as a placeholder. To see why the last oe holds we ote that this is x x x x }{{} times x x x x }{{} times Traslatig this ito a balls ad bis problem, the coefficiet of x would correspod to the umber of solutios of e e e where each e i 0. We have already solved this problem usig bars ad stars, amely this ca be doe i ways, givig us the desired result. We will eed oe more tool, ad that is a way to multiply fuctios. We have the followig which is a simple exercise i expadig ad groupig coefficiets. Puttig it altogether we are tryig to fid the coefficiet of x for the fuctio x x 3 x 4 x 5 x 3 x 4 x x 3 x 5. Give fx a 0 a x a x, gx b 0 b x b x. Of course traslatig oe problem to aother is oly good if we have a techique to solve the secod problem. So our ext atural step is to fid ways to determie the coefficiet of fuctios of this type. To help us do this we will mae use of the followig idetities. The fxgx a 0 b 0 a 0 b a b 0 x a 0 b a b a b 0 x.

5 The ey is that the coefficiet of x is foud by combiig elemets a i x i b i x for i 0,...,. We ow show how to use these various rules to solve a combiatorial problem. Example: How may ways are there to put 5 balls ito 7 boxes so that each box has betwee ad 6 balls? Solutio: First we traslate this ito a polyomial problem. We are looig for 5 balls so we will be looig for a coefficiet of x 5. The costrait o each box is the same, amely the betwee ad 6 balls so that the fuctio we will be cosiderig is gx x x 3 x 4 x 5 x 6 7 x 4 x x x 3 x 4 7. Here we pulled out a x out of the iside which the becomes x 4 i frot. Now sice we are looig for the coefficiet of x 5 ad we have a factor of x 4 i frot, our problem is equivalet to fidig the coefficiet of x i g x x x x 3 x 4 7. Combiatorially, this is also what we would do. Namely distribute two balls ito each bi, fourtee balls i total, ad the decide how to deal with the remaiig. Usig our idetities we have g x x x x 3 x 4 7 x 5 7 x 5 7 x x x 5 x 0 x 0 0 Fially, we use the rule for multiplyig fuctios together to fid the coefficiet of x. I particular ote that i the first part of the product oly three terms ca be used, as all the rest will ot cotribute to the coefficiet of x. So multiplyig we have that the coefficiet to x is , 055. This ca also be doe combiatorially. The first term couts the umbers without restrictio. The secod term the taes off the exceptioal cases, but it taes off too much ad so the third term is there to add it bac i. Lecture 8 April 5 We ca use the rule for multiplyig fuctios together to prove various results. For istace, let fx x m ad gx x, the fxgx x m. We ow compute the coefficiet of x r of fxgx i two differet ways to get r 0 m r m r. The left had side follows by usig the rule of multiplyig fx ad gx while the right had side is the biomial theorem for fxgx. We tur to a problem of fidig a geeratig fuctio. Example: Show that the geeratig fuctio for the umber of iteger solutios to e e e 3 e 4 with 0 e e e 3 e 4 is Solutio: fx as fx x x x 3 x 4. First ote that we ca rewrite the fuctio fx x x x 3 x x 4 x 6 x 3 x 6 x 9 x 4 x 8 x. Note that traslatig this bac ito a balls ad bis problem this says that we have four bis. I the first bi we have ay umber of balls, i the secod bi we have a multiple of two umber of balls, i the third bi we have a multiple of three umber of balls ad i the fourth bi we have a multiple of four umber of balls. I other words the geeratig fuctio fx couts the umber of solutios to f f 3f 3 4f 4 with f, f, f 3, f 4 0. We ow eed to show that these two problems have the same umber of solutios. To do this, let us start with a solutio of e e e 3 e 4 ad pictorially represet our solutio by puttig four row of s with e 4 stars i the first row, e 3 stars i the secod row, e stars i the third row ad e stars i the fourth row. Fially, let f 4 be the umber of colums with four s, f 3 the umber of colums with three s, f the umber of colums with two s ad f the umber of colums with oe. This gives a solutio to f f 3f 3 4f 4. This gives a oe-to-oe correspodece betwee solutios to the two differet

6 problems, so they have a equal umber of solutios givig the result. As a example of the last step, suppose we start with The pictorially this correspods to the followig picture we have mared the umber of s i each row/colum So traslatig this gives us the solutio to the secod problem This problem says that the umber of ways to brea ito a sum of at most four parts is the same as the umber of ways to brea ito a sum of parts where each part has size at most four. We will geeralize this idea i the ext lecture. For ow we start by looig at partitios. A partitio of a umber is a way to brea up as a sum of smaller pieces. For istace 3 4 are the ways to brea four ito pieces. We do ot care about the order of the pieces so that ad ad are cosidered the same partitio. This correspods to the umber of ways to distribute idetical balls amog idetical bis for ow we will suppose that we have a ulimited umber of bis so that we ca have ay umber of parts i the partitio. Let p deote the umber of partitios of. partitios of p 0 0, 3,, 3 3 4,, 3 5, 4 5, 7 3,, 4 3, 5 6, 3, 4, 3 5,, 4, 3 3, 6 The fuctio p has bee well studied, but is highly o-trivial. Oe of the greatest mathematical geiuses of the twetieth cetury was the Idia mathematicia Sriivasa Ramauja. Origially a cler i Idia he was able through persoal study discover dozes of previously uow relatioships about partitios. He set these alog with other discoveries to mathematicias i Eglad ad oe of them, G. H. Hardy, was able to recogize the writigs as somethig ew ad excitig which caused him to brig Ramauja to Eglad which resulted i oe of the great mathematical collaboratios of the twetieth cetury. Examples of facts that Ramauja discovered iclude that p5 4 is divisible by 5; p7 5 is divisible by 7; ad p 6 is divisible by. Oe atural questio is to whether the exact value of p is ow. There is a ice formula for p, amely sih π x ] 4 p π A [ d dx x 4 x where A is a specific sum of 4th roots of uity. Provig this is beyod the scope of our class! We ow tur to the much easier problem of fidig the geeratig fuctio for p, that is we wat to fid px. The first thig is to observe that a partitio of size correspods to a solutio of e e 3e 3 e, where each e i represets the umber of parts of size i. Notice i particular the the cotributio of e will be a multiple of. So turig this bac ito balls ad bis we have balls ad ifiitely may bis. So the umber of solutios usig what we did last time is Sice px xx x 3 }{{} e term x x 4 x 6 x 3 x 6 x 9 }{{}}{{} e term e 3 term x x x 3. }{{} e term z z z 3 z, we ca rewrite the geeratig fuctio as px x x x 3 x x.

7 We ca do variatios o this. For istace we ca loo at partitios of where o part is repeated. I other words we wat to restrict our partitios so that e 0 or for each. If we deote these partitios by p d the p d x x x x x Similarly we ca cout partitios with each part odd. I other words we wat to restrict out partitios so that e 0 for each. If we deote these partitios by p o the it is easy to modify our costructio for p to get p o x. We ca combie these two geeratig fuctios to derive a remarable result. For we have p d p o. Remarably, we do ot ow i geeral what p d or p o is, evertheless we do ow that they are equal! As a example of this we have that p o 6 4 because of the partitios, 3, 5, 3 3, while p d 6 4 because of the partitios 3, 5, 4, 6 It suffices for us to show that the two geeratig fuctios are idetical. If the fuctios are the same the the coefficiets must be the same givig us the desired result. So we have that x p d x x x x x4 x6 x8 x x x3 x 3 x x 3 x 5 x p o x. With the curret partitio, if ay two parts are equal combie them ito a sigle part. For example for the partitio 4 4 This rule taes a partitio with odd parts ad produces a partitio with distict parts. To go i the opposite directio this cae be doe by applyig the followig rule util it caot be doe aymore: With the curret partitio, if ay part is eve the split it ito two equal parts. For example for the partitio These two rules give a bijective relatioship betwee partitios with a odd umber of parts ad partitios with distict parts. Lecture 9 April 7 We ca visually represet a partitio as a series of rows of dots much lie we did i the example i the previous lecture. This represetatio is ow as a Ferrer s diagram. The umber of dots i each row is the size of the part ad we arrage the rows i wealy decreasig order. The partitio 4 3 would be represeted by the followig diagram. We ca use Ferrer s diagrams to gai isight ito properties of partitios. For istace oe simple operatio that we ca do is to tae the traspose or the cojugate of the partitio. This is doe by flippig across the lie i the picture below, so that the colums become rows ad rows become colums. We ca also give a bijective proof. To do this we eed to give a way to tae a partitio with odd parts ad produce a uique partitio with distict parts, ad vice versa. This ca be doe by repeatedly applyig the followig rule util it caot be doe aymore: Sice we do ot chage the umber of dots by taig the traspose this taes a partitio ad maes a ew

8 partitio. So i this case we get the ew partitio 4. Note that if we tae the traspose of the traspose that we get bac the origial diagram. There are some partitios such that the traspose of the partitio gives bac the partitio, such partitios are called self cojugate. For istace there are two self cojugate partitios for 8, amely 4 ad 3 3. A famous result says that the umber of self cojugate partitios is equal to the umber of partitios with distict odd parts. So for example for 8 there are two partitios with distict odd parts, amely 7 ad 5 3. Oe proof of this relatioship is based o usig Ferrer s diagrams. We will ot fill i all of the details here but give the followig hit for 8. Also ote that whe we tae the traspositio that the umber of rows becomes the size of the largest part i the traspositio while the size of the largest part becomes the umber of rows. So mappig a partitio to its traspose establishes the followig fact. There is a oe-to-oe correspodece betwee the umber of partitios of ito exactly m parts ad the umber of partitios of ito parts of size at most m with at least oe part of size m. It is easy to cout partitios with parts of size at most m with at least oe part of size m. I particular if we let m deote the umber of partitios of ito exactly m parts the we have usig the techiques from the last lecture x m m x x x x m. Aother object associated with a Ferrer s diagram is the Durfee square which is the largest square of dots that is i the upper left corer of the Ferrer s diagram. For example for the partitio we have a 3 3 Durfee square as show i the followig picture. ad the two partitios, oe o the right of the Durfee square ad oe below the Durfee square as illustrated below. i i square partitio with parts at most i partitio with at most i parts If we group partitios accordig to the size of the largest Durfee square ad the use geeratig fuctios for partitios with at most i parts ad partitios with parts at most i which by traspositio are the same geeratig fuctio we get the followig idetity: px x }{{} square x i } {{ } left partitio x i } {{ } bottom partitio x x x x. There are a lot more fasciatig ad iterestig thigs that we ca tal about i regards to partitios. We fiish our discussio with the followig. x i z z z 5 z 7 z z 5 <j< i z 3j j/ This is very similar to what we ecoutered i the previous lecture, amely x i. We saw that this last expressio was used to cout the umber of partitios with distict parts. I the curret expressio somethig similar is goig o, the differece is that previously every partitio of with distict parts cotributed to the coefficiet of x ; ow the cotributio of a partitio of with distict parts depeds o whether the umber of parts is eve or odd. Namely, if the umber of parts is eve the the cotributio is ad if the umber of parts is odd the cotributio is. I particular, it is ot hard to see that E O x, x i It should be oted that every partitio ca be decomposed ito three parts. Namely a Durfee square where E is the umber of partitios of ito a eve umber of distict parts ad O is the umber of partitios of ito a odd umber of distict

9 parts. So to determie the product we eed to determie E O, we will see that this value is, 0 or. To do this we give a bijectio betwee the umber of partitios of ito a eve umber of distict parts ad the umber of partitios of ito a odd umber of distict parts. This is doe by taig the Ferrer s diagram of a partitio ito distict parts ad comparig the size of the smallest part call it q ad the size of the largest 45 ru i the upper left corer call it r. If q > r the tae the poits from the 45 ru ad mae it ito a ew part. If q r the tae the smallest part ad put it i at the ed as the 45 ru. A example of what this is doig is show below. I the partitio o the left we tae the smallest part ad put it at the ed of the first few rows while i the partitio o the right we tae the small 45 ru at the ed of the first few rows ad tur it ito the smallest part. We have p p p p 5 p 7 where the terms o the right had side are those from the petagoal theorem. This follows by otig that x x px j x 3j j/ <j< Now usig the rule for multiplyig fuctios together ad comparig coefficiets o the left ad right had sides we see that for that 0 p p p p 5p 7, rearragig ow gives the desired result. I particular this gives a ivolutio betwee partitios of ito a odd umber of distict parts ad partitios of ito a eve umber of distict parts. All that remais is to fid the fixed poits of the ivolutio, amely those partitios ito distict parts where this operatio fails. It is ot too hard to see that the oly way that this ca fail is if we have oe of the two followig types of partitios. I the oe o the left we have q r but we caot move q dow to the 45 lie because of the overlap. I the oe o the right we have q < r but if we tae the poits o the 45 lie it will create a partitio which does ot have distict parts. Coutig these exceptioal cases are partitios of size 3j ± j. Puttig all of this together alog with a little bit of playig with terms explais the form of the product. This is ow as Euler s petagoal theorem because the umbers that come up i the powers with ozero coefficiets are petagoal umbers which ca be formed by stacig a square o top of a triagle. The petagoal theorem has a ice applicatio which is useful for quicly geeratig the umber of partitios of. We ow retur to geeratig fuctios. The type of geeratig fuctios that we have ecoutered are what are called ordiary geeratig fuctios which are useful i problems ivolvig combiatios. But there is aother class of geeratig fuctios ow as expoetial geeratig fuctios which are useful i problems ivolvig arragemets. Type of geeratig fuctio ordiary expoetial form a x a x! used to cout combiatios arragemets The term expoetial comes from the fact that if a a the the resultig fuctio is x! ex. A typical problem ivolvig combiatios loos at coutig the umber of oegative solutios to a equatio of the form where we have restrictios o the i. I particular for each solutio that we fid we add to the cout for the umber of combiatios. For arragemets we start with the same basic problem, amely we must first choose what to arrage so we have to have a oegative solutio to a equatio similar to the form

10 where agai we have restrictio o the i. But ow after we have chose our terms we still eed to arrage or order them. From a previous lecture if we have objects of type, objects of type, ad so o the the umber of ways to arrage them is!!!!, so ow for each solutio to this is what will be cotributed to the cout for the umber of arragemets. The expoetial fuctio is perfectly set up to cout the umber of expoetial fuctios. To see this cosider, l S xl l! l S xl l! xl l! l S where S i is the possible values for i. The a typical product will be of the form x! x! x! if the the cotributio to x /! is! x!!!!, the umber of arragemets, just lie we wated! Let us loo at two related problems to compare the differeces of the two types of geeratig fuctios. Example: Fid a geeratig fuctio which couts the umber of ways to pic digits from 0s, s ad s so that there is at least oe 0 chose ad a odd umber of s. Solutio: Usig techiques from previous lectures we have that the ordiary geeratig fuctio is xx x 3 xx 3 x 5 xx x x x x x x x 3 x. Example: Fid a geeratig fuctio which couts the umber of terary sequeces sequeces formed usig the digits 0, ad of legth with at least oe 0 ad a odd umber of s. Solutio: This problem is related to the previous, except that after we pic out which digits we will be usig we still eed to arrage the digits. So we will use a expoetial geeratig fuctio to cout these. I our case we have x x! x3 3! x x3 3! x5 5! x x! e x e x e x e x e3x e x e x. Note that the approach is early idetical for both problems, the oly differece i the iitial setup for the geeratig fuctios is the itroductio of the factorial terms. So the same techiques ad sills that we leared i settig up problems before still holds. The oly real tric is to decide whe to use a expoetial geeratig fuctio ad whe ot to use it. I the ed it boils dow for ow to whether or ot we eed to accout for orderig i what we choose. If we do t the go with a ordiary geeratig fuctio, if we do go with a expoetial geeratig fuctio. Lecture 0 April 0 I the last lecture we saw how to set up a expoetial geeratig fuctio to solve a arragemet problem. As with ordiary geeratig fuctios, if we wat to use expoetial geeratig fuctios we eed to be able to fid coefficiets of the fuctios that we set up. To do this we list some idetities. e x e x e x e x e x e lx x! 0 0 x! l x! x! x4 4! x6 6! x x3 3! x5 5! x7 7! e x x x! x3 3! x! Example: Fid the umber of terary sequeces sequeces formed usig the digits 0, ad of legth with at least oe 0 ad a odd umber of s. Solutio: From last time we ow that the expoetial geeratig fuctio gx which couts the umber of terary sequeces is gx e3x e x e x 3 x! x! x! 3 x!. So readig off the coefficiet of x /! we have that the umber of such solutios is 3 / for ad 0 for 0.

11 Example: Fid the expoetial geeratig fuctio where the th coefficiet couts the umber of letter words which ca be formed usig letters from the word BOOKKEEPER. Readig off the coefficiets we see that the aswer to our questio with people is 3 3. I a later lecture we will see a alterative way to derive this expressio usig the priciple of iclusio-exclusio. Solutio: We will use a expoetial geeratig fuctio sice we are iterested i coutig the umber of words ad the arragemet of letters i words gives differet words. Sice order is importat we will go with a expoetial geeratig fuctio. We group by letters. There are 3 Es, Ks ad Os ad B, P ad R. So puttig this together the desired expoetial geeratig fuctio is fx x x! x3 x 3 x x 3!! 6x 33 x! 66x3 3! 758x4 4! 300x5 5! 30 x6 6! 3400x7 7! 84000x8 8! 500 x9 9! 500x0 0!. The last step is doe by multiplyig the polyomial out, lettig the computer do all of the heavy liftig. So we ca read off the coefficiets ad see, for example, that there are 3400 words of legth 7 that ca be formed usig the letters from BOOKKEEPER. Example: How may ways are there to place distict people ito three differet rooms with at least oe perso i each room? Solutio: This does ot loo lie a arragemet problem, so we might ot automatically thi of usig expoetial geeratig fuctios. However, we previously saw that the umber of ways of arragig objects of type, objects of type,..., ad objects of type is the same as the umber of ways to distribute distict objects ito bis so that bi gets objects, bi gets objects,..., ad bi gets objects. So this problem is perfect for expoetial geeratig fuctios. I particular, sice there are three room we ca set this up as 3 where i are the umber people i room i. Sice each room must have at least oe perso we have that i. So the expoetial geeratig fuctio is hx x x! x3 3! x4 4! 3 e x 3 e 3x 3e x 3e x 3 x! 3 x! x. x!