Notes on exponential generating functions and structures.


 Jordan Wright
 3 years ago
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1 Notes o expoetial geeratig fuctios ad structures. 1. The cocept of a structure. Cosider the followig coutig problems: (1) to fid for each the umber of partitios of a elemet set, (2) to fid for each the umber of permutatios of a elemet set, (3) to fid for each the umber of subsets of a elemet set, (4) to fid for each the umber of labelled trees o a set of vertices. The precedig problems are of a differet ature tha those we typically solve usig ordiary geeratig fuctios. The differece is that the thigs to be couted are tied to a specified  elemet set for each. For istace, if we wish to list all partitios of a 5elemet set, we ca do it oly by itroducig a particular 5elemet set to work with. Thus we might use the set {a, b, c, d, e} ad begi listig the partitios {{a, b}, {c, d, e}}, {{c}, {a, b, d, e}}, ad so o. O the other had if we wish to list all partitios of the umber 5, there is o uderlyig set of 5 elemets to be itroduced; we just list the possibilities: (5), (4, 1) ad so o. What is commo to the problems listed at the outset is that i each of them, we begi for each with a set of elemets, ad the we are asked to cout all possible ways of imposig a certai structure upo that give set. Thus for a partitio, we are to structure the set by arragig its elemets ito blocks. For a permutatio, we are to arrage them i some order. For a subset, we are to arrage them ito those chose to belog to the subset, ad those ot. For a tree, we are to arrage the elemets ito the vertices of a tree. The purpose of expoetial geeratig fuctios is to solve coutig problems like these, i which some kid of structure is specified ad we are asked to fid for each the umber of ways to arrage the elemets of a elemet set ito such a structure. We defie the expoetial geeratig fuctio associated with a coutig problem for structures to be F (x) = f() x!, where f() is the umber of structures o a elemet set. For the momet, we will have to accept this o faith as a good defiitio. As we proceed further, however, this defiitio will be justified by the fact that it leads to additio ad multiplicatio priciples that are appropriate to structurecoutig problems. As we shall see, it also leads to importat ew coutig priciples that go beyod what we ca do with additio ad multiplicatio. 2. Trivial structures Before it is possible to work out ay sigificat examples of expoetial geeratig fuctios it will be ecessary to build a small repetoire of geeratig fuctios for some seemigly stupid structures. We call them trivial structures because the structures themselves have little cotet. Nevertheless we will fid that their geeratig fuctios are ot i ay sese trivial, but may be iterestig fuctios. Furthermore these structures are ot at all trivial i their usefuless, for every structure we cosider will have these as buildig blocks at some level. We begi with the trivial structure, which is the structure of beig a set. What we mea is that to impose the trivial structure o a give set is to give the elemets oly that structure which they already have: the structure of a set. The poit is, there is just oe such structure o every set. Thus the umber of trivial beig a set structures o a elemet set is f() = 1 for every. The expoetial geeratig fuctio for this trivial structure is therefore F (x) = =0 1 x! = ex.
2 Note that the geeratig fuctio for this most trivial structure is ot at all a trivial fuctio. More geerally, we ca cosider trivial structures that express restrictios o the size of a set. The simplest such restrictio is the structure of beig a 1elemet set. To impose this structure o a 1elemet set is to give it oly the structure it already has but to impose it o aythig but a oe elemet set is impossible. I short, there is by defiitio exactly oe structure of 1 elemet set o every 1elemet set, ad oe o ay other set. The umber of trivial 1elemet set structures o a elemet set is therefore f() = 1 if = 1, ad f() = 0 otherwise. The correspodig geeratig fuctio is F (x) = f() x! = x. I a similar way, correspodig to ay restrictio we might care to place o the size of a set we ca defie a trivial structure reflectig that restrictio. By defiitio there is to be oe such structure o a set if it obeys the restricitio, ad oe if it does t. The umber f() of our trivial structures o a elemet set is thus, by defiitio, the idicator fuctio that is 1 for good values of ad 0 for bad values. The expoetial geeratig fuctio F (x) = f()x /! for our trivial structure is the simply the sum of x /! take over all allowed values of. Fortuately, i may cases this is simple to express i closed form, as i the two examples we just did. Here are some examples of trivial structures. Although the first four examples below are quite trivial ideed, they are also the oes that are most ofte useful. (1) The trivial structure of set : F (x) = e x. (2) The trivial 1elemet set structure: F (x) = x. (3) The trivial empty set strucutre: F (x) = 1. (4) The trivial oempty set structure: F (x) = e x 1. (5) The trivial evesize set structure: F (x) = cosh(x) = (e x + e x )/2. (6) The trivial oddsize set structure: F (x) = sih(x) = (e x e x )/2. As a exercise to be sure you uderstad this so far, I suggest you pause here ad work out i each of the above examples why the geeratig fuctio F (x) is what I said it is. 3. Additio ad multiplicatio priciples To cout aythig iterestig usig expoetial geeratig fuctios, we eed to kow what it meas to add ad multiply them. To state the relevat priciples, let s suppose that we have give ames to some kids of structures, say fstructures, gstructures, ad hstructures. We let f() be the umber of fstructures o a elemet set, so the expoetial geeratig fuctio coutig fstructures is F (x) = f()x /!. Similarly for g ad h. Additio priciple for expoetial geeratig fuctios: Suppose that the set of fstructures o each set is the disjoit uio of the set of gstructures ad the set of hstructures. The F (x) = G(x) + H(x). The additio priciple is rather obvious ad is o differet from the additio priciples for other types of eumeratio, such as straight coutig, or ordiary geeratig fuctios. As a example, we ca cosider f = trivial structure, g = empty structure, h = oempty structure. Every set is either empty or oempty, ad these possibilities are mutually exclusive, so we have F (x) = G(x) + H(x). I this case, we kow that F (x) = e x ad G(x) = 1, which yields the formula H(x) = e x 1 for the expoetial geeratig fuctio of the oempty set structure. The power of expoetial geeratig fuctios begis to appear whe we cosider the multiplicatio priciple. We begi by defiig a otio of product of two structures.
3 Defiitio: Let g ad h deote two types of structures o fiite sets. A g h structure o a set A cosists of (i) a ordered partitio of A ito disjoit subsets A = A 1 A 2, (ii) a gstructure o A 1, ad (iii) a hstructure o A 2, where the structures i (ii) ad (iii) are chose idepedetly. Multiplicatio priciple for expoetial geeratig fuctios: If G(x) ad H(x) are the expoetial geeratig fuctios for gstructures ad hstructures, respectively, the the expoetial geeratig fuctio for g h structures is F (x) = G(x)H(x). There is a atural geeralizatio of this priciple to the product of three or more geeratig fuctios. Namely, a g 1 g 2 g r structure o A cosists of a ordered partitio of A ito r disjoit subsets A = A 1 A 2 A r, ad idepedetly chose g i structures o each subset A i. With this defiitio, to give a g 1 g 2 g r structure is equivalet to givig a g 1 (g 2 g r ) structure, ad it the follows by iductio o r that the geeratig fuctio for g 1 g 2 g r structures is F (x) = G 1 (x)g 2 (x) G r (x). 4. Examples (1) Let us fid the expoetial geeratig fuctio for the umber of subsets of a elemet set. To choose a subset of A is equivalet to choosig a ordered partitio of A ito A 1 = the subset, ad A 2 = its complemet. No further structure is imposed o A 1 or A 2, which is to say that we have the trivial structure o each of them. We apply the multiplicatio priciple with G(x) ad H(x) both equal to the trivial structure geeratig fuctio e x, obtaiig the geeratig fuctio for subsets as F (x) = e x e x = e 2x = (2x) = 2 x!!. Of course this accords with our prior kowledge that a elemet set has 2 subsets. (2) Modifyig the previous example a bit, let us fid the geeratig fuctio coutig oempty subsets of a elemet set. We still have trivial structure o A 2, but we are ow imposig the structure of oempty set o A 1. Thus we get F (x) = (e x 1)e x = e 2x e x = (2 1) x!. Agai we get the expected aswer, sice we already kew that there are 2 1 oempty subsets of a elemet set. (3) Let us fix a umber k ad cout fuctios from a elemet set to {1,..., k}. As a structure o the elemet set A, to give a fuctio A {1,..., k} is equivalet to givig the ordered partitio A = A 1 A k i which A i is the set of elemets of A that are mapped by the fuctio to i. I other words, our structure is equivalet to a product of k trivial structures. Therefore we should multiply together k trivialstructure geeratig fuctios to get F (x) = (e x ) k = e kx = k x! as the geeratig fuctio coutig fuctios A {1,..., k} o a elemet set A. Oce agai this accords with our prior kowledge that there are k such fuctios.
4 (4) The previous example might seem rather elemetary, but cosider this slight variatio: we will cout surjective fuctios from a elemet set A to {1,..., k}. The oly differece from the previous example is that ow each block A i i the ordered partitio that describes the fuctio must be oempty. I other words our structure is equivalet to the product of k trivial oempty set structures. We ca solve this problem as easily as the last oe, replacig e x for the trivial structure with e x 1 for the oempty set structure. This yields the geeratig fuctio F (x) = (e x 1) k for surjective fuctios A {1,..., k} o a elemet set A. From our study of distributio problems, we kow that the umber of partitios of a elemet set A ito k blocks is (by defiitio) the Stirlig umber S(, k). The umber of ordered partitios ito k blocks, or equivaletly of surjectios A {1,... k}, is k!s(, k). Therefore we have the idetity or (e x 1) k = (e x 1) k k! = k!s(, k) x!, S(, k) x!. So we have come up with a expoetial geeratig fuctio for the Stirlig umbers S(, k) as varies, usig oly extremely simple combiatorial cosideratios. By cotrast a ordiary geeratig fuctio for these same umbers ca oly be arrived at through a fairly subtle bijectio. (5) I the last part of example (4) we have foud the expoetial geeratig fuctio coutig uordered partitios of a elemet set ito k blocks, sice the umber of these is S(, k). Summig over all values of k we ca cout all partitios of a elemet set. Their expoetial geeratig fuctio is therefore B x! = k (e x 1) k k! = e ex 1 where the Bell umber B is defied to be the umber of all partitios of a elemet set. Here we ca take ote of a curious fact: the Bell umber geeratig fuctio we just foud turs out to have the form G(H(x)) with G(x) = e x the geeratig fuctio for the trivial structure ad H(x) = e x 1 the geeratig fuctio for the trivial oempty structure. We shall explai this later o. (6) We ow cout permutatios of a elemet set. I this example we will idetify permuatios of A with liear orderigs of A, rather tha with bijectios from A to itself. We will see later o what happes if we take the latter poit of view istead. If = 0, so the set is empty, we shall agree that there is a uique empty permutatio. Otherwise, for a oempty set, we ca choose a permutatio recursively as follows: first decide the first elemet of the permutatio, the decide how to permute the remaiig elemets. Note that this is correct for = 1 because of our covetio that the empty set has oe permutatio. We ca describe the choices ivolved icely i terms of product structures ad trivial structures. To choose a first elemet ad a permutatio of the remaiig elemets is to choose a product structure of the form (oeelemet set) (permutatio). More precisely, such a product structure cosists of a ordered partitio A = A 1 A 2 i which A 1 is required to cosist of a sigle elemet, together with a permutatio of the elemets of A 2. Our chose first elemet becomes the elemet of A 1 here.
5 If F (x) stads for the expoetial geeratig fuctio coutig permutatios we apply the additio ad multiplicatio priciples to get the idetity F (x) = 1 + xf (x). The first term here is the trivial empty set structure coutig the empty permutatio. The term xf (x) couts permutatios of oempty sets by the multiplicatio priciple, the factor x beig the geeratig fuctio for the trivial oeelemet set structure. Solvig for F (x), we have F (x) = 1/(1 x) = x =! x!, i agreemet with our prior kowledge that there are! permutatios of a elemet set. Note that this computatio has othig to do with the repetitio priciple for ordiary geeratig fuctios. We are dealig here with 1/(1 x) as a expoetial geeratig fuctio, ad it does t cout repetitios of aythig. (7) We have ow solved all but oe of the coutig problems metioed at the begiig of these otes. The problem ot yet solved is that of coutig trees o a elemet set of vertices. There are may possible versios of this problem, some of which we will cosider i detail later o. For ow, we solve oly oe versio of the problem. We will cout biary trees whose vertices, icludig the root, are labelled by the give elemet set. We shall agree that the empty tree couts as a biary tree. To choose a biary tree o a oempty vertex set A, we partitio A ito three parts: the root A 1, the left subtree A 2, ad the right subtree A 3. Sice there is oly oe root vertex, the structure o A 1 is the trivial 1elemet set structure. The other parts A 2 ad A 3 are agai give the structure of biary tree. Note that our agreemet that the empty tree couts as a tree allows for the possibility that A 2 ad/or A 3 are empty. Let us deote the expoetial geeratig fuctio for labelled biary trees by T (x). The the descriptio we have just give leads to the idetity Solvig for T (x), we get T (x) = 1 + xt (x) 2. T (x) = 1 1 4x = C x = x!c 2x!, Where C is the th Catala umber. At first it might appear that this example is essetially the same as our earlier determiatio of the umber of biary trees o vertices, usig ordiary geeratig futios. I fact, however, there is a profoud differece. What we couted here was ot abstract ulabelled biary trees, but labelled biary tree structures o a give set of vertices. Ideed, we do ot fid that the umber of them is C, but rather!c, sice T (x) is a expoetial geeratig fuctio. The cofusio results from the accidetal fact that the expoetial geeratig fuctio for!c ad the ordiary geeratig fuctio for C are the same. This accidet occurred because biary trees have a iheret order, so there are exactly! ways to label a ulabelled biary tree o verticies. Later, we will use expoetial geeratig fuctios to cout uordered trees, ad the this accidetal pheomeo will ot occur. 5. The fuctioal compositio priciple We are ow goig to discover the real power of expoetial geeratig fuctios. This power comes from the fact that if G(x) ad H(x) are expoetial geeratig fuctios, the the fuctioal compositio G(H(x)) has a combiatorial iterpretatio.
6 I order for G(H(x)) to make sese as a formal power series, H(x) must satisfy H(0) = 0, that is, the costat term of the geeratig fuctio H(x) must be zero. Combiatorially, this meas that H(x) must cout structures that oly exist o oempty sets, sice h(0), the umber of hstructures o the empty set, is required to be zero. Defiitio: Let g ad h be types of structures, ad assume that there are o hstructures o the empty set. A composite g h structure o a set A cosists of (i) a partitio of A ito ay umber of blocks, (ii) idepedetly chose hstructures o each block of the partitio, ad (iii) a gstructure o the set of blocks. Here the structure i (iii) is also chose idepedetly from the structures i (ii). Compositio priciple: If the expoetial geeratig fuctios for gstructures ad hstructures are G(x) ad H(x), respectively, where H(0) = 0, the the expoetial geeratig fuctio for composite g h structures is give by F (x) = G(H(x)). There is a importat poit to be made about the way i which the partitio i a composite structure is chose. A priori, the partitio should be chose freely, without ay restrictio o its umber of blocks or o their sizes. However, we ca always impose implicit restrictios o these by usig suitable h ad gstructures i parts (ii) ad (iii) of the defiitio of composite structure. I particular, we ca ofte impose useful restrictios by takig trivial structures for h or g. A secod poit relates to the requiremet that H(0) = 0. Notice that i formig a composite structure, you choose hstructures o the blocks, which are oempty by defiitio. I the defiitio of composite structure it therefore makes o differece how may hstructures there are o the empty set. If we wat to use for h a type of structure which does ot have H(0) = 0, we ca always compesate by subtractig the costat term of H(x) from H(x), modifyig it to obtai a ew series Ĥ(x) with Ĥ(0) = 0. This should oly be doe with care, however. I most cases, whe we have aalyzed the problem correctly, the H(x) that goes ito a applicatio of the compositio priciple will aturally have H(0) = 0. If it does ot, it may be a sig of error i our logic. 6. Compositio priciple examples (1) Cosider agai the Bell umber geeratig fuctio e ex 1 which couts all partitios of a elemet set. Before, we had to get this by summig the geeratig fuctio for Stirlig umbers S(, k) over all values of k. Usig the compositio priciple, we ca get the result directly. The structure of a partitio of A is merely a composite structure i which the structures o each block ad o the set of blocks are all trivial. However, ote that the structure take o each block is required to be oe which oly exists o oempty sets. Therefore we should take g to be the trivial structure ad h to be the oempty set structure. This gives G(x) = e x, H(x) = e x 1 ad F (x) = G(H(x)) = e ex 1. (2) The previous example admits a multitude of variatios. For istace, we ca cout partitios of a elemet set ito blocks with more tha oe elemet by replacig H(x) with the geeratig fuctio for the trivial structure of set of more tha oe elemet, which is e x x 1. This gives the geeratig fuctio F (x) = e ex 1 x I a similar vei, we see that the geeratig fuctio for partitios with oly oddsize blocks is F (x) = e sih(x),
7 while the geeratig fuctio for partitios with oly evesize blocks is F (x) = e cosh(x) 1. (3) Suppose you are asked to rate your prefereces amog differet foods, or sports, or whatever. You may prefer some to others, but there might also be oes you like equally well. Thus your rakig may ot be a total orderig of all the items, but rather a partitio of the items ito idifferece classes, with a liear orderig o the set of classes. Such a structure o the set of items is called a preferece orderig. We ca regard a preferece orderig as a composite structure cosistig of a partitio, a liear orderig o the set of blocks, ad a trivial oempty set structure o each block. We have already foud the geeratig fuctio G(x) = 1/(1 x) for liear orderigs. Applyig the compositio priciple with H(x) = e x 1, we get the geeratig fuctio for preferece orderigs F (x) = 1 1 (e x 1) = 1 2 e x. Sice this is a ew cocept we have ot ecoutered before, it is iterestig to evaluate the first few terms of the geeratig fuctio. This ca be doe by substitutig for e x the first few terms of its Taylor series ad usig log divisio to get 1 2 e x = 1 1 x x 2 /2 x 3 /6 = 1 + x + 3x2 /2 + 13x 3 /6 +. Thus there is 1 preferece order o the empty set, 1 preferece order o a set of oe item, 3 possible preferece orders o a set of 2 items, ad 13 o a set of 3 items. For two items, the three possibilities are ({a, b}), ({a}, {b}), ad ({b}, {a}). For three items, you may wish to pause ad list the 13 possibilities yourself. (4) Here is a a approach to coutig permutatios differet tha the oe we used earlier. Before, we idetified permutatios with liear orderigs. This time, we idetify permutatios of a set A with bijectios from A to A, ad make use of the decompositio of a permutatio as a product of disjoit cycles. We defie a cyclic orderig of a elemet set to be a permutatio of its elemets cosistig of a sigle cycle of legth. Note that there are ( 1)! cyclic orderigs of a elemet set. We shall agree that there is o cyclic orderig of the empty set. Let L(x) be the expoetial geeratig fuctio for cyclic orderigs. Sice we kow how may cyclic orderigs there are, we ca write this out explicitly as L(x) = ( 1)! x! = x /. =1 Next cosider ay permutatio of A = {1,..., } expressed i cycle otatio. To choose such a permutatio we ca first choose a partitio of A, the arrage each block ito a cycle. O the set of cycles there is o further structure that is, we have a trivial structure. Applyig the compositio priciple, with G(x) = e x ad H(x) = L(x), we fid that the geeratig fuctio for permutatios is F (x) = e L(x). However, we already kow that the permutatio geeratig fuctio is equal to 1/(1 x), so we get the idetity ad hece e L(x) = 1/(1 x), =1 L(x) = log 1/(1 x) = log(1 x).
8 I this case, we already kew the aswer to the coutig problem, ad we have used it to derive a idetity. Namely, we have foud the power series expasio log(1 x) = x / = x + x2 2 + x3 3 + by purely combiatorial meas, without usig calculus. Of course, this is a formal power series expasio, ad tells us othig about the domai of covergece o the right had side. 7. The derivative priciple Derivative priciple: Suppose a fstructure o a set A is equivalet to a hstructure o the set A { } formed by augmetig A with a extra elemet. The Here are some examples: F (x) = H (x). (1) Suppose we liearly order the elemets of A plus a extra elemet. The positio of i the orderig cuts A ito two subsets, so such a structure is equivalet to partitioig A ito blocks A 1 ad A 2, ad puttig a liear order o each block. Writig F (x) for the geeratig fuctio for liear orderigs, we obtai the differetial equatio F (x) = F (x) 2. Sice F (0) = 1, we ca solve this to get the idetity F (x) = 1/(1 x) i yet aother way. (2) We will cout alteratig permutatios of {1,..., } for every. By this we mea a permutatio which has the form a 1 < a 2 > a 3 < a 4 > < a 1 > a i oerow otatio. Note that we ca oly have the above patter if is odd. I particular, we agree that there is o empty alteratig permutatio, ad that the uique permutatio of {1} is alteratig. Let F (x) be the expoetial geeratig fuctio coutig alteratig permutatios. By listig them, you ca verify that there are 2 alteratig permutatios of {1, 2, 3} (amely 132 ad 231) ad 16 of {1, 2, 3, 4, 5}, so the first few terms of F (x) are F (x) = x + 2x 3 /6 + 16x 5 / Now cosider the problem of choosig a alteratig permutatio o umbers {1,..., +1}, which we thik of as A = [] = {1,..., } augmeted with the extra elemet + 1. Of course alteratig permuatios o A { + 1} oly exist whe is eve. By the derivative priciple, the geeratig fuctio for alteratig permutatios o A { + 1} is F (x). Note that sice F (x) cotais oly odd powers of x, it follows that F (x) cotais oly eve powers of x, as it should. Whe > 0, to choose a alteratig permuatio o A { + 1}, you must put some of the umbers 1,..., before + 1 ad the remaiig umbers after + 1. Sice a alteratig permutatio begis with a icrease ad eds with a decrease, + 1 ca t be at either ed. Thus the portios before ad after + 1 must both be oempty. I the portio before + 1 the last step must be a decrease, sice the step to + 1 is automatically a icrease. Thus the first portio is a alteratig permutatio. By similar reasoig, so is the secod portio. Thus the set 1,..., receives a product structure: it is partitioed ito two parts, each arraged ito a alteratig permutatio. The correspodece betwee such structures o 1,..., ad alteratig permutatos o 1,..., + 1 is clearly bijective. By the multiplicatio priciple, the geeratig fuctio for such structures is F (x) 2, ad addig 1 for the case = 0 we arrive at the differetial equatio F (x) = F (x)
9 Together with the iitial value F (0) = 0, this equatio determies F (x). The solutio is F (x) = ta x. Thus we have foud a combiatorial iterpretatio of the coefficiets of the power series for ta x it is the expoetial geeratig fuctio for alteratig permutatios. By similar reasoig you ca show that sec x is the expoetial geeratig fuctio for eve alteratig permutatios (which start ad ed with a icreasig step, ad iclude the empty permutatio). 8. Proof of the priciples I this sectio we will prove the coutig priciples for expoetial geeratig fuctios. The additio priciple is obvious, so we begi with the multiplicatio priciple. Multiplicatio priciple: Let f() be the umber of g h structures o a elemet set A. To choose a g hstructure is to choose a partitio of A ito A 1 ad A 2, with gstructure o A 1 ad hstructure o A 2. If A 1 = k, the there are ( k) choices of the partitio, g(k) choices of gstructure, ad h( k) choices of hstructure. Summig over all k we have Hece f() = k=0 ( ) g(k)h( k) = k f()/! = k=0 k=0 g(k) h( k) k! ( k)!.! g(k) h( k) k! ( k)!. This last sum is precisely the coefficiet of x i G(x)H(x), showig that F (x) = G(x)H(x). Compositio priciple: We will use the additio ad multiplicatio priciples. As part of a g h structure o A we have a partitio with some umber of blocks, say k. We will fid the geeratig fuctio for g h structures with exactly k blocks, the sum over all k. Fixig the umber of blocks to be k, let us cosider the structure o A cosistig of a ordered partitio A = A 1 A k, together with a hstructure o each block A i ad a gstructure o the set of blocks. There are always g(k) ways to choose the g structure, ad the remaiig part of the structure is just the product of k hstructures. Hece the geeratig fuctio for such ordered composite structures with exactly k blocks is g(k)h(x) k. Note that we have used here the hypothesis that H(0) = 0, that is, that there are o hstructures o the empty set. Without this hypothesis, some of the parts A i i i the product of hstructures might be empty, so that we would ot ecessarily be coutig structures with exactly k blocks. Now, there are k! ordered structures as above for each composite g h structure with k blocks, so the geeratig fuctio for the latter is g(k) H(x) k. k! Summig over all k we obtai the geeratig fuctio for composite structures F (x) = k g(k) H(x)k k! = G(H(x)).
10 Derivative priciple: The derivative of x /! is x 1 /( 1)!. Therefore if F (x) = f()x /! the the coefficiet of x /! i F (x) is f( + 1), the umber of fstructures o a elemet set A augmeted by a extra elemet.
CS103X: Discrete Structures Homework 4 Solutions
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