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 {,,, }? The aswer to this questio is deoted by (, which is typically read as choose To obtai a formula for (, we first cosider a differet coutig problem: how may ways are there to choose a sequece of distict elemets of the set {,, }? I other words, how may ijective fuctios f are there from the set {,, } to the set {,,, }? This is easy to determie: there are possible values for f(), ( ) possible values for f(), ad so forth, so the umber of such fuctios is give by ( )( ) ( + ) = ( )( ) (3)()() ( ( ) (3)()() =! (! Of course, this is differet from the aswer to Questio, because we are coutig ordered sequeces of legth, rather tha elemet subsets Every elemet subset of {,,, } ca be ordered i precisely! differet ways We therefore obtai the idetity! ( ) =! (!, or ( )! =!(! Remar The above formula maes sese for ; we obviously have ( = for > Let us ow cosider aother method of determiig the itegers (, which proceeds ot by coutig directly but istead by establishig a recurrece relatio Let X, deote the collectio of all subsets of {,, } havig size, so that X, = ( We ca partitio the set X, ito two subsets Let X +, deote the collectio of all -elemet subsets of {,, } which cotai the iteger Such a subset is give by the uio of {} with a ( )-elemet subset of {,, } It follows that ( ) X +, = X, = Let X, deote the collectio of -elemet subsets of {,, } which do ot cotai the iteger We the have X, = X,, so that ( ) X, = We therefore have at least for > ( ) ( ) = X, = X +, + X, = + ( ),
Exercise 3 Prove the idetity ( ) ( = ) ( + ) ( directly from the formula ) =! It may be istructive to orgaize the itegers ( ito a table ( )!(! ( ) ( ) ( ) ( ) ( ) ( 3 3) ( 4 3 4), which is called Pascal s triagle The recurrece relatio ( ) ( = ( ) + dictates how to fill this table i: each etry is the sum of the two etries diagoally above it More specifically, we obtai 3 3 4 6 4, Let s ow see what we ca lear usig the method of geeratig fuctios itroduced i the last lecture Fix a iteger, ad defie a power series F (x) by the formula F (x) = ( ) x Sice ( = for >, this is actually a fiite sum: that is, F (x) is a polyomial fuctio of x For
example, whe =, we obtai F = For >, we ca use our recurrece relatio to obtai F (x) = ( ) x = ( ) x + ( ) x = x( ( ) x l + ( ) x l = (x + )F (x) It follows that F (x) = (x + ) Now tae variables y ad z, ad write (y + z) = y ( + z y ) = y F ( z y ) = y ( ) ( z y ) = ( ) y z This is a idetity of polyomials, ad therefore remais valid after substitutig ay umbers we lie for y ad z We have prove: Theorem 4 (Biomial Theorem) For ay quatities y ad z (belogig to the itegers, or to the real umbers, or more geerally ay commutative rig) ad ay, we have (y + z) = ( ) y z Because of Theorem 4, the itegers ( are ofte called biomial coefficiets I the last lecture, we metioed that geeratig fuctios should geerally be viewed as formal power series: that is, we geerally do ot care whether or ot they coverge However, we ca sometimes get iformatio by evaluatig a geeratig fuctio at a poit Let us close this lecture by givig a illustratio of this priciple Questio 5 Fix a iteger Compute the sum ( We ow describe a few differet approaches to this questio First, let s use the method of geeratig fuctios Let F (x) be defied as above, so that ( is the coefficiet of x i F (x) Let F (x) deote the derivative of F (x), so that the coefficiet of x i F (x) is ( ) It follows that ( is the sum of the coefficiets of the polyomial F (x): that is, it is the iteger F () We saw above that F (x) = (x+) It follows that F (x) = (x + ), so that F () = Let us ow try to arrive at the same aswer by combiatorial meas The idea is to iterpret ( ) as the solutio to a coutig problem Questio 6 Suppose we are give a group of people How may ways are there to select a committee (that is, a subset of the collectio of people) together with a leader of that committee (who we require belogs to the committee) Let s try aswer Questio 6 i two differet ways Suppose first that we are iterested i committees of size I this case, there are ( choices for the members of the committee, ad choices for its leader: a total of ( choices i all The aswer to Questio 6 is the give by summig over : that is, by ( Here is aother way to solve the coutig problem posed i Questio 6 First, there are choices for the leader of the committee Oce the leader is fixed, the remaiig people ca each be assiged to the committee or ot, for a total of choices We ca therefore give as a aswer to Questio 6 We coclude that ( ) = 3
Let us describe oe more way of arrivig at the aswer The idea is to iterpret the sum appearig i Questio 5 ot the solutio to a coutig problem, but as the solutio to a expected value problem Questio 7 Suppose that a subset S of {,, } is chose at radom What is the expected umber of elemets of S? The aswer is evidetly O the other had, we ca compute the aswer as S {,,} S Here there ( terms i the umerator where the associated summad is, so we ca rewrite the umerator as ( ) We therefore obtai ( ) = = Recall that a deragemet of the set {,, } is a permutatio of {,, } with o fixed poits: that is, a permutatio π such that π(i) i for all i Let s begi with the followig questio: Questio 8 How may deragemets are there of the set {,, }? permutatio chose at radom is a deragemet? What is the probability that a To fix ideas, let D deote the umber of deragemets of the set {,, } We have D = (the idetity permutatio of the empty set has o fixed poits), D = (the idetity permutatio of {} does have a fixed poit), D = (there is a uique deragemet of {, }, give by the o-idetity permutatio), D 3 = (the deragemets of the set {,, 3} are precisely the cyclic permutatios) Our goal ow is to obtai a formula for the itegers D As a startig poit, we ow that the total umber of permutatios of the set {,, } is! We therefore have! = D + X, where X is the set of permutatios of {,, } which have at least oe fixed poit I fact, we ca say somethig more refied For >, let X deote the collectio of all permutatios of {, } havig exactly fixed poits The X ca be writte as the disjoit uio of the subsets X, X,, X We therefore have! = D + X + X + X 3 + + X Remar 9 We ca write this equatio more aturally as! = X + X + + X, where X deotes the set of permutatios with o fixed poits: that is, the set of deragemets To determie D from the above formula, we eed to ow how big the sets X are How ca we describe a permutatio π with exactly fixed poits? First, we eed to specify which elemets of {,, } are fixed poits of π: there are ( possibilities for this i all Secod, we eed to specify what our permutatio does o the remaiig elemets This restricted permutatio must be fixed poit free (otherwise, π would have more tha fixed poits), so the umber of choices is give by D We therefore obtai the formula ( ) X = D so that our earlier equatio reads! = ( ) D 4
Usig our formula for the biomial coefficiets ( ) =!!(!, we see that both sides of this expressio are divisible by! Dividig out, we get = D!(! The aalysis above gives a series of equatios (oe for each ) which ca be used to successively solve for the itegers D Rather tha treatig all of these equatios separately, it will be coveiet to thi about them all at oce, usig the method of geeratig fuctios For this, let us itroduce a formal variable x Multiplyig our previous equatio by x, we get x = D x!(! Summig over all, we obtai a idetity of formal power series x = D x!(! It is ow coveiet to rearrage the sum o the right had side: ote that givig a iteger ad aother iteger betwee ad is equivalet to givig a pair of oegative itegers ad l, with = +l We therefore obtai x =,l D l x +l! The right had side of this expressio factors as a product ( x! )( l =,l ) x! The first factor should be familiar: it the power series expasio for the expoetial fuctio e x Let us itroduce a otatio for the secod factor: F (x) = l = + x + x3 3 + The power series F (x) is called the expoetial geeratig fuctio for the sequece of itegers {D l } l It differs from the geeratig fuctios we have met so far because of the additio of a auxiliary factor of o the coefficiet of x l We ca ow rewrite our equatio as x = ex F (x), which is easy to solve: we get F (x) = e x x 5