Counting Principles and Generating Functions

Transcription

1 Uit- Coutig Piciples ad Geeatig Fuctios. THE RULES OF SUM AND PRODUCT.. Sum Rule. (The Piciple of disjuctive Coutig) If a fist task ca be doe i ways ad a secod task i ways, ad if these tasks caot be doe at the same time, the thee ae + ways to do oe of these tasks. I othe wods. If a set X is the uio of disjoit o empty subsets S, S,..., S, the X S + S S... Poduct Rule. (The piciple of Sequetial Coutig) Suppose that a pocedue ca be boke dow ito a sequece of two tasks. If thee ae ways to do the fist task ad ways to do the secod task afte the fist task has bee doe, the thee ae ways to do the pocedue. I othe wods, If S, S,..., S ae o empty sets, the the umbe of elemets i the catesia poduct S S... S is the poduct i S. That is, S S... S i i S. Poblem.. A studet ca choose a compute poject fom oe of thee lists. The thee lists cotai 3, 5 ad 9 possible pojects, espectively. How may possible pojects ae thee to choose fom? Solutio. The studet ca choose a poject fom the fist list i 3 ways, fom the secod list i 5 ways, ad fom the thid list i 9 ways. Hece, thee ae pojects to choose fom. Poblem.. Suppose that eithe a membe of the mathematics faculty o a studet who is a mathematics majo is chose as a epesetative to a uivesity committee. How may diffeet choices ae thee fo this epesetative if thee ae 37 membes of the mathematics faculty ad 83 mathematics majos? Solutio. The fist task, choosig a membe of the mathematics faculty, ca be doe i 37 ways. The secod task, choosig a mathematics majo, ca be doe i 83 days. Fom the sum ule it follows that thee ae possible ways to pick this epesetative. i

2 COMBINATORICS AND GRAPH THEORY Poblem.3. What is the value of k afte the followig code has bee eecuted? k : 0 fo i : to k : k + fo i : to k : k fo i m : to m k : k +. Solutio. The iitial value of k is zeo. This block of code is made up of m diffeet loops. Each time a loop is tavesed, is added to k. Let T i be the task of tavesig the i th loop. The task T i ca be doe i i ways, sice the i th loop is tavesed i times. Sice o two of these tasks ca be doe at the same time, the sum ule shows that the fial value of k, which is the umbe of ways to do oe of the tasks T i, i,,..., m, is m. Poblem.4. I a vesio of the compute laguage BASIC, the ame of a vaiable is a stig of oe o two alpha umeic chaactes, whee uppe case ad lowe case lettes ae ot distiguished. (A alpha umeic chaacte is eithe oe of the 6 Eglish lettes o oe of the 0 digits). Moeove, a vaiable ame must begi with a lette ad must be diffeet fom the five stig of two chaactes that ae eseved fo pogammig use. How may diffeet vaiable ames ae thee i this vesio of BASIC? Solutio. Let V equal the umbe of diffeet vaiable ames i this vesio of BASIC. Let V be the umbe of these that ae oe chaacte log ad V be the umbe of these that ae two chaacte log. The by the sum ule, V V + V. Note that V 6, sice a oe-chaacte vaiable ame must be a lette. Futhemoe, by the poduct ule thee ae stigs of legth two that begi with a lette ad ed with a alphaumeic chaacte. Howeve, five of these ae ecluded, so that V Hece, thee ae V V + V diffeet ames fo vaiables i this vesio of BASIC. Poblem.5. Each use o a compute system has a passwod, which is si to eight chaactes log, whee each chaacte is a uppe case lette o a digit. Each passwod must coati at least oe digit. How may possible passwods ae thee? Solutio. Let P be the total umbe of possible passwods, ad let P 6, P 7 ad P 8 deote the umbe of possible passwods of legth 6, 7, ad 8 espectively.

3 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 3 By the sum ule, P P 6 + P 7 + P 8 We will ow fid P 6, P 7, ad P 8. Fidig P 6 diectly is difficult. To fid P 6 it is easie to fid the umbe of stigs of uppe case lettes ad digits that ae si chaactes log, icludig those with o digits, ad subtact fom this the umbe of stigs with o digits. By the poduct ule, the umbe of stigs of si chaactes is 36 6 ad the umbe of stigs with o digits is 6 6. Hece, P , 76, 78, , 95, 776, 867, 866, 560. Similaly, it ca be show that P , 364, 64, 096 8, 03, 80, 76 70, 33, 353, 90. ad P , 8, 09, 907, , 87, 064, 576, 6, 8, 84, 880. Cosequetly, P P 6 + P 7 + P 8, 684, 483, 063, 360. Poblem.6. I how may ways ca we daw a heat o a spade fom a odiay deck of playig cads? A heat o a ace? A ace o a kig? A cad umbeed though 0? A umbeed cad o a kig? Solutio. Sice thee ae 3 heats ad 3 spades we may daw a heat o a spade i ways. We may daw a heat o a ace i ways, sice thee ae oly 3 aces that ae ot heats. We may daw a ace o a kig i ways. These ae 9 cads umbeed though 0 i each of 4 suits, clubs, diamods, heats, o spades. So we may choose a umbeed cad i 36 ways. Thus, we may choose a umbeed cad o a kig i ways. Poblem.7. How may ways ca we get a sum of 4 o of 8 whe two distiguishable dice (say oe die is ed ad the othe is white) ae olled? How may ways ca we get a eve sum? Solutio. Let us label the outcome of a o the ed die ad a 3 o the white die as the odeed pai (, 3). The we see that the out comes (, 3), (, ), ad (3, ) ae the oly oes whose sum is 4. Thus, thee ae 3 ways to obtai the sum. Likewise, the obtai the sum 8 fom the outcomes (, 6), (3, 5), (4, 4), (5, 3), ad (6, ). Thus, thee ae outcomes whose sum is 4 o 8. The umbe of ways to obtai a eve sum is the same as the umbe of ways to obtai eithe the sum, 4, 6, 8, 0 o.

4 4 COMBINATORICS AND GRAPH THEORY Thee is way to obtai the sum, 3 ways to obtai the sum 4, 5 ways to obtai 6, 5 ways to obtai a 8, 3 ways to obtai a 0, ad way to obtai a. Theefoe, thee ae ways to obtai a eve sum. Poblem.8. How may ways ca we get a sum of 8 whe two idistiguishable dice ae olled? A eve sum? Solutio. Had the dice bee distiguishable, we should obtai a sum of 8 by the outcomes (, 6), (3, 5), (4, 4), (5, 3) ad (6, ), but sice the dice ae simila, the outcomes (, 6) ad (6, ) ad, as well, (3, 5) ad (5, 3) caot be diffeetiated ad thus we obtai the sum of 8 with the oll of two simila dice i oly 3 ways. Likewise, we ca get a eve sum i ways. Poblem.9. If thee ae 4 boys ad gils i a class, fid the umbe of ways of selectig oe studet as class epesetative. Solutio. Usig sum ule, thee ae ways of selectig oe studet (eithe a boy o a gil) as class epesetative. Poblem.0. If a studet is gettig admissio i 4 diffeet Egieeig Colleges ad 5 Medical Colleges, fid the umbe of ways of choosig oe of the above colleges. Solutio. Usig sum ule, thee ae ways of choosig oe of the colleges. Poblem.. I how may ways ca you get a total of si whe ollig two dice? Solutio. The evet get a si is the uio of the mutually eclusive subevets. A : two 3 s A : a ad a 4 A 3 : a ad a 5 Evet A ca occu i oe way, A ca occu i two ways (depedig o which die lads 4), ad A 3 ca occu i two ways, so the umbe of ways to get a si is Poblem.. The chais of a auditoium ae to be labeled with a lettes ad a positive itege ot eceedig 00. What is the lagest umbe of chais that ca be labeled diffeetly? Solutio. The pocedue of labelig a chai cosists of two tasks, amely, assigig oe of the 6 lettes ad the assigig oe of the 00 possible iteges to the seat. The poduct ule shows that thee ae diffeet ways that a chai ca be labeled. Theefoe, the lagest umbe of chais that ca be labeled diffeetly is 600. Poblem.3. Thee ae 3 mico computes i a compute cete. Each mico compute has 4 pots. How may diffeet pots to a mico compute i the cete ae thee? Soltuio. The pocedue of choosig a pot cosists of two tasks, fist pickig a mico compute ad the pickig a pot o this mico compute. Sice thee ae 3 ways to choose the mico compute ad 4 ways to choose the pot o matte which mico compute has bee selected, the poduct ule shows that thee ae pots. Poblem.4. How may diffeet bit stigs ae thee of legth seve? Solutio. Each of the seve bits ca be chose i two ways, sice each bit is eithe 0 o. Theefoe, the poduct ule shows thee ae a total of 7 8 diffeet bit stigs of legth seve.

5 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 5 Poblem.5. How may diffeet licese plates ae available if each cotais a sequece of thee lettes followed by thee digits (ad o sequeces of lettes ae pohibited, eve if they ae abscece)? Solutio. Thee ae 6 choices fo each of the thee lettes ad te choices fo each of the thee digits. Hece, by the poduct ule thee ae a total of ,576,000 possible licese plates. Poblem.6. How may fuctios ae thee fom a set with m elemets to oe with elemets? (Coutig Fuctios). Solutio. A fuctio coespods to a choice of oe of the elemets i the co-domai fo each of the m elemets i the domai. Hece, by the poduct ule thee ae.... m fuctios fom a set with m elemets to oe with elemets. Fo eample, thee ae 5 3 diffeet fuctios fom a set with thee elemets to a set with 5 elemets. Poblem.7. How may oe-to-oe fuctios ae thee fom a set with m elemets to oe with elemets? (Coutig oe-to-oe Fuctios) Solutio. Fist ote whe m > thee ae o oe-to-oe fuctios fom a set with m elemets to a set with elemets. Now let m. Suppose the elemets i the domai ae a, a,..., a m. Thee ae ways to choose the value of the fuctio at a. Sice the fuctio is oe-to-oe, the value of the fuctio at a ca be picked i ways. (Sice the value used fo a caot be used agai). I geeal, the value of the fuctio at a k ca be choose i k + ways. By the poduct ule, thee ae ( )( )... ( m + ) oe-to-oe fuctios fom a set with m elemets to oe with elemets. Fo eample, thee ae oe-to-oe fuctios fom a set with elemets to a set with 5 elemets. Poblem.8. The fomat of telephoe umbes i Noth Ameica is specified by a umbeig pla. A telephoe umbe cosists of 0 digits, which ae split ito a 3-digit aea code, a 3-digit office code, ad a 4-digit statio code. Because of sigatig cosideatios, these ae cetai estictios o some of these digits. To specify the allowable fomat, let X deote a digit that ca take ay of the values 0 though 9, let N deote a digit that ca take ay of the values though 9, ad let Y deote a digit that must be a 0 o a. Two umbeig plas, which will be called the old pla ad the ew pla, will be discussed. (The old pla, i use i the 960s, has bee eplaced by the ew pla, but the ecet apid gowth i demad fo ew umbes will make eve this ew pla obsolete). As will be show, the ew pla allows the use of moe umbes. I the old pla, the fomats of the aea code, office code, ad statio code ae NYX, NNX ad XXXX, espectively, so that telephoe umbes had the fom NYX-NNX-XXXX. I the ew pla, the

6 6 COMBINATORICS AND GRAPH THEORY fomats of these codes ae NXX, NXX ad XXXX, espectively, so that telephoe umebs have the fom NXX-NXX-XXXX. How may diffeet Noth Ameica telephoe umbes ae possible ude the old pla ad ude the ew pla? (The telephoe Numbeig plae) Solutio. By the poduct ule, thee ae aea codes with fomat NYX ad aea codes with fomat NXX. Similaly, by the poduct ule, thee ae office codes with fomats NNX. The poduct ule also shows that thee ae ,000 statio codes with fomat XXXX. Cosequetly, applyig the poduct ule agai, it follows that ude the old pla thee ae ,000,04,000,000 diffeet umbes available i Noth Ameica. Ude the ew pla thee ae ,000 6,400,000,000 diffeet umbes available. Poblem.9. What is the value of k afte the followig code has bee eecuted? k : 0 fo i : to fo i : to fo i m : to m k : k +. Solutio. The iitial value of k is zeo. Each time the ested loop is tavesed, is added to k. Let T i be the task of tavesig the i th loop. The the umbe of times the loop is tavesed is the umbe of ways to do the tasks T, T,..., T m. The umbe of ways to cay out the task T j, j,,..., m, is j, sice the j th loop is tavesed oce fo each itege i j with i j j. By the poduct ule, it follows that the ested loop is tavesed... m times. Hece, the fial value of k is... m. Poblem.0. Use the poduct ule to show that the umbe of diffeet subsets of a fiite set S is s. (Coutig subsets of a Fiite Set). Solutio. Let S be a fiite set. List the elemets of S i abitay ode. Recall that thee is a oe-to-oe coespodece betwee subsets of S ad bit stigs of legth S. Namely, a subset of S is associated with the bit stig with a i the i th positio if the i th elemet i the list is i the subset, ad a 0 i this positio othewise.

7 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 7 By the poduct ule, thee ae S bit stigs of legth S. Hece, P(S) S. Poblem.. Licece plates i the caadia povice of Otaio cosist of fou lettes followed by thee of the digits 0 9 (ot ecessaily distict). How may diffeet licece plates ca be made i otaio? Solutio. Thee ae 6 ways i which the fist lette ca be chose, 6 ways i which the secod ca be chose. Similaly, fo the thid ad fouth. By the multiplicatio ule, the umbe of ways i which the thee lettes ca be chose is By the same easoig thee ae 0 3 ways i which the fial thee digits of a otaio licece plate ca be selected ad, all i all ,976,000 diffeet licece plates which ca be maufactued by the govemet of otaio ude its cuet system. Poblem.. How may umbes i the age do ot have ay epeated digits? Solutio. Imagie eumeatig all umbes of the desied type i the spiit of Fig... Fig... Thee ae ie choices fo the fist digit (ay of 9). Oce this has bee chose, thee emais still ie choices fo the secod (the chose fist digit caot be epeated but 0 ca ow be used). Thee ae ow eight choices fo the thid digit ad seve fo the fouth. Altogethe, thee ae possible umbes. Poblem.3. How may eve umbes i the age have o epeated digits? Solutio. The questio is equivalet to askig fo the umbe of ways i which oe ca wite dow a eve umbe i the age without epeatig digits. This evet ca be patitioed ito two mutually eclusive cases. Case. The umbe eds i 0. I this case, thee ae ie choices fo the fist digit ( 9) ad the eight fo the secod (sice 0 ad the fist digit must be ecluded). So thee ae umbes of this type.

8 8 COMBINATORICS AND GRAPH THEORY Case. The umbe does ot ed i 0. Now thee ae fou choices fo the fial digit (, 4, 6 ad 8), the eight choices fo the fist digit (0 ad the last digit ae ecluded), ad eight choices fo the secod digit (the fist ad last digits ae ecluded). Thee ae umbes of this type. By the additio ule, thee ae eve umbes i the age with o epeated digits. Poblem.4. A typesette (log ago) has befoe him 6 tays, oe fo each lette of the alphabet. Each tay cotais te copies of the same lette. I how may ways ca he fom a thee lette wod which equies atmost two diffeet lettes? By wod, we mea ay sequece of thee lettes pt, fo eample ot ecessaily a eal wod fom the dictioay. Two ways ae diffeet uless they use the idetical pieces of type. Solutio. The evet atmost two diffeet lettes is compised of two mutually eclusive cases : Case. The fist two lettes ae the same. Hee, the thid lette ca be abitay, that is, ay of the 58 lettes which emai afte the fist two ae set ca be used. So the umbe of ways i which this case ca occu is 60 9 (60 ) 603,70. Case. The fist two lettes ae diffeet. I this case, the thid lette must match oe of the fist two, so it must be oe of the 8 lettes emaiig i the two tays used fo the fist two lettes. The umbe of ways i which this case occus is ,70,000. By the additio ule, the umbe of ways to fom a wod usig at most two diffeet lettes is 603,70,70,000,773,70. Theoem.. A set of cadiality cotais subsets (icludig the empty set ad the etie set itself). Poof. Thee ae seveal ways to pove this fudametal esult. We peset oe hee which uses the ideas of this sectio. Give objects a, a,..., a, each subset coespods to a sequece of choices. Is a i the subset. Is a i the subset. Fially, is a i the subset. Thee ae two aswes to the fist questio, two fo the secod, ad so o. I all, thee ae factos Ways i which all choices ca be made. Thus, these ae subsets. Poblem.5. If distiguishable dice ae olled, i how may ways ca they fall? If 5 distiguishable dice ae olled, how may possible outcomes ae thee? How may if 00 distiguishable dice ae tossed?

9 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 9 ways. Solutio. The fist dice ca fall (evet E ) i 6 ways ad the secod ca fall (evet E ) i 6 Thus, thee ae outcomes whe dice ae olled. Also, the thid, fouth, ad fifth die each have 6 possible outcomes so thee ae possible outcomes whe all 5 dice ae tossed. Likewise thee ae 6 00 possible outcomes whe 00 dice ae tossed. Suppose that the licece plates of a cetai state equie 3 Eglish lettes fol- Poblem.6. lowed by 4 digits. (a) How may diffeet plates ca be maufactued if epetitio of lettes ad digits ae allowed? (b) How may plates ae possible if oly the lettes ca be epeated? (c) How may ae possible if oly the digits ca be epeated? (d) How may ae possible if o epetitios ae allowed at all? Solutio. (a) sice thee ae 6 possibilities fo each of the 3 lettes ad 0 possibilities fo each of 4 digits. 9? (b) (c) (d) Poblem.7. (a) How may 3-digit umbes ca be fomed usig the digits, 3, 4, 5, 6, 8 ad (b) How may ca be fomed if o digit ca be epeated? Solutio. Thee ae 7 3 such 3-digit umbes i (a) Sice each of the 3-digits ca be filled with 7 possibilities. Likewise, the aswe to questio. (b) is sice these ae 7 possibilities fo the hudeds digit but oce oe digit is used it is ot available fo the tes digit (sice o digit ca be epeated i this poblem). Thus, thee ae oly 6 possibilities fo the tes digit, ad the fo the same easo thee ae oly 5 possibilities fo the uits digit. How may diffeet licece plates ae thee that ivolve, o 3 lettes fol- Poblem.8. lowed by 4 digits? Solutio. We ca fom plates with lette followed by 4 digits i ways, plates with lettes followed by 4 digits i ways, ad plates with 3 lettes followed by 4 digits i ways. These sepaate evets ae mutually eclusive so we ca apply the sum ule to coclude that thee ae ( ) 0 4 plates with, o 3 lettes followed by 4-digits. Poblem.9. How may diffeet plates ae thee that ivolve, o 3 lettes followed by,, 3 o 4 digits? Solutio. We see that thee ae ( ) 0 ways to fom plates of, o 3 lettes followed by digit. ( ) 0 plates of, o 3 lettes followed by digits, ( )0 3 plates of, o 3 lettes followed by 3 digits, ad ( )0 4 plates of, o 3 lettes followed by 4 digits.

10 0 COMBINATORICS AND GRAPH THEORY Thus, we ca apply the sum ule to coclude that thee ae ( )0 + ( )0 + ( )0 3 + ( )0 4 ( )( ) ways to fom plates of, o 3 lettes followed by,, 3, o 4 digits. Poblem.30. (a) How may digit o 3-digit umbes ca be fomed usig the digits, 3, 4, 5, 6, 8 ad 9 if o epetitio is allowed? (b) How may umbes ca be fomed usig the digits, 3, 4, 5, 6, 8 ad 9 if o epititio ae allowed? Solutio. (a) Thee ae 7.6.5, thee-digit umbes possible. Likewise, we ca apply the poduct ule to see that these ae 7.6 possible -digit umbes. Hece, these ae possible two-digit o thee-digit umbes. (b) The umbe of digits ae ot specified i this poblem so we ca fom oe-digit umbes, two-digit umbes, o thee digit umbes, etc. But sice o epetitios ae allowed ad we have oly the 7 iteges to wok with, the maimum umbe of digits would have to be 7. Applyig the poduct ule, we see that we may fom 7 oe-digit umbes, two digit umbes thee digit umbes, fou digit umbes, five digit ubmes, si-digit umbes, ad seve-digit umbes. The evets of fomig oe-digit umbes, two digit umbes, thee digit umbes, etc., ae mutually eclusive evets so we apply the sum ule to see that thee ae diffeet umbes we ca fom ude the estictios of this poblem. Poblem.3. How may thee-digit umbes ae thee which ae eve ad have o epeated digits? (Hee we ae usig all digits 0 though 9). Solutio. Fo a umbe to be eve it must ed i 0,, 4, 6 o 8. Thee ae two cases to coside. Fist, suppose that the umbe eds i 0 ; the thee ae 9 possibilities fo the fist digit ad 8 possibilities fo the secod sice o digit ca be epeated. Hece thee ae 9.8 thee-digit umbes that ed i 0. Now suppose the umbe does ot ed i 0. The thee ae 4 choices fo the last digit (, 4, 6 o 8). Whe this digit is specified, the thee ae oly 8 possibilities fo the fist digit, sice the umbe caot begi with 0. Fially, thee ae 8 choices fo the secod digit ad theefoe thee ae umbes that do ot ed i 0. Accodigly sice these two cases ae mutually eclusive, the sum ule gives eve thee-digit umbes with o epeated digits. Poblem.3. Suppose that we daw a cad fom a deck of 5 cads ad eplace it befoe the et daw. I how may ways ca 0 cads be daw so that the teth cad is a epetitio of a pevious daw? Solutio. Fist we cout the umbe of ways, we ca daw 0 cads so that the 0 th cad is ot a epetitio. Fist choose what the 0 th cad will be. This ca be doe i 5 ways. If the fist 9 daws ae diffeet fom this, the each of the 9 daws ca be chose fom 5 cads. Thus thee ae 5 9 ways to daw the fist 9 cads diffeet fom the 0 th cad.

11 COUNTING PRINCIPLES AND GENERATING FUNCTIONS Hece, thee ae (5 9 )(5) ways to choose 0 cads with the 0 th cad diffeet fom ay of the pevious 9 daws. Hece, thee ae 5 0 (5 9 )(5) ways to daw 0 cads whee the 0 th is a epetitio sice thee ae 5 0 ways to daw 0 cads with eplacemets. Poblem.33. I how may ways ca 0 people be seated i a ow so that a cetai pai of them ae ot et to each othe? Solutio. Thee ae 0! ways of seatig all 0 people. Thus, by idiect coutig, we eed oly cout the umbe of ways of seatig the 0 people whee the cetai pai of people (say, A ad B) ae seated et to each othe. If we teat the pai AB as oe etity, the thee ae 9 total etities to aage i 9! ways. But A ad B ca be seated et to each othe i diffeet odes, amely AB ad BA. Thus, thee ae ()(9!) ways of seatig all 0 people whee A ad B ae et to each othe. The aswe to ou poblem the is 0! ()(9!). Poblem.34. I how may ways ca oe select two books fom diffeet subjects fom amog si distict compute sciece books, thee distict mathematics books, ad two distict chemisty books? Solutio. Usig poduct ule oe ca select two books fom diffeet subjects as follows : (i) oe fom compute sciece ad oe fom mathematics i ways. (ii) oe fom compute sciece ad oe fom chemisty i 6. ways. (iii) oe fom mathematics ad oe fom chemisty i 3. 6 ways. Sice these sets of selectios ae paiwise disjoit oe ca use the sum ule to get the equied umbe of ways which is Poblem.35. Fo a set of si tue o false questios, fid the umbe of ways of asweig all questios. Solutio. The umbe of ways of asweig the fist questio is. The secod questio ca also be asweed i ways ad similaly fo othe 4 questios. Hece, the total umbe of ways of asweig all the questios is Poblem.36. Thee pesos ete ito ca, whee thee ae 5 seats. I how may ways ca they take up thei seats? Solutio. The fist peso has a choice of 5 seats ad ca sit i ay oe of those 5 seats. So, thee ae 5 ways of occupyig the fist seat. The secod peso has a choice of 4 seats. Similaly, the thid peso has a choice of 3 seats. Hece, the equied umbe of ways i which all the thee pesos ca seat is Poblem.37. Thee ae fou oads fom city X to Y ad five oads fom city Y to Z, fid (i) how may ways is it possible to tavel fom city X to city Z via city Y. (ii) how diffeet oud tip outes ae thee fom city X to Y to Z to Y ad back to X. Solutio. (i) I goig fom city X to Y, ay of the 4 oads may be take. I goig fom city Y to Z, ay of the 5 oads may be take. So by the poduct ule, thee ae ways to tavel fom city X to Z via city Y.

12 COMBINATORICS AND GRAPH THEORY (ii) A oud tip jouey ca be pefomed i the followig fou ways. () Fom city X to Y () Fom city Y to Z (3) Fom city Z to Y (4) Fom city Y to X. () Ca be pefomed 4 ways, 5 ways to pefom, 5 ways to pefom 3 ad 4 ways to pefom 4. By poduct ule, thee ae oud tip outes. Poblem.38. How may bit stigs of legth eight eithe stat with a bit o ed with the two bits 00? Solutio. The fist task, costuctig a bit stig of legth eight begiig with a bit, ca be doe i 7 8 ways. This follows by the poduct ule, sice the fist bit ca be chose i oly oe way ad each of the othe seve bits ca be chose i two ways. The secod task, costuctig a bit stig of legth eight edig with the two bits 00, ca be doe i 6 64 ways. This follows by the poduct ule, sice each of the fist si bits ca be chose i two ways ad the last two bits ca be chose i oly oe way. Both tasks, costuctig a bit stig of legth eight that begis with a ad eds with 00, ca be doe i 5 3 ways. This follows by the poduct ule, sice the fist bit ca be chose i oly oe way, each of the secod though the sith bits ca be chose i two ways, ad the last two bits ca be chose i oe way. Cosequetly, the umbe of bit stigs of legth eight that begi with a o ed with a 00, which equals the umbe of ways to do eithe the fist task o the secod task, equals PERMUTATIONS Suppose that, we have thee lettes a, b ad c. The, all possible aagemets of ay two lettes out of these thee lettes ca be eumeated as : ab, ac, bc, ba, ca ad cb. If we make a aagemet of all thee lettes out of these thee, the we have abc, acb, bac, bca, cab ad cba as possible aagemets. Each of the distict ode of aagemets of a give set of distict objects, takig some o all of them at a time (with o without epetitio), is called a pemutatio of the objects. The total umbe of pemutatios of distict objects, take at a time ( ), is equal to the total ubme of ways of placig objects i boes. This is deoted as P(, ) o P. This umbe is equal to ( )... ( + ). Notatio P : We kow that P is the umbe of pemutatio of distict objects take at a time, ad this is equal to ( )( )... ( + ). Theefoe, P ( )( )... ( + ) * ( )! ( )!! ( )!

13 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 3 Theefoe P! ( )!! P Q 0! ( )! 0! The umbe of pemutatios of distict objects, take at a time, is give by P!. Pemutatio of objects whe all ae ot distict, i this case, the umbe of pemutatios is give by! p! q!! The umbe of distiguishable pemutatios that ca be fomed fom a collectio of objects, take all at a time, i which the fist object appeas k times, the secod object k times, ad so o, is give by! k! k! k!... k! 3 whee k + k k. The umbe of pemutatios of distict objects, take at a time, whe epetitios ae allowed, is give by. The umbe of ways of aagig objects ude some estictio the umbe of aagemets of the same umbe of object without estictio the umbe of aagemets of the same umbe of aagemets with the opposite estictio. Geeatig fuctio fo pemutatio : The coefficiet of i a polyomial P() is P ad P is the umbe of pemutatios of! objects take at a time. Theoem.. The umbe of -pemutatios of a set with distict elemets is P(, ) ( )( )... ( + ) P! ( )! Poof. The fist elemet of the pemutatio ca be chose i ways, sice thee ae elemets i the set. Thee ae ways to choose the secod elemet of the pemutatio, sice thee ae elemets left i the set afte usig the elemet picked fo the fist positio. Similaly, thee ae ways to choose the thid elemet ad so o, util thee ae eactly ( ) + ways to choose the th elemet. Cosequetly, by the poduct ule, thee ae ( )( )... ( + ) -pemutatios of the set.

14 4 COMBINATORICS AND GRAPH THEORY It follows that P(, ) ( )( )... ( + )!. ( )! Eample. Let A be {,, 3, 4}. The the sequeces 4, 4, 34 ad 43 ae same pemutatios of A take 3 at a time. The sequeces, 43, 3, 4, ad ae eamples of diffeet pemutatios of A take two at a time. The total umbe of pemutatios of A take thee at a time is 4P 3 o 4.3. o 4. The total umbe of pemutatios of A take two at a time is 4P o 4.3 o. Note. Whe, we ae coutig the distict aagemets of the elemets of A, with A, ito sequeces of legth. Such a sequece is simply called a pemutatio of A. The umbe of pemutatios of A is thus P o. ( ). ( )...., if. This umbe is also witte! ad is ead factoial. Both P ad! ae built i fuctios o may calculatos. Eample. Let A be {a, b, c}. The the possible pemutatios of A ae the sequeces abc, acb, bac, bca, cab ad cba. Fo coveiece, we defie 0! to be. The fo evey 0 the umbe of pemutatios of objects is!. If ad. Eample 3. Let A cosist of all 5 cads i a odiay deck of playig cads. Suppose that this deck is shuffled ad a had of five cads is dealt. A list of cads i this had, i the ode i which they wee dealt, is a pemutatio of A take five at a time. Eamples would iclude AH, 3D, 5C, H, JS, H, 3H, 5H, QH, KD ; JH, JD, JS, 4H, 4C ; ad 3D, H, AH, JS, 5C. Note that the fist had ad last hads ae the same, but they epeset diffeet pemutatios sice they wee dealt i a diffeet ode. The umbe pemutatios of A take five at a time is 5 P 5 5! 47! o o 3, 875, 00. This is the umbe of five-cad hads that ca be dealt if we coside the ode i which they wee dealt. Eecise 4. The umbe of distiguishable wods that ca be fomed fom the lettes of! MISSISSIPPI is o 34, 650.!4!4!! Theoem.3. Suppose that two tasks T ad T ae to be pefomed i sequece. If T ca be pefomed i ways, ad fo each of these ways T ca be pefomed i ways, the the sequece T T ca be pefomed i ways. (Multiplicatio piciple of coutig). Poof. Each choice of a method of pefomig T will esult i a diffeet way of pefomig the task sequece. Thee ae such methods, ad fo each of these we may choose ways of pefomig T.

15 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 5 Thus, i all, thee will be ways of pefomig the sequece T T. See Fig.. fo the case whee is 3 ad is 4. Fig... Theoem.4. Suppose that tasks T, T,... T k ae to be pefomed i sequece. If T ca be pefomed i ways, ad fo each of these ways T ca be pefomed i ways, ad fo each of these ways of pefomig T T i sequece, T 3 ca be pefomed i 3 ways, ad so o, the the sequece T T... T k ca be pefomed i eactly... k ways. (Eteded Multiplicatio piciple of coutig). Theoem.5. Give atual umbes ad with, the umbe of ways to place mables of diffeet colous ito umbeed boes, at most oe mable to a bo, is P(, ). Notice that ( )( )... ( + ) P(, ) ( )( )... (3)()() ! ( )! Thus, P(, ) 0!.! a fomula which holds also fo 0 ad because P(, 0) ad ( )! Theoem.6. The umbe of pemutatios of symbols is!. The umbe of -pemutatios of symbols is P(, ). Poblem.39. How may pais of dace pates ca be selected fom a goup of wome ad 0 me?

16 6 COMBINATORICS AND GRAPH THEORY Solutio. The fist woma ca be paied with ay of 0 me, the secod woma with ay of the emaiig 9 me, the thid with ay of the emaiig 8, ad so o. These ae P(0, ) possible couples. Poblem.40. Thee ae 7! 5040 ways i which seve people ca fom a lie. I how may ways ca seve people fom a cicle? Solutio. A cicle is detemied by the ode of the people to the ight of ay oe of the idividuals, say Eic. Thee ae si possibilities fo the peso i Eic s ight, the five possibilities fo the et peso, fou fo the et, ad so o. The umbe of possible cicles is 6! 70. Poblem.4. I how may ways ca the lettes of the Eglish alphabet be aaged so that thee ae eactly te lettes betwee a ad z? Solutio. Thee ae P(4, 0) aagemets of the lettes of the alphabet (ecludig a ad z) take te at a time, ad hece.p(4, 0) stigs of lettes, each begiig ad edig with a a ad az(eithe lette comig fist i a stig). Fo each of these stigs, thee ae 5! ways to aage the 4 emaiig lettes ad the stig. So thee ae altogethe.p(4, 0). 5! aagemets of the desied type. Poblem.4. I how may ways ca te adults ad five childe stad i a lie so that o two childe ae et to each othe? Solutio. Imagie a lie of te adults amed A, B,..., j, XD XJ XH XC XI XE XB XA XG XF X, the X s epesetig the possible locatios fo the childe. Fo each such lie, the fist child ca be positioed i ay of the spots, the secod child i ay of the emaiig 0, ad so o. Hece, the childe ca be positioed i P(, 5) ways. Fo each such positioig, thee ae 0! ways of odeig the adults A,..., J, so, by the multiplicatio ule, the umbe of lies of adults ad childe is 0! P(, 5). Poblem.43. A label idetifie, fo a compute system, cosists of oe lette followed by thee digits. If epetitios ae allowed, how may distict label idetifies ae possible? Solutio. Thee ae 6 possibilities fo the begiig lette ad thee ae 0 possibilities fo each of thee digits. Thus, by the eteded multiplicatio piciple, thee ae o 6,000 possible label idetifies. Poblem.44. Let A be a set with elemets. How may subsets does A have? Solutio. We kow that, each subset of A is detemied by its chaacteistic fuctio, ad if A has elemets, this fuctio may be descibed as a aay of 0 s ad s havig legth. The fist elemet of the aay ca be filled i two ways (with a 0 o a ), ad this is tue fo all succeedig elemets as well. Thus, by the eteded multiplicatio piciple, thee ae factos ways of fillig the aay, ad theefoe subsets of A.

17 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 7 Poblem.45. How may diffeet sequeces, each of legth, ca be fomed usig elemets fom A if (a) elemets i the sequece may be epeated? (b) all elemets i the sequece must be distict? Solutio. Fist we ote that ay sequece of legth ca be fomed by fillig boes i ode fom left to ight with elemets of A. I case (a), we may use copies of elemets of A. Let T be the task fill bo, let T be the task fill bo, ad so o. The combied task T T... T epesets the fomatio of the sequece. Case (a). T ca be accomplished i ways, sice we may copy ay elemet of A fo the fist positio of the sequece. The same is tue fo each of the tasks T, T,... T. The by the eteded multiplicatio piciple, the umbe of sequeces that ca be fomed is factos Now we coside case (b), hee also T ca be pefomed i ways, sice ay elemet of A ca be chose fo the fist positio. Which eve elemet is chose, oly ( ) elemets emai, so that T ca be pefomed i ( ) ways, ad so o, util fially T ca be pefomed i ( ) o ( + ) ways. Thus, by the eteded piciple of multiplicatio, a sequece of distict elemets fom A ca be fomed i ( )( )... ( + ) ways. Poblem 46. How may thee-lette wods ca be fomed fom lettes i the set {a, b, y, z}. If epeated lettes ae allowed? Solutio. Hee is 4 ad is 3, so the umbe of such wods is 4 3 o 64. Poblem 47. wod MAST? How may wods of thee distict lettes ca be fomed fom the lettes of the Solutio. The umbe is 4 P 3 Poblem 48. thee? We begi by taggig the A s ad N s i ode to distiguish betwee them tempoaily. Solutio. 4! (4 3)! o 4!! o 4. How may distiguishable pemutatios of the lettes i the wod BANANA ae Fo the lettes B, A, N, A, N, A 3, thee ae 6! o 70 pemutatios. Some of these pemutatios ae idetical ecept fo the ode i which the N s appea. Fo eample, A A A 3 BN N ad A A A 3 BN N.

18 8 COMBINATORICS AND GRAPH THEORY I fact, the 70 pemutatios ca be listed i pais whose membes diffe oly i the ode of the two N s. This meas that if the tags ae dopped fom the N s oly 70 o 360 distiguishable pemutatios emai. Reasoig i a simila way we see that these ca be guoped i goups of 3! o 6 that diffe oly i the ode of the thee A s. Fo eample, oe goup of 6 cosists of BNNA A A 3, BNNA A 3 A, BNNA A A 3, BNNA A 3 A, BNNA 3 A A, BNNA 3 A A. Doppig the tags would chage these 6 ito the sigle pemutatio BNNAAA. Thus, thee ae 360 o 60 distiguishable pemutatios of the lettes of BANANA. 6 Poblem 49. A ma, a woma, a boy, a gil, a dog, ad a cat ae walkig dow a log ad widig oad oe afte the othe. (a) I how may ways ca this happe? (b) I how may ways ca this happe if the dog comes fist? (c) I how may ways ca this happe if the dog immediately follows the boy? (d) I how may ways ca this happe if the dog (ad oly the dog) is betwee the ma ad the boy? Solutio. (a) Thee ae 6! 70 ways fo si ceatues to fom a lie. (b) If the dog comes fist, the othes ca fom 5! lies behid. (c) If the dog immediately follows the boy, the the dog-boy pai should be thought of as a sigle object to be put ito a lie with fou othes. Thee ae 5! 0 such lies. (d) If the ma, dog, ad boy appea i this ode, the thikig of ma-dog-boy as a sigle object to be put ito a lie with thee othes, we see that thee ae 4! possible lies. Similaly, thee ae 4! lies i which the boy, dog, ad ma appea i this ode. So, by the additio ule, thee ae 4! + 4! 48 lies i which the dog (ad oly the dog) is betwee the ma ad the boy. Poblem.50. I how may ways ca te adults ad five childe stad i a cicle so that o two childe ae et to each othe? Solutio. Aage the adults ito a cicle i oe of 9! ways. Thee ae the 0 locatios fo the fist child, 9 fo the secod, 8 fo the thid, 7 fo the fouth, ad 6 fo the fifth. The aswe is 9! ( ) 9! P(0, 5). Poblem.5. Fid (a) P(5, 3), P(4, 4) ad P(7, ) (b) 0! 7!, 00! ad P(7, 0). 98! Solutio. (a) P(5, 3) , P(4, 4) , P(7, )

19 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 9 (b) 0! 7! , 00! 98! P(7, 0) 7! 7!. Poblem.5. A getlema has 6 fieds to ivite. I how may ways ca he sed ivitatio cads to them, if he has thee sevats to cay the cads? Solutio. A cad ca be sed to ay oe fied by ay oe of the thee sevats. Let us take the tasks of sedig cads to si fieds as T, T, T 3, T 4, T 5 ad T 6. Each of the tasks ca be completed i thee distict ways accodig to the umbe of sevats to cay the cads. Thus, by the multiplicatio piciple of coutig the tasks T T T 3 T 4 T 5 T 6 ca be pefomed i ways. Poblem.53. How may umbes of thee digits ca be fomed with the digits,, 3, 4 ad 5 if the digits i the same umbe ae ot epeated? How may such umbes ae possible betwee 00 ad 0,000? Solutio. Hee we have to fid the umbe of pemutatios of 5 distict objects (digits) take 3 at a time. This is give by 5 P The umbes betwee 00 ad 0,000 ae umbes of thee digits ad of fou digits. The total umbe of thee digits umbes, fomed with the give digits, is calculated above ad it is equal to 60. Similaly, the total umbe of fou digits umbes, fomed with,, 3, 4 ad 5, is give by 0. 5 P Thus, the equied umbe is Poblem.54. A telegaph has 5 ams ad each am is capable of 4 distict positios, icludig the positio of est. What is the total umbe of sigals that ca be made? Solutio. Thee ae five ams say T, T, T 3, T 4 ad T 5. Each am ca be i ay oe of the fou positios. Fo each of the positio of am T, thee ae fou possible positios fo the secod am T, fo each of the possible positios fo T T, thee ae fou possible positios fo the thid am T 3 ad so o. Thus, by the multiplicatio piciple of coutig the total possible positios fo T T T 3 T 4 T 5 is Sice each distict positio is a distict sigal, so total umbe of possible sigals is 04 icludig the sigal (meaigless) coespodig to the situatio whe all ams ae i est positio. Theefoe, the total umbe of sigals that ca be geeated is

20 0 COMBINATORICS AND GRAPH THEORY Poblem.55. Fid the umbe of positive iteges geate tha a millio that ca be fomed with the digits, 3, 0, 3, 4, ad 3. Solutio. The umbes geate tha a millio must be of 7 digits. I the give set of digits, appea twice, 3 appea thice ad all othes ae distict. 7! Thus, the total umbe of seve digit umbes that ca be fomed with give digits is!3! 40. The set of these 40 positive iteges, iclude some umbes which begi with 0. Clealy, these ubmes ae less tha a millio ad they must ot be couted i ou aswe. The umbe of such umbes is give by the pemutatios of 6 o-zeo digits ad is equal to 6!!3! 60. Theefoe, the umbe of positive iteges geate tha a millio that ca be fomed with give digits is equal to Poblem.56. How may distiguishable pemutatios of the lettes i the wod BANANA ae thee? Solutio. The wod BANANA has 6 lettes. All the lettes ae ot distict. Let us use subscipt to distiguish them tempoaily. Let the lettes be B, A, N, A, N, A 3. Thus, the umbe of pemutatios ae 6! 70. Some of the pemutatios ae idetical like A A A 3 BN N ad A A A 3 BN N ecept the ode i which the N s appea ! This meas that if we dop the subscipts, the total umbe of pemutatios will be 6!! 360. Similaly, if we dop subscipt with A s the total umbe of distiguishable pemutatios ae 60. Poblem.57. Thee ae 0 stalls fo aimals i a ehibitio. Thee aimals ; lio, pussycat ad hose ae to be ehibited. Aimals of each kid ae ot less tha 0 i umbe. What is the possible umbe of ways of aagig the ehibitio. Solutio. Thee ae thee types of aimals ad 0 stalls. Oe stall ca be filled by ay of the thee aimals. Oce the fist stall is filled, the secod stall ca be filled agai i thee ways by placig ay of the thee aimals i it. We have to fill 0 such stalls ad umbe of each aimal is geate tha equal to 0, so we have ways to fill the stalls. Thus, we ca aage the ehibitio i ways. Poblem.58. I how may ways ca 0 diffeet eamiatio papes be scheduled so that (i) the best ad the wost always come togethe (ii) the best ad the wost eve come togethe?

21 COUNTING PRINCIPLES AND GENERATING FUNCTIONS Solutio. object. (i) Let us coside the best ad the wost pape togethe ad coside them as oe We have, ow, 9 objects (papes). These 9 objects ca be aaged i 9! ways. Ad i each of these 9! aagemets, the best ad the wost papes ca be aaged i! ways. Theefoe, the umbe of ways i which the 0 papes ca be scheduled i this situatio! * 9! (ii) Without ay estictio, the 0 papes ca be scheduled i 0! ways. We have just calculated i pat (i) that total umbe of ways i which the 0 papes ca be scheduled so that the best ad the wost always come togethe Theefoe, the umbe of ways of schedulig 0 papes so that the best ad the wost eve come togethe 0! Poblem.59. I how may diffeet ways ca 5 me ad 5 wome sit aoud a table, if (i) thee is o estictio (ii) o two wome sit togethe? Solutio. The poblem is elated to cicula pemutatio of 0 objects (5 me ad 5 wome). If thee is o estictio the the umbe of pemutatios is (0 )! 9! Notice hee the diffeece i aagemet betwee clockwise ad aticlockwise. I the secod case, thee is a estictio that o two wome ae allowed to sit side by side. To meet this estictio each woma should occupy a sit betwee two me. The umbe of ways five me ca sit aoud a table is 4! 4. Oce these five me have sat o alteate chais, the five wome ca occupy the 5 empty chais i 5! ways. Thus, total umbe of ways, i this case, will be 4 * 5! 4 * Poblem.60. Fid the sum of all the fou-digit umbes that ca be fomed with the digits 3,, 3 ad 4. Solutio. Oe thig is woth oticig hee that a fou-digit umbe so fomed does ot coati a epeated digit ecept the digit 3. This is implied fom the questio, because if it wee ot so, 3 should ot have bee epeated i the list of the digits. To fid the sum of the fou-digit umbes fomed with 3,, 3 ad 4, we have to calculate the sum of digits at uit place i all such umbes. The sum of the digits at te, huded ad thousad place will be the same, oly theis place value will chage. The umbe of fou-digit umbes i which appeas at uit place is detemied by the umbe of pemutatios of digits 3, 3 ad 4 to fill te, huded ad thousad place. Ad this is 3!! 3. Similaly, the umbe of fou-digit umbes i which 3 appeas at uit place is 3! 6.

22 COMBINATORICS AND GRAPH THEORY The umbe of fou-digit umbes i which 4 appeas at uit place is 3!! 3. Theefoe, sum of the digits i the uit place of all the umbes As stated above, the sum of the digits i all such umbes at te, huded ad thousad places is 36 each. Thus, the sum of all such umbes Poblem.6. We ae asked to make slips fo all umbes up to five-digit. Sice the digits 0,, 6, 8 ad 9 ca be ead as 0,, 9, 8 ad 6 whe they ae ead upside dow, thee ae pais of umbes that ca shae same slip if the slips ae ead upside dow o ight sideup (e.g ad 9968). Fid the umbe of slips equied fo all five-digit umbes. Solutio. We have 0 digits. We have to make all five digit umbes. The total such umbes is equal to 0 5. Hee we have to make slips fo these may umbes. The umbes made of digits 0,, 6, 8 ad 9 ca be ead upside dow o ight side up. Ad, thee ae 5 5 may such five-digit umbes (all those five-digit umbes made of digits 0,, 6, 8 ad 9). Out of these 5 5 may umbes, howeve, thee ae some umbes that ead the same eithe upside dow o ight side up. Fo eample, 986, ad thee ae 3 5 such umbes (cete place filled with 0, o 8). Thus, thee ae umbes that ca be ead upside dow o ight side up. Ad, fo these umbes we eed oly (55 3 * 5 ) umbe of slips. Theefoe, umbe of slips equied to be made is 0 5 (55 3 * 5 ). Poblem.6. How may biay sequeces of -bits log have eve umbe of s? Solutio. Thee will be possible biay sequeces of -bits log. This ca be veified by the pemutatio of objects whe epetitios ae allowed. Thee ae places ad two objects (0 ad ). The fist place ca be filled i ways, fo each of these, the secod place ca be filled i ways, so we have ways to fill fist two places. Etedig the sequece upto the th steps, we have possible aagemets ad hece possible biay sequeces. We ca ow make pais of biay sequeces i such a way that two sequeces diffe oly at th place. Thee will such pais, ad i each pai oe sequece will have eve umbe of s. Thus, umbe of biay seqeuces of bits log havig eve umbe of s.

23 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 3 Poblem.63. Solutio. How may diffeet wods ca be made fom lettes of the wod committee. The wod committee cotais lettes c, o ad i oce ad m, t ad e twice. Whe a wod is fomed fom these lettes, a lette may appea at the most the umbe of time it appea i the wod committee o ot at all. So geeatig fuctio fo c, o ad i is give by ( + ) each, wheeas, fo c, m ad e is give by + +! each. Thus, geeatig fuctio fo the poblem is the give by ( + ) 3 + +! 3 ( ) + +! If wods ae to be fomed takig all the lettes at oce, the the umbes of such wods is give by the coefficiet of ad this is equal to 9! 9 9!!!!. Poblem.64. A fai si-sided die is tossed fou times ad the umbes show ae ecoded i a sequece. How may diffeet sequeces ae thee? Solutio. Let us assume, hee, that each toss is a object ad the umbe appeaig o the face of a die is the umbe of times it occus. Thus each object occus at least oce ad at the most 6 times. Also the ode of appeaig of diffeet s (coceptually) to make it o 3 o 4 o 5 o 6 is fied ad is oe ad oly oe way. The geeatig fuctio fo fist toss is give as The sum of all the coefficiets, hee, is 6 ad umbe of possible sequeces of umbes, oe face of a die i oe toss, is 6 oly, die is tossed fou times, Theefoe, geeatig fuctio fo the poblem is { } 4. The umbe of sequece is the sum of coefficiets of all the tems i this geeatig fuctio. This is equal to 6 4. Poblem.65. Fid the geeatig fuctio, also called eumeato, fo pemutatios of objects with the followig specified coditios : (a) each object occus at the most twice (b) each object occus at least twice (c) each object occus at least oce ad at the most k times. Solutio. (a) Each object occus at the most twice implies that a object may occu 0, o times. The epoetial geeatig fuctio fo a object ude this coditio is give as !

24 4 COMBINATORICS AND GRAPH THEORY Thee ae objects, so the geeatig fuctio fo the poblem is witte as + +!. (b) I this case, a objcet occus at least twice. This implies that a object may appea o moe times. The epoetial geeatig fuctio fo a object ude this coditio is ! 3! 4! Theefoe, fo the poblem dealig with objects, the geeatig fuctio ca be witte as ! 3! 4!. (c) Hee, each object occus at least oce ad at the most k times. That is to say that a object may occu o o 3 o... o k times. The epoetial geeatig fuctio fo a object ude this coditio is ! 3! 3 Sice the poblem fo objects, the geeatig fuctio fo the poblem is witte as k k! 3 k ! 3! k!. Poblem.66. How may ways ae thee to select a fist pize wie, a secod pize wie, ad a thid-pize wie fom 00 diffeet people who have eteed a cotest? Solutio. Because it mattes which peso wis which pize, the umbe of ways to pick the thee pize wies is the umbe of odeed selectios of thee elemets fom a set of 00 elemets, that is, the umbe of 3-pemutatios iof a set of 00 elemets. Cosequetly, the aswe is P(00, 3) ,00. Poblem.67. Suppose that a saleswoma has to visit eight diffeet cities. She must begi he tip i a specified city, but she ca visit the othe seve cities i ay ode she wishes. How may possible odes ca the saleswoma use whe visitig these cities? Solutio. The umbe of possible paths betwee the cities is the umbe of pemutatios of seve elemets, sice the fist city is detemied, but the emaiig seve ca be odeed abitaily. Cosequetly, thee ae 7! ways fo the saleswoma to choose he tou. If, fo istace, the saleswoma wishes to fid the path betwee the cities with miimum distace, ad she computes the total distace fo each possible path, she must coside a total of 5040 paths. 3 4

25 COUNTING PRINCIPLES AND GENERATING FUNCTIONS 5 Poblem.68. Suppose that thee ae eight ues i a ace. The wie eceives a gold medal the secod-place fiishe eceives a silve medal, ad the thid-place fiishe eceives a boze medal. How may diffeet ways ae thee to awad these medals, if all possible outcomes of the ace ca occu ad thee ae o ties? Solutio. The umbe of diffeet ways to awad the medals is the umbe of 3-pemutatios of a set with eight elemets. Hece, thee ae P(8, 3) possible ways to awad the medals. Poblem.69. How may pemutatios of the lettes ABCDEFGH cotai the stig ABC? Solutio. Because the lettes ABC must occu as a block, we ca fid the aswe by fidig the umbe of pemutatios of si objects, amely, the block ABC ad the idividual lettes D, E, F, G ad H. Because these si objects ca occus i ay ode, thee ae 6! 70 pemutatios of the lettes ABCDEFGH i which ABC occus as a block..3 COMBINATIONS Let A be a set with A, ad let. The the umbe of combiatios of the elemets of A, take at a time that is the umbe of -elemet subsets of A is!.!( )! The umbe of combiatios of A, take at a time, does ot deped o A, but oly o ad. This umbe is ofte witte C ad is called the umbe of combiatios of objects take at a time.! We have C.!( )! Suppose k selectios ae to be made fom items without egad to ode ad that epeats ae allowed, assumig at least k copies of each of the items. The umbe of ways these selectios ca be made is ( + k ) C k. Poblem.70. Show that C C. Solutio. We have!! C!( )! ( ( ))!( )! C. Poblem.7. Compute the umbe of distict five-cad hads tha ca be dealt fom a deck of 5 cads. Solutio. This umbe is 5 C 5 because the ode i which the cads wee dealt is ielevat. 5! 5C 5 o,598,960. 5!47! Poblem.7. I how may ways ca a pize wie choose thee CDs fom the top te list if epeats ae allowed? Solutio. Hee is 0 ad k is 3. Thee ae (0 + 3 )C 3 o C 3 ways to make the selectios. The pize wie ca make the selectio i 0 ways.