月日旅游與酒店管理學院酒店Q

DiscreteMathematics

Chapter5Counting大葉大學資訊工程系黃鈴玲Ch5-2Acountingproblem:(Example15)Eachuseronacomputersystemhasapassword,whichissixtoeightcharacterslong,whereeachcharactersisanuppercaseletteroradigit.Eachpasswordmustcontainatleastonedigit.Howmanypossiblepasswordsarethere?Thissectionintroducesavarietyofothercountingproblemsthebasictechniquesofcounting.§5.1TheBasicsofcountingCh5-3BasiccountingprinciplesThesumrule:Ifafirsttaskcanbedoneinn1waysandasecondtaskinn2ways,andifthesetaskscannotbedoneatthesametime.thentherearen1+n2

waystodoeithertask.Example11Supposethateitheramemberoffacultyorastudentischosenasarepresentativetoauniversitycommittee.Howmanydifferentchoicesarethereforthisrepresentativeifthereare37membersofthefacultyand83students?n1n2n1+n2waysCh5-4Example12Astudentcanchooseacomputerprojectfromoneofthreelists.Thethreelistscontain23,15and19possibleprojectsrespectively.Howmanypossibleprojectsaretheretochoosefrom?

Sol:

23+15+19=57projects.Theproductrule:Supposethataprocedurecanbebrokendownintotwotasks.Iftherearen1waystodothefirsttaskandn2waystodothesecondtaskafterthefirsttaskhasbeendone,thentherearen1n2waystodotheprocedure.n1n2n1×n2waysCh5-5Example2Thechairofanauditorium(大禮堂)is

tobelabeledwithaletterandapositiveinteger

notexceeding100.Whatisthelargestnumberof

chairsthatcanbelabeleddifferently?Sol:

26×100=2600waystolabelchairs.

letter

Example4Howmanydifferentbitstringsarethereoflengthseven?Sol:

1

2

3

4

5

6

7□□□□□□□↑↑↑↑↑↑↑

0,10,10,1......0,1→27

種Ch5-6Example5Howmanydifferentlicenseplates(車牌)areavailableifeachplatecontainsasequenceof3lettersfollowedby3digits?Sol:□□□□□□→263．103letterdigitExample6Howmanyfunctionsaretherefromasetwithmelementstoonewithnelements?Sol:

f(a1)=?

可以是b1～bn,共n種

f(a2)=?

可以是b1～bn,共n種：

f(am)=?

可以是b1～bn,共n種∴nma1a2．．．a(chǎn)mb1b2．．．bnfCh5-7Example7Howmanyone-to-onefunctionsaretherefromasetwithm

elementstoonewithnelement?

(mn)Sol:

f(a1)=?

可以是b1～bn,共n種

f(a2)=?

可以是b1～bn,但不能=

f(a1),共n-1種

f(a3)=?

可以是b1～bn,但不能=

f(a1),也不能=f(a2),

共n-2種

：：

f(am)=?

不可=f(a1),f(a2),...,f(am-1),故共n-(m-1)種∴共n．(n-1)．(n-2)．...．(n-m+1)種1-1function#Ch5-8Example15Eachuseronacomputersystemhasapasswordwhichis6to8characterslong,whereeachcharacterisanuppercaseletteroradigit.Eachpasswordmustcontainatleastonedigit.

Howmanypossiblepasswordsarethere?Sol:

Pi:

#ofpossiblepasswordsoflengthi,i=6,7,8

P6=366-266

P7=367-267

P8=368-268

∴P6

+P7

+P8

=366+367+368-

266

-267-268種Ch5-9Example14InaversionofBasic,thenameofavariableisastringofoneortwoalphanumericcharacters,whereuppercaseandlowercaselettersarenotdistinguished.Moreover,avariablenamemustbeginwithaletterandmustbedifferentfromthefivestringsoftwocharactersthatarereservedforprogramminguse.HowmanydifferentvariablenamesarethereinthisversionofBasic?Sol:

LetVi

bethenumberofvariablenamesoflengthi.

V1=26

V2=26．36–5 ∴26+26．36–5differentnames.Ch5-10※TheInclusion-ExclusionPrinciple(排容原理)ABExample17

Howmanybitstringsoflengtheighteitherstartwitha1bitorendwiththetwobits00?Sol:

１２３４５６７８□□□□□□□□↑↑......①1

0,10,1→共27種②............00→共26種③1...........00→共25種27+26-25種Ch5-1101bit1※TreeDiagrams

Example18Howmanybitstringsoflengthfourdo

nothavetwoconsecutive1s?

Sol:

Exercise:11,17,23,27,38,39,47,5300000111001(0000)(0001)(0010)(0100)(0101)(1000)(1001)(1010)∴8bitstrings00110bit3Ch5-12Ex38.Howmanysubsetsofasetwith100elementshavemorethanoneelement?Sol:

Ex39.Apalindrome(迴文)isastringwhosereversalisidenticaltothestring.Howmanybitstringsoflengthnarepalindromes?

(abcdcba是迴文,abcd不是)Sol:Ifa1a2

...anisapalindrome,then

a1=an,a2=an-1,

a3=an-2,…Thm.4of§4.3

放不放

放不放

放不放空集合及

只有1個元素的集合Ch5-13§5.2ThePigeonholePrinciple(鴿籠原理)Theorem1(ThePigeonholePrinciple)If

k+1ormoreobjectsareplacedintokboxes,thenthereisatleastoneboxcontainingtwoormoreoftheobjects.ProofSupposethatnoneofthe

kboxescontainsmorethan

oneobject.Thenthetotalnumberofobjectswouldbeatmostk.Thisisacontradiction.Example1.Amongany367people,theremustbeatleasttwowiththesamebirthday,becausethereareonly366possiblebirthdays.Ch5-14Example2Inanygroupof27Englishwords,theremustbeatleasttwothatbeginwiththesameletter.Example3Howmanystudentsmustbeinaclasstoguaranteethatatleasttwostudentsreceivethesamescoreonthefinalexam?(0~100points)Sol:

102.(101+1)Theorem2.(Thegeneralizedpigeonholeprinciple)IfN

objectsareplacedintokboxes,thenthereisatleastoneboxcontainingatleastobjects.e.g.21objects,10boxestheremustbeonebox

containingatleastobjects.Ch5-15Example5Among100peoplethereareatleastwhowereborninthesamemonth.(100objects,12

boxes)Ch5-16Example10Duringamonthwith30daysabaseballteamplaysatleast1gameaday,butnomorethan45games.Showthattheremustbeaperiodofsomenumberofconsecutivedaysduringwhichtheteammustplayexactly14games.存在一段時間的game數(shù)和=14(跳過)Ch5-17Sol:Letaj

bethenumberofgamesplayedonorbeforethejthdayofthemonth.(第1天～第j天的比賽數(shù)和)ThenisanincreasingsequenceofdistinctintegerswithMoreover,isalsoanincreasingsequenceofdistinctintegerswithThereare60positiveintegersbetween1and59.Hence,suchthat(跳過)Ch5-18Def.

Supposethatisasequenceofnumbers.Asubsequenceofthissequenceisasequenceof

theformwhere

e.g.sequence:8,11,9,1,4,6,12,10,5,7subsequence:8,9,12()9,11,4,6()Def.Asequenceiscalledincreasing

(遞增)ifAsequenceiscalled

decreasing

(遞減)ifAsequenceiscalledstrictlyincreasing

(嚴格遞增)

ifAsequenceiscalled

strictlydecreasing

(嚴格遞減)

ifCh5-19Theorem3.Everysequenceofn2+1

distinctrealnumberscontainsasubsequenceoflengthn+1thatiseitherstrictlyincreasingorstrictlydecreasing.Example12.Thesequence8,11,9,1,4,6,12,10,5,7contains10=32+1terms(i.e.,n=3).

Thereisastrictlyincreasingsubsequenceoflengthfour,namely,1,4,5,7.Thereisalsoadecreasingsubsequenceoflength4,namely,11,9,6,5.Exercise21Constructasequenceof16positiveintegersthathasnoincreasingordecreasingsubsequenceof5terms.Sol:Exercise:5,13,15,31(跳過)Ch5洪-20§5.3鏡P奇ermu標tati妹ons(排列)an孤dCo云mbin患atio惕ns(組合)Def覽.Aperm帳utat障ionofa繁set騾of強dist榮inct檢obj摘ects數(shù)is德ano蓄rder恰eda挺rran旬geme臺nto異fth共ese鴿obje瓜cts.廢An遺orde途red野arra柏ngem燥ent坐ofrelem投ents員of宴ase帳tis淋cal避led知anr-per悼muta傷tion.Exam鳥ple侍2.LetS={1,賴2,3掙}.The裕ar襯ran泰gem震ent3,1,尚2is吊ap蛾erm鋸uta脅tio敏no緒fS.燃The認ar混ran辛gem兇ent3,2isa養(yǎng)2-p叫ermu丘tati純ono醋fS.The親ore銷m1喪.The谷nu催mbe仙ro捎fr-pe墻rmu花tat傅ion好of孝a斧set經(jīng)wi完thndist陷inct冷ele峽ment襲sis位置：1狠2羊3亡…貍r□□匯□…□放法：…Ch5-21Exam信ple鉗4.How戲many銳dif盼fere翻ntw染ays譯are閘ther知eto易sel決ectafi脹rst-苦priz堡ewi棕nner司(第一名),送as朵eco確nd-傻pri鏟ze隔win久ner奶,a詠nd芒a艙th溝ird約-pr偏ize竿wi槽nne惕rf弊rom伍10汗0d扇iff牌ere思nt皮peo壘ple上wh閘oh役ave龜ent膜ere穴da夫co抹nte嗚st刪?Sol逝:Exam堅ple弄6.Supp故ose蘭that建as研ales殿woma仗nha泉sto挺vis辨it8di撥ffer柿ent鐵citi養(yǎng)es.健She郊must揚beg淘inh政ert鍬rip痛ina疫spe果cifi董ed廈city愛,bu扶tsh兄eca掌nvi義sit趙the充othe甩rci笨ties株in傳any鍛orde司rsh屑e啊wish巴es.紀How務many勢pos貍sibl湖eor桶ders丈can啦the始sal恢eswo情man垃use今whe壞nvi鄭siti徹ngt寬hese賣cit答ies欺?Sol衰:Ch5-22Def.Anr-co瓣mbi忍nat泳ionofe朱leme夫nts繳ofa掠set才is北anuno代rde喂red違se臺lec扎tio圓no絡frelem咽ents秒fro戰(zhàn)mth漢ese巴t.Exa之mpl把e9LetSbe順the只se鴨t{1,肢2,3罵,4}.鉗The贈n{1,旬3,延4}is狂a3-co千mbi圓nat趴ion蠅fr擁omS.The幻玉ore驢m2The照nu勝mbe襪ro考fr-co授mbi卡nat最ion俘so漫fa味se魔t崗wit太hnele貝men裹ts,書wh古erenis煎ap窄osi托tiv給ei片nte修ger歸an感drisan悄int糞ege破rw席ith特,攤eq萌ual貫spf:稱為binomialcoefficientCh5-23Exam徐ple住10.We仰see壯th仗atC(4,2馳)=6,sin醒ce宏the2-co伶mbin侄atio傲nso旅f{a,b,c,d}are且the斗six局subs臣ets{a,b},全{a,c},聾{a,d},{b,c},{b,d}and{c,d}Coro吼llar呈y2.Letnandrbe歲non貢neg筑ati搶ve布int夾ege概rs邊wit繩hrn.The川nC(n,r)=C(n,n-r)pf:FromThm霞2.組合意仰義：選r個拿走,相當於競是選n-r個留下.Ch5體-24Exam廈ple愉12.How炎many筑way蜘sar芒eth妨ere必tos昨elec幣t5圍play臘ers隱from裹a1窩0-me巖mber羽ten居nis曠team諷to城make五at遺rip繼toa憑mat額cha君tan奔othe劣rsc仰hool哥?Sol:C(10販,5)紅=25捐2Exam擠ple輩15.Supp骨ose疏ther克ear擱e9惜facu復lty辱memb稍ers右int港hem立ath龍depa烤rtme賤nta襯nd1招1in挪the比com緒pute桃rsc確ienc淚ede卡part晴ment匪.Ho給wma晃nyw外ays走are蘋ther拿eto鈴sel茅ect嚼a評co氣mmit謊tee宮ift佳hec狐ommi謹ttee準is篇toc葛onsi畝sto薯f3謠facu營lty示memb湊ers執(zhí)from替the蒙mat逼hde疤part挪ment斥and據(jù)4f炊rom咐the配comp醋uter隙sci啟ence賀dep協(xié)artm嚴ent?Sol:Exer錦cise私:3,覽11,再13懇,2作1,內(nèi)33,乘34慕.Ch5-25§5.烏4型B己ino環(huán)mia屠lC性oef形fic禮ien萄ts泛(二項式係趨數(shù))Exam荷ple箏1.要產(chǎn)生xy2項時，晝需從三個史括號中選位兩個括號懶提供y，剩下一醒個則提供x(注意：同研一個括號筋中的x跟y不可能相勸乘)∴共有徐種笨不同來占源的xy2∴xy2的係數(shù)=Theo椒rem螺1.(Th下eB固ino效mia荒lT君heo錯rem醬,二項式繡定理)Letx,ybev毛aria漸bles劇,an尤dle撓tnbea攜pos刃itiv認ein投tege激r,t縮慧henCh5及-26Exam凱ple橡4.What折is邪the氏coef沃fici谷ent季ofx12y13inthe互expa反nsio滑nof后?Sol泥:∴Cor竿1.Letnbea騙pos移itiv拔ein慰tege揀r.T這henpf:By駕Thm踩1,目l魄etx=y=1Cor堂2.Letnbea板pos窯itiv鐵ein的tege扭r.貿(mào)Thenpf:by綿Thm斃1.(1-1)n=0Ch5-27Theo達rem糧2.(Pa炒sca支l’s膚id姑ent視ity索)Letnandkbe記pos啞iti派ve纖int