in-how-many-ways-can-you-distribute-20-different-books-to-5-students-such-that-each-student-gets-at-least-2-books-

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Question Number 179818 by mr W last updated on 02/Nov/22

$$inhowmanywayscanyoudistribute$$$$20differentbooksto5studentssuch$$$$thateachstudentgetsatleast2books?$$

Answered by a.lgnaoui last updated on 03/Nov/22

$$$$$$3444543445$$$$3445443454$$$$3454443544$$$$3544453444$$$$$$$$4434544435$$$$4435444534$$$$4534445434$$$$5434454434$$$$$$$$44453$$$$44543total=20ways$$$$45443$$$$54443$$$$$$

Commented by a.lgnaoui last updated on 03/Nov/22

$$inmyanswerIconsidered$$$$nodifferencebetwenbooks$$$$order12345\left(forsudents\right)$$$$3,4,5Numberofbooksfor$$$$eachstudent.$$

Commented by mr W last updated on 03/Nov/22

$$atleast2books\ne morethan2books$$$$besidesthebooksaredifferentand$$$$thestudentsarealsodifferent$$$$(logically!)$$

Answered by Acem last updated on 04/Nov/22

$$\mathit{P}\mathit{r}\mathit{i}\mathit{m}\mathit{i}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{d}\mathit{r}\mathit{a}\mathit{f}\mathit{t}$$$$$$$$A)distributing2dif.bks.toeachstu.$$$$A=C_2^{20}{C}_2^{18}{C}_2^{16}{C}_2^{14}{C}_2^{12}5!=2514159648000$$$$$$$$B)distributingtheother10dif.bks.to5stu.$$$$B=C_{10}^{10}\times 5+C_9^{10}\times 5.C_1^1\times 4+C_8^{10}\times 5C_2^2\times 4$$$$\dots .+C_2^{10}\times 5C_8^8\times 4+C_1^{10}\times 5C_9^9\times 4$$$$NumWays=A+B$$

Answered by mr W last updated on 03/Nov/22

$$thenumberofwaystodistribute20$$$$booksto5studentssuchthateach$$$$studentgetsatleast2booksisthe$$$$coef.of{x}^{20}intheexpansionof$$$$20!{(e^x-1-x)}^{5}whichis$$$$64341775001400.$$

Commented by mr W last updated on 03/Nov/22

Commented by mr W last updated on 03/Nov/22

$$isolvedinawaywhichseemstobea$$$$goodwaytome.i{can}^{\prime }tand{won}^{\prime }t$$$$answerwhyididitdifferentlyas$$$$you.$$$$mythoughtwas:instep1ileteach$$$$studentgetonebookandinstep2$$$$eachgetsatleastonebook.$$$$iconsideredagainandfoundoutthat$$$$icanalsosolvethequestionwithmy$$$$method\left(usinggeneratingfunction\right)$$$$evenwithoutstep1.soihavemodified$$$$mysolutioncorrespondingly.$$$$$$$$i{couldn}^{\prime }ttotallyfollowyoursolution,$$$$butithinkitisnotcorrect.youcan$$$$checkyoursolutionforexamplethat$$$$wedistribute5booksto2students$$$$andeachgetsatleast2books.the$$$$correctanswershouldbe20.butyour$$$$methodgetsalreadyinstep1$$$$A=C_2^5{C}_2^{3}2!=60.thatmeansthe$$$$methodiswrong.sopleaserecheck$$$$yourownsolution.$$

Commented by Acem last updated on 03/Nov/22

$$Hello,$$$$Forthe1ststep,whyonly5booksnot10$$$$wecanchooserandomly2booksandgive$$$$themtosomeoneasfollow$$$$$$$$A)distributing2dif.bks.toeachstu.$$$$A=C_2^{20}{C}_2^{18}{C}_2^{16}{C}_2^{14}{C}_2^{12}5!=2514159648000$$$$$$

Commented by Acem last updated on 03/Nov/22

$$Thankyoumyfriendforeverything,infactiam$$$$surethat{i}^{\prime }vesomethingmissingwiththe2step$$$$thatiswhy{didn}^{\prime }tcomptitresult,andwilltry$$$$tofixitlater,butnowamshockedaboutthe1st$$$$step,okiwillrecheckitwhenfinishmywork.$$$$Haveanicetime$$

Commented by Acem last updated on 03/Nov/22

$$$$$$Helloagain,$$$$Igotthemistake$$$$\text{Prime causes double exponent: use braces to clarify}$$$$andthetrueis:$$$$A=C_2^{20}{C}_2^{18}{C}_2^{16}{C}_2^{14}{\mathit{C}}_{12}^{12}5!=38093328000✓$$$$Now,dealingwith12,11,10,\dots booksiseasier$$$$thandealingwith15,14,\dots books$$$$$$$$Anywayigottheresult:62174055528000$$$$Iwillfeedbackmymethodverysoon$$$$$$$$Ifyoursistruethatmeanistillmissingsome$$$$casesofdistrubtion$$$$$$$$P.s:pleasemakethequestionwithlessnumber$$$$as9booksforexampleaslongastheideaof$$$$thesolutioniswellknown$$$$$$$$Thankyou$$$$$$

Commented by mr W last updated on 03/Nov/22

$$asihavesaid,youcancheckyour$$$$methodwiththemostsimplecase:$$$$5booksto2student,atleast2books$$$$foreachstudent.$$$$igottheanswer:20ways.thisis$$$$trueandcanbeeasilyproved.$$

Commented by Acem last updated on 03/Nov/22

$$Iknowthat,butasthenumberofitemsincreases$$$$thebranchesincrease,andhereifweneglect$$$$onecaseofdistrubtionwewouldloseitsbranches$$$$$$

Commented by mr W last updated on 03/Nov/22

$$thenyoucanputanexampleby$$$$yourselfwhichyouthinkisgood.$$

Answered by Acem last updated on 04/Nov/22

$$\mathit{T}\mathit{h}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{i}\mathit{n}\mathit{o}\mathit{n}\mathit{e}\mathit{s}\mathit{t}\mathit{a}\mathit{g}\mathit{e}$$$$NumWays=$$$$5![C_{12}^{20}{C}_2^8{C}_2^6{C}_2^4{C}_2^2+C_{11}^{20}{C}_3^9{C}_2^6{C}_2^4{C}_2^2+$$$$C_{10}^{20}(C_4^{10}{C}_2^6{C}_2^4{C}_2^2+C_3^{10}{C}_3^7{C}_2^4{C}_2^2+$$$$C_9^{20}(C_5^{11}{C}_2^6{C}_2^4{C}_2^2+C_4^{11}{C}_3^7{C}_2^4{C}_2^2+$$$$+C_3^{11}{C}_3^8{C}_3^5{C}_2^2)+$$$$C_8^{20}(C_6^{12}{C}_2^6{C}_2^4{C}_2^2+C_5^{12}{C}_3^7{C}_2^4{C}_2^2+$$$$C_4^{12}{C}_4^8{C}_2^4{C}_2^2+C_4^{12}{C}_3^8{C}_3^5{C}_2^2+$$$$C_3^{12}{C}_3^9{C}_3^6{C}_3^3)+$$$$C_7^{20}(C_7^{13}{C}_2^6{C}_2^4{C}_2^2+C_6^{13}{C}_3^7{C}_2^4{C}_2^2+$$$$C_5^{13}{C}_4^8{C}_2^4{C}_2^2+C_5^{13}{C}_3^8{C}_3^5{C}_2^2$$$$C_4^{13}{C}_4^9{C}_3^5{C}_2^2+C_4^{13}{C}_3^9{C}_3^6{C}_3^3)]$$$$$$$$NumWays=62174055528000$$$$$$$$Thingsareveryclearinfrontofyou$$$$andsofarihavenotfeltthat{i}^{\prime }veoverlooked$$$$oneofthecases$$$$$$

Commented by Acem last updated on 04/Nov/22

$$A)C_3^5{C}_2^{2}2!=20$$$$B)(C_4^6{C}_2^2+C_3^6{C}_3^3)2!=70not50$$$$$$$$C)Ourmainquestion“20booksto5stu{.}^{″}$$$$Canyounoteyourmethodtocompareour$$$$distributionmethodswithoutusing$$$$specialtables$$

Commented by mr W last updated on 04/Nov/22

$$ifyouarenotconvincedthatyour$$$$resultiswrong,whynotverifyyour$$$$solutionwithsomesimplecases$$$$whichcanbeeasilychecked$$$$“{manuallly}^{″}?forexample$$$$example1:5booksto2students,$$$$answershouldbe20.$$$$example2:6booksto2students,$$$$answershouldbe50.$$

Commented by mr W last updated on 04/Nov/22

$$A)5booksto2students:$$$$answeris20.thiscanbechecked:$$$$C_2^5\times 2=20$$$$icangetthesameresultviacoef.of$$$$x^{5}inexpansionof5!{(e^x-1-x)}^2.$$$$$$$$B)6booksto2students:$$$$answeris50.thiscanbechecked:$$$$C_2^6\times 2+C_3^6=15\times 2+20=50,not70!$$$$infactonecanenumerateallthese$$$$50caseseasily.$$$$icangetthesameresultviacoef.of$$$$x^{6}inexpansionof6!{(e^x-1-x)}^2.$$$$$$$$C)20booksto5students:$$$$theresultisthecoef.of{x}^{20}inthe$$$$expansionof20!{(e^x-1-x)}^5,which$$$$is64341775001400.$$$$icangetthisresult“{manually}^{″}inan$$$$otherwayasfollowing.$$$$todistribute20booksinto5groups$$$$suchthateachgroupgetsatleast$$$$2books,therearefollowing30$$$$possibilities:$$$$12+2+2+2+2\Rightarrow \frac{20!}{12!{(2!)}^{4}4!}\times 5!$$$$11+3+2+2+2\Rightarrow \frac{20!}{11!3!{(2!)}^{3}3!}\times 5!$$$$10+4+2+2+2\Rightarrow \frac{20!}{10!4!{(2!)}^{3}3!}\times 5!$$$$10+3+3+2+2\Rightarrow \frac{20!}{10!{(3!2!)}^{2}2!2!}\times 5!$$$$9+5+2+2+2\Rightarrow \frac{20!}{9!5!{(2!)}^{3}3!}\times 5!$$$$9+4+3+2+2\Rightarrow \frac{20!}{9!4!3!{(2!)}^{2}2!}\times 5!$$$$9+3+3+3+2\Rightarrow \frac{20!}{9!{(3!)}^{3}2!3!}\times 5!$$$$8+6+2+2+2\Rightarrow \frac{20!}{8!6!{(2!)}^{3}3!}\times 5!$$$$8+5+3+2+2\Rightarrow \frac{20!}{8!5!3!{(2!)}^{2}2!}\times 5!$$$$8+4+4+2+2\Rightarrow \frac{20!}{8!{(4!2!)}^{2}2!2!}\times 5!$$$$8+4+3+3+2\Rightarrow \frac{20!}{8!4!2!{(3!)}^{2}2!}\times 5!$$$$8+3+3+3+3\Rightarrow \frac{20!}{8!{(3!)}^{4}4!}\times 5!$$$$7+7+2+2+2\Rightarrow \frac{20!}{{(7!)}^2{(2!)}^{3}2!3!}\times 5!$$$$7+6+3+2+2\Rightarrow \frac{20!}{7!6!3!{(2!)}^{2}2!}\times 5!$$$$7+5+4+2+2\Rightarrow \frac{20!}{7!5!4!{(2!)}^{2}2!}\times 5!$$$$7+5+3+3+2\Rightarrow \frac{20!}{7!5!2!{(3!)}^{2}2!}\times 5!$$$$7+4+4+3+2\Rightarrow \frac{20!}{7!3!2!{(4!)}^{2}2!}\times 5!$$$$7+4+3+3+3\Rightarrow \frac{20!}{7!4!{(3!)}^{3}3!}\times 5!$$$$6+6+4+2+2\Rightarrow \frac{20!}{{(6!)}^{2}4!{(2!)}^{2}2!2!}\times 5!$$$$6+6+3+3+2\Rightarrow \frac{20!}{2!{(6!)}^2{(3!)}^{2}2!2!}\times 5!$$$$6+5+5+2+2\Rightarrow \frac{20!}{6!{(5!)}^2{(2!)}^{2}2!2!}\times 5!$$$$6+5+4+3+2\Rightarrow \frac{20!}{6!5!4!3!2!}\times 5!$$$$6+5+3+3+3\Rightarrow \frac{20!}{6!5!{(3!)}^{3}3!}\times 5!$$$$6+4+4+4+2\Rightarrow \frac{20!}{6!2!{(4!)}^{3}3!}\times 5!$$$$6+4+4+3+3\Rightarrow \frac{20!}{6!{(4!)}^2{(3!)}^{2}2!2!}\times 5!$$$$5+5+5+3+2\Rightarrow \frac{20!}{3!2!{(5!)}^{3}3!}\times 5!$$$$5+5+4+4+2\Rightarrow \frac{20!}{2!{(5!)}^2{(4!)}^{2}2!2!}\times 5!$$$$5+5+4+3+3\Rightarrow \frac{20!}{4!{(5!)}^2{(3!)}^{2}2!2!}\times 5!$$$$5+4+4+4+3\Rightarrow \frac{20!}{5!3!{(4!)}^{3}3!}\times 5!$$$$4+4+4+4+4\Rightarrow \frac{20!}{{(4!)}^{5}5!}\times 5!$$$$\mathrm{\Sigma }:64341775001400$$

Commented by mr W last updated on 04/Nov/22

Commented by Acem last updated on 05/Nov/22

$$Helloandgoodeveningmybestfriend$$$$Atfirstandforthe$$$$C_2^6\times 2+C_3^6=15\times 2+20=50,not70!$$$$yousay\dots ..or3booksforeachstudent{C}_3^6{C}_3^3$$$$youknowaswellthatwehave2waysfordistrubting$$$$causewehavedifferentbooksthatiswhy$$$$imultiply{C}_3^{6}by2togetfinally70ways$$$$waitingforyourreplySir,andsureiwill$$$$bebackfortheimportantissueaboutmainqusetion$$$$Thanksforyourefforts$$

Commented by mr W last updated on 05/Nov/22

$$asyouwrote{C}_3^{6}youhavealready$$$$countedallcases.example:saythe$$$$booksarea,b,c,d,e,f.whenweselect$$$$booksa,b,cforthefirstperson,the$$$$secondpersongetsautomactically$$$$thebooksd,e,f.whenthefirstperson$$$$getsd,e,f,thesecondpersonsgets$$$$automaticallya,b,c.thatmeansboth$$$$casesarecompletelyincludedin{C}_3^6.$$$$thisisbecausebothpersonsget$$$$equalnumberofbooks.{it}^{\prime }sdifferent$$$$whenapersongets2booksandthe$$$$otherperson4books,thenwemust$$$$write2\times C_2^6.thinkaboutit,shouldbe$$$$clear.$$$$example4booksfor2persons:$$$$a+bcdorbcd+a✓$$$$b+acdoracd+b✓$$$$c+abdorabd+c✓$$$$d+abcorabc+d✓$$$$ab+cdorcd+ab(?!)$$$$ac+bdorbd+ac(?!)$$$$ad+bcorbc+ad(?!)$$$$bc+adorad+bc(?!)$$$$bd+acorac+bd(?!)$$$$cd+aborab+cd(?!)$$$$youcanseewecannotmultiplyby2$$$$forthecaseswhenbothpeoplegetthe$$$$samenumberofbooks.$$

Commented by Acem last updated on 05/Nov/22

$$Yesiagreewithyou,andwiththisinmind,{let}^{\prime }s$$$$askthefollowingessentialquestion:$$$$Howdoesthewholeprocesswork?$$$$$$$$A)Mr.wpleasecometoget3books,$$$$“Booksselectionprocess{C}_3^6{}^{″}heretheyare.$$$$therestbooksforAcem\dots \dots {it}^{\prime }sok!$$$$$$$$B)“Booksselectionprocess{C}_3^6{}^{″}$$$$\{Math,Fluidmechanics,Steamgenerators\}$$$$Nowmeandyoulikethesesciencefields,therest$$$$booksareaboutotherthings.$$$$andhere,iseethat{there}^{\prime }sstilladifferent,$$$${}^{\prime }infactiwantthesebooksformenot{you}^{\prime }$$$$so,{don}^{\prime }twehavedifferentwaystodistribute?$$$$withdiferentbooksandpeople$$

Commented by mr W last updated on 05/Nov/22

$$i{can}^{\prime }tfollowyousir.$$$$thereare50waystodistribute6$$$$differentbooksto2studentssuch$$$$thateachgetsatleast2books.$$$$doyouagreeornot?$$$$ifyouagree,{that}^{\prime }sfine,wecanclose$$$$thecase.$$$$ifyou{don}^{\prime }tagree,youarewrong,and$$$$ialsoclosethecase,becausei{can}^{\prime }t$$$$contributeanymore.youonlyneedto$$$$enumerateallpossibilitiesandthen$$$${you}^{\prime }llseethatthereareindeedonly$$$$50differentways.$$

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