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Question Number 84993 by mr W last updated on 18/Mar/20

$$100applesshouldbepackedinthree$$$$boxesandeachboxshouldcontain$$$$atleast10apples.inhowmanyways$$$$canthisbedone?$$

Commented by john santu last updated on 18/Mar/20

$$\mathrm{let}\mathrm{the}\mathrm{box}\mathrm{we}\mathrm{call}\mathrm{it}\mathrm{ABC}$$$$[(10,10,80)(10,11,79),(10,12,78),$$$$...(10,80,10)]\times 2$$

Commented by john santu last updated on 18/Mar/20

$$\Rightarrow 71\times 2=142\mathrm{ways}$$

Commented by mr W last updated on 18/Mar/20

$$youtreattheboxesasdifferent?$$$${what}^{\prime }sthemeaningof\times 2?$$

Commented by john santu last updated on 18/Mar/20

$$\mathrm{yes}.\mathrm{i}\mathrm{assume}\mathrm{the}\mathrm{boxes}\mathrm{are}\mathrm{different}.$$$$\times 2\mathrm{is}\mathrm{mean}\mathrm{has}2\mathrm{case}$$$$\mathrm{case}2$$$$(11,10,79)(12,10,78)...(80,10,10)$$

Commented by mr W last updated on 18/Mar/20

$$asexample(11,11,78)isnotincluded$$$$inyourconsideration.$$

Commented by mr W last updated on 19/Mar/20

$${it}^{\prime }swrongsir.$$$$you{didn}^{\prime }ttakeintoaccountthateach$$$$boxshouldhaveatleast10apples.$$$$theformulayouappliedisforthecase$$$$thataboxmaygetzeroapple.besides$$$$inourquestiontherearethreeboxes,$$$$not4asyoutook.$$

Commented by Rio Michael last updated on 18/Mar/20

$$\mathrm{okay}\mathrm{i}\mathrm{just}\mathrm{read}\mathrm{on}\mathrm{this}.\mathrm{thanks}\mathrm{mr}\mathrm{Wfor}\mathrm{this}\mathrm{question}$$$$\mathrm{what}\mathrm{i}\mathrm{got}\mathrm{is}{}^{\prime }\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{dividing}n\mathrm{identical}$$$$\mathrm{objects}\mathrm{into}r\mathrm{groups}\mathrm{is}{}^{n+r-1}{C}_{r-1}\mathrm{ways}.$$$$\mathrm{lets}\mathrm{apply}\mathrm{it}\mathrm{to}\mathrm{the}\mathrm{question}$$$$\mathrm{i}\mathrm{get}\mathrm{number}\mathrm{of}\mathrm{ways}={}^{100+4-1}{\mathrm{C}}_{4-1}\mathrm{such}\mathrm{that}\mathrm{each}\mathrm{gets}\mathrm{at}\mathrm{least}10$$$$={}^{103}{\mathrm{C}}_3=357760\mathrm{ways}$$

Answered by Rio Michael last updated on 18/Mar/20

$$\mathrm{this}\mathrm{is}\mathrm{my}\mathrm{approach}\mathrm{sir}$$$$\ast \mathrm{the}\mathrm{first}\mathrm{bag}\mathrm{can}\mathrm{take}10\mathrm{apples}\mathrm{in}{}^{100}{\mathrm{C}}_{10}\mathrm{ways}.$$$$\ast \mathrm{the}\mathrm{next}\mathrm{bag}\mathrm{can}\mathrm{take}10\mathrm{apples}\mathrm{in}{}^{90}{\mathrm{C}}_{10}\mathrm{ways}.$$$$\ast \mathrm{the}\mathrm{third}\mathrm{can}\mathrm{take}10\mathrm{apples}\mathrm{in}{}^{80}{\mathrm{C}}_{10}\mathrm{ways}$$$$\Rightarrow \mathrm{total}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{for}90\mathrm{apples}={}^{100}{\mathrm{C}}_{10}\times {}^{90}{C}_{10}\times {}^{80}{C}_{10}\times 2\mathrm{as}\mathrm{the}\mathrm{number}$$$$\mathrm{of}\mathrm{apples}\mathrm{drop}\mathrm{down}\mathrm{to}90.$$$$\mathrm{now}\mathrm{we}\mathrm{can}\mathrm{share}\mathrm{the}\mathrm{remainder}\mathrm{in}{}^{10}{\mathrm{C}}_3\mathrm{ways}$$$$\Rightarrow \mathrm{total}\mathrm{number}\mathrm{of}\mathrm{ways}=$$$${}^{100}{\mathrm{C}}_{10}\times {}^{90}{C}_{10}\times {}^{80}{C}_{10}\times 2\times {}^{10}{C}_3$$$$$$$$$$

Commented by mr W last updated on 18/Mar/20

$$theapplesareidentical.$$

Commented by Rio Michael last updated on 18/Mar/20

$$\mathrm{ofcourse}\mathrm{sir}\mathrm{we}\mathrm{just}\mathrm{select}\mathrm{at}\mathrm{random}$$

Commented by mr W last updated on 18/Mar/20

$$butyourworkingsayssomething$$$$different:$$$$\ast \mathrm{the}\mathrm{first}\mathrm{bag}\mathrm{can}\mathrm{take}10\mathrm{apples}\mathrm{in}{}^{100}{\mathrm{C}}_{10}\mathrm{ways}.$$$$\Rightarrow thismeanstheapplesaredistinct.$$$$sinceiftheapplesareidenticaland$$$$thefirstbagtakes10apples,then$$$$thereisonlyonepossibility,namely$$$$itobtains10apples.youcannotgive$$$$ihm10applesindifferentways.$$

Commented by Rio Michael last updated on 18/Mar/20

$$$$$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}\mathrm{sir}$$

Answered by Joel578 last updated on 18/Mar/20

$$\mathrm{This}\mathrm{problem}\mathrm{is}\mathrm{same}\mathrm{with}\mathrm{how}\mathrm{many}\mathrm{integer}$$$$\mathrm{solutions}\mathrm{of}$$$$x_1+x_2+x_3=100,10⩽x_i⩽80,i=1,2,3$$$$\mathrm{and}\mathrm{the}\mathrm{problem}\mathrm{above}\mathrm{is}\mathrm{same}\mathrm{with}\mathrm{how}$$$$\mathrm{many}\mathrm{integer}\mathrm{solutions}\mathrm{of}$$$$y_1+y_2+y_3=70,0⩽y_i⩽70$$$$\mathrm{It}\mathrm{has}\left(\begin{array}{c}70+3-1\\ 3-1\end{array}\right)=\left(\begin{array}{c}72\\ 2\end{array}\right)=\frac{72\left(71\right)}{2}=2556\mathrm{ways}$$

Commented by jagoll last updated on 18/Mar/20

$$\mathrm{how}\mathrm{to}\mathrm{get}\left(\begin{array}{c}70+3-1\\ 3-1\end{array}\right)?$$

Commented by mr W last updated on 18/Mar/20

$${that}^{\prime }scorrectsir!$$$$whatiftheboxesareidentical?that$$$$meansweonlydividetheapplesinto$$$$threeparts,eachpartwithatleast$$$$10apples.$$

Commented by jagoll last updated on 18/Mar/20

$$\mathrm{if}50\mathrm{candies}\mathrm{are}\mathrm{divided}\mathrm{to}4\mathrm{children}$$$$\mathrm{with}\mathrm{each}\mathrm{child}\mathrm{getting}\mathrm{a}\mathrm{minimum}$$$$\mathrm{of}10\mathrm{candies}$$

Commented by jagoll last updated on 18/Mar/20

$$\left(\begin{array}{c}50+4-1\\ 4-1\end{array}\right)=\left(\underset{3}{\stackrel{53}{}}\right)=\frac{53.52.51}{3.2.1}?$$

Commented by mr W last updated on 18/Mar/20

$$no.eachkidgets9atfirst.thereare$$$$14remaining.tosharethese14pieces$$$$thereare\left({}_{4-1}^{14-1}\right)=\left({}_3^{13}\right)ways.$$$$seestarsandbarsmethod.$$

Commented by jagoll last updated on 18/Mar/20

$$\mathrm{why}\mathrm{not}\left(\underset{4-1}{\stackrel{10-1}{}}\right)\mathrm{sir}?\mathrm{each}\mathrm{kid}\mathrm{get}\mathrm{minimum}$$$$10\mathrm{candies}.\mathrm{there}\mathrm{are}10\mathrm{remaining}$$$$\mathrm{to}\mathrm{share}\mathrm{these}10\mathrm{pieces}$$

Commented by Joel578 last updated on 18/Mar/20

$$\mathrm{To}\mathrm{sir}\mathrm{mr}\mathrm{W}$$$$\mathrm{Sir},\mathrm{please}\mathrm{elaborate}\mathrm{your}\mathrm{question}\mathrm{again}.$$$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{any}\mathrm{differences}\mathrm{with}\mathrm{the}\mathrm{earlier}\mathrm{question}.$$$$\mathrm{Maybe}\mathrm{some}\mathrm{example}\mathrm{will}\mathrm{help}$$

Commented by Joel578 last updated on 18/Mar/20

$$\mathrm{To}\mathrm{sir}\mathrm{Jagoll}$$$$\mathrm{Here}\mathrm{I}\mathrm{send}\mathrm{a}\mathrm{clear}\mathrm{explanation}\mathrm{of}\mathrm{the}\mathrm{formula}$$$$\mathrm{I}\mathrm{used}.\mathrm{Hope}\mathrm{this}\mathrm{will}\mathrm{help}.$$

Commented by mr W last updated on 18/Mar/20

$$iftheysharethe10remaining,it$$$$meansoneortwokidmaygetzero$$$$candy,becauseeachofthehasalready$$$$10candiesbefore.toshare10candies$$$$among4kids,zeroisallowed,there$$$$are\left({}_{4-1}^{10+4-1}\right)=\left({}_3^{13}\right)ways.$$$$$$$$formula:tosharenobjectsamong$$$$rpersons,eachwith⩾1object,there$$$$are\left({}_{r-1}^{n-1}\right)ways.$$$$tosharenobjectsamong$$$$rpersons,eachwith⩾0object,there$$$$are\left({}_{r-1}^{n+r-1}\right)ways.$$

Commented by john santu last updated on 18/Mar/20

$$\mathrm{does}\mathrm{means}\mathrm{that}\mathrm{all}\mathrm{three}\mathrm{boxes}$$$$\mathrm{considered}\mathrm{identical}$$

Commented by Joel578 last updated on 18/Mar/20

Commented by Joel578 last updated on 18/Mar/20

Commented by mr W last updated on 18/Mar/20

$$tojoelsir:$$$$when100applesaredividedinto3$$$$parts,itcanbeforexample$$$$20,30,50$$$$thisisonlyonepossibility.butwhen$$$$theseapplesareputinto3distinctboxes,$$$$thereare6differentways,e.g.$$$$A=20,B=30,C=50$$$$A=20,B=50,C=30$$$$A=30,B=20,C=50$$$$A=30,B=50,C=20$$$$A=50,B=20,C=30$$$$A=50,B=30,C=20$$$$thisiswhatwehadtillnow.$$$$whatiasknowishowmanyways$$$$aretheretodivide100applesinto3$$$$parts$$

Commented by jagoll last updated on 19/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by jagoll last updated on 19/Mar/20

$$\mathrm{thank}\mathrm{you}$$

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