Five-balls-are-to-be-placed-in-three-boxes-Each-can-hold-all-the-five-balls-In-how-many-different-ways-can-we-place-the-balls-so-that-no-box-remains-empty-if-balls-are-different-but-boxes-are-ident

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Question Number 21802 by Tinkutara last updated on 04/Oct/17

$$\mathrm{Five}\mathrm{balls}\mathrm{are}\mathrm{to}\mathrm{be}\mathrm{placed}\mathrm{in}\mathrm{three}$$$$\mathrm{boxes}.\mathrm{Each}\mathrm{can}\mathrm{hold}\mathrm{all}\mathrm{the}\mathrm{five}\mathrm{balls}.$$$$\mathrm{In}\mathrm{how}\mathrm{many}\mathrm{different}\mathrm{ways}\mathrm{can}\mathrm{we}$$$$\mathrm{place}\mathrm{the}\mathrm{balls}\mathrm{so}\mathrm{that}\mathrm{no}\mathrm{box}\mathrm{remains}$$$$\mathrm{empty},\mathrm{if}\mathrm{balls}\mathrm{are}\mathrm{different}\mathrm{but}\mathrm{boxes}$$$$\mathrm{are}\mathrm{identical}?$$

Commented by mrW1 last updated on 06/Oct/17

$$25\mathrm{ways}$$

Answered by ajfour last updated on 06/Oct/17

$$coeff.of{x}^{5}in\frac{5!}{3!}{[x+\frac{x^2}{2!}+\frac{x^3}{3!}]}^3$$$$=coeff.of{x}^{2}in20{(1+\frac{x}{2}+\frac{x^2}{6})}^3$$$$=20(\frac{3}{6}+\frac{3}{4})=20\times \frac{5}{4}=25.$$

Commented by Tinkutara last updated on 06/Oct/17

$$\mathrm{But}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{why}\mathrm{to}\mathrm{find}$$$$\mathrm{such}\mathrm{coefficients}\mathrm{in}\mathrm{these}\mathrm{expansions}?$$$$\mathrm{I}\mathrm{know}\mathrm{this}\mathrm{formula}\mathrm{but}{\mathrm{don}}^{\prime }\mathrm{t}$$$$\mathrm{understand}\mathrm{the}\mathrm{reason}.$$

Commented by ajfour last updated on 09/Oct/17

$$if{x}_1+x_2+x_3=9$$$$placing9ballsinthreeboxessuch$$$$thattheycanremainempty,then$$$$itisequivalenttochoosing{x}^{9}from$$$$theexpansionof$$$${(1+x+x^2+\dots )}^3$$$$=(1+x+x^2+\dots )(1+x+x^2+\dots )(1+x+x^2+..)$$$$becausepowersofxfromeach$$$$bracketshalladdandsimilarly$$$$no.ofballsineachboxshalladd.$$$$thesumhastobe9,sowechoose$$$$no.ofwaysofobtaining{x}^9=$$$$coefficientof{x}^9.$$$$=coeff.of{x}^{9}intheexpansion$$$$of{(1+x+x^2+\dots )}^3$$$$orin{(1-x)}^{-3}={}^{9+3-1}{C}_{3-1}{=}^{11}{C}_2.$$

Commented by mrW1 last updated on 06/Oct/17

$$\mathrm{nice}\mathrm{explained}!$$$$$$$$\mathrm{as}\mathrm{result}\mathrm{one}\mathrm{can}\mathrm{use}\mathrm{the}\mathrm{notation}\mathrm{of}$$$$\mathrm{stirling}\mathrm{number}\left\{{}_3^5\right\}\mathrm{which}\mathrm{means}$$$$\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{to}\mathrm{put}5\mathrm{different}$$$$\mathrm{objects}\mathrm{into}3\mathrm{identical}\mathrm{boxes}.$$$$\left\{{}_3^5\right\}=25$$

Commented by Tinkutara last updated on 06/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by mrW1 last updated on 09/Oct/17

$$\mathrm{to}\mathrm{mr}.\mathrm{ajfour}:$$$$\mathrm{you}\mathrm{said}:$$$$=coeff.of{x}^{9}intheexpansion$$$$of{(1+x+x^2+\dots )}^3$$$$orin{(1-x)}^{-3}={}^{9-3+1}{C}_3{=}^7{C}_3.$$$$$$$$\mathrm{should}\mathrm{the}\mathrm{last}\mathrm{line}\mathrm{not}\mathrm{be}:$$$$orin{(1-x)}^{-3}={}^{9+3-1}{C}_9{=}^{11}{C}_9.$$$$?$$

Commented by ajfour last updated on 09/Oct/17

$$yesitisinfactasyousay..$$$$idintremembercorrectly,n$$$$thoughtidid.$$

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