10-couples-are-invited-to-a-dinner-and-should-be-seated-at-a-round-table-in-how-many-ways-can-the-host-do-this-1-generally-2-if-the-husband-and-the-wife-of-each-couple-should-sit-together-3

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Question Number 218310 by mr W last updated on 06/Apr/25

$$10couplesareinvitedtoadinner$$$$andshouldbeseatedataroundtable.$$$$inhowmanywayscanthehostdo$$$$this,$$$$1)generally.$$$$2)ifthehusbandandthewifeof$$$$eachcoupleshouldsittogether.$$$$3)ifnotwomenshouldsitnext$$$$toeachother.$$$$4)as3),buttwospeicialcouples$$$$shouldnotsitnexttoeachother.$$$$\underset{―}{itmeans}thatthehusbandfrom$$$$couple1maynotsitnexttothe$$$$wifefromcouple2andthe$$$$husbandfromcouple2maynot$$$$sitnexttothewifefromcouple1.$$$$butcertainlythehusbandsof$$$$bothcouplesmaysitnexttotheir$$$$ownwifes.$$

Commented by nikif99 last updated on 08/Apr/25

$$\underset{―}{Generalsolution}$$$$ifpersons=N\Rightarrow couples=N/2=C$$$$1)N!$$$$2)C!\times 2^{C+1}$$$$3)C!\times 2^2$$$$4)(C-2)!\times N\times (N-3)\times 2^2$$

Answered by nikif99 last updated on 08/Apr/25

$$Itryananswer\left(figurefollows\right).$$$$10couplesare20persons.$$$$1)Arrangementsof20people,evenin$$$$circle,withdifferentplacesare20!$$$$2)Couplesmustsitinadjacentplaces.$$$$Consider10twinseats\left(seeTableA\right).$$$$Thereare10!waysfor10couples.$$$$Also,couple1caninterchangeseats$$$$\left(\times 2\right),couple2candothesame\left(\times 2\right),\dots$$$$tillcouple10\left(\times 2\right),e.g.2^{10}.Also,all$$$$couplescanshiftoneplace(seered$$$$bracket)\left(\times 2\right).Total10!\cdot 2^{11}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\dots Notwomeninadjacentplaces.$$$$Ifthepattern(mw-mw-mw-mw\dots )$$$$\{wheremman,wwoman\}changesto$$$$(mw-wm-wm-wm\dots )thecircle$$$${doesn}^{\prime }tclosenormally.$$$$10couplesaresittingin10twinseats$$$$in10!ways\left(seeTableB\right).Also,the$$$$patterncanchangeto(wm-wm-wm\dots )$$$$\left(\times 2\right).Also,allcouplescanshiftone$$$$place\left(redbracket\right)\left(\times 2\right).Total10!\cdot 2^2$$$$4)Restriction\left(3\right)plusnottwocertain$$$$couplesnexttoeachother.$$$$Considertwocouples,redandblue,not$$$$allowedinadjacentseats\left(seeTableC\right).$$$$Positions\mathit{n}\mathit{o}\mathit{t}allowedforthosetwo$$$$couplesare10-1\left({}^a\right).Remaining8$$$$couplesaresittedin8!ways\left({}^b\right).Also$$$$thesetwocouplesarenotallowedat$$$$firstandlastpositions,wheretable$$$$ends\left(seeredbracket\right),soremaining$$$$8coupleshaveanother8!waysof$$$$sitting\left({}^c\right).Also,couplescan$$$$interchangeseats\times 2\left({}^d\right).$$$$Knowingallarrangementsare10!,$$$$aboveexceptionsarededucted.$$$$10!-\underset{d}{\underbrace{2}}\cdot \underset{a}{\underbrace{(10-1)}\underset{b}{(10-2)!}}-\underset{d}{\underbrace{2}}\cdot \underset{c}{\underbrace{(10-2)!}}=$$$$10!-(10-2)![2(10-1)+2]=$$$$10!-2\cdot 10(10-2)!=10\cdot (10-3)(10-2)!$$$$Finally,casesallowedmust\times 2for$$$$interchangingseats,\times 2forshifting$$$$oneplace.Total8!\cdot 10\cdot 7\cdot 2^2$$

Commented by nikif99 last updated on 08/Apr/25

Commented by mr W last updated on 19/Apr/25

$$thanks!$$$$weshouldconsiderthatthetable$$$$isroundandtheseatsareidentical.$$$$$$$$asfor4){let}^{\prime }slookatthecasewith$$$$only4couples.i^{\prime }llget$$$$4!\times 3!-2\times 2\times 3!\times 3!+2\times 2!\times 3!=24.$$$$itshouldbeabletocheckthis$$$$manuelly.$$

Answered by mr W last updated on 19/Apr/25

$$1)$$$$arrange20personsaroundround$$$$table:$$$$(20-1)!=19!$$$$$$$$2)$$$$arrange10couplesaroundround$$$$table,thereare(10-1)!ways.each$$$$couplehas2waystochangethe$$$$positionsofhusbandandwife:$$$$(10-1)!\times 2^{10}=9!\times 2^{10}$$$$$$$$3)$$$$toarrange10wivesaroundround$$$$tablethereare(10-1)!ways.to$$$$arrangethe10husbandsthereare$$$$10!ways:$$$$(10-1)!\times 10!=9!\times 10!=90\times 8!\times 9!$$$$$$$$4)$$$$10!\times 9!-2\times 2\times 9!\times 9!+2\times 8!\times 9!$$$$=56\times 8!\times 9!$$

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