A-class-has-13-children-To-play-a-game-one-child-is-the-referee-and-the-other-children-are-divided-in-three-teams-with-four-children-in-each-team-In-how-many-ways-can-the-class-play-the-game-

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Question Number 142765 by mr W last updated on 05/Jun/21

$$Aclasshas13children.Toplaya$$$$gameonechildistherefereeandthe$$$$otherchildrenaredividedinthree$$$$teamswithfourchildrenineachteam.$$$$Inhowmanywayscantheclassplay$$$$thegame?$$

Commented by mr W last updated on 06/Jun/21

$$generallyifndistinctobjectsare$$$$dividedinseveralidenticalboxes$$$$with{n}_1,n_2,\dots ,n_{m}objectsrespectively.$$$$n_1+n_2+\dots +n_m=n.thenumberofways$$$$is\frac{n!}{n_1!n_2!\dots n_m!}.$$$$incurrentcaseitis\frac{13!}{{(4!)}^{3}1!}=450450.$$$$$$$$onecanalsoget:$$$$forthefirstgroupthereare{C}_4^{13}$$$$ways.forthesecondgroup{C}_4^{9}ways$$$$andthethirdgroup{C}_4^{5}ways.the$$$$remainingoneisautomaticallthe$$$$referee.$$$$\Rightarrow C_4^{13}\times C_4^9\times C_4^5=450450.$$$$ortoselecttherefereethereare$$$$C_1^{13}ways.forthefirstgroup{C}_4^{12}ways$$$$andthesecondgroup{C}_4^{8}ways,the$$$$remainingchildrenareautomaticall$$$$thethirdgroup.$$$$\Rightarrow C_1^{13}\times C_4^{12}\times C_4^8=450450$$

Answered by gsk2684 last updated on 05/Jun/21

$$\mathrm{selecting}\mathrm{a}\mathrm{referee}\mathrm{in}13\mathrm{ways}.$$$$\mathrm{remaining}12\mathrm{can}\mathrm{be}\mathrm{grouped}\mathrm{into}\mathrm{three}\mathrm{each}\mathrm{contain}$$$$4\mathrm{children}.\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{done}\mathrm{in}\frac{12!}{{(4!)}^{3}3!}$$$$\mathrm{simultaneously}\left(13\right)\left(\frac{12!}{{(4!)}^3.3!}\right)$$

Commented by mr W last updated on 05/Jun/21

$$canyouexplainhowyougetthe$$$$numberofwaystodivide12children$$$$inthreegroupswith4childreneach?$$

Commented by gsk2684 last updated on 06/Jun/21

$$\mathrm{selecing}4\mathrm{from}12\mathrm{in}{12}_{{\mathrm{C}}_4}\mathrm{ways}\mathrm{for}\mathrm{first}\mathrm{group}$$$$\mathrm{selecing}4\mathrm{from}\mathrm{remain}8\mathrm{in}{8}_{{\mathrm{C}}_4}\mathrm{ways}\mathrm{for}\mathrm{second}\mathrm{group}$$$$\mathrm{selecing}4\mathrm{from}\mathrm{remain}4\mathrm{in}{4}_{{\mathrm{C}}_4}\mathrm{ways}\mathrm{for}\mathrm{third}\mathrm{group}$$$$\mathrm{but}\mathrm{groups}\mathrm{have}\mathrm{equal}\mathrm{members}$$$$\mathrm{so}\mathrm{they}\mathrm{can}\mathrm{be}\mathrm{interchange}\mathrm{entirely}$$$$3\mathrm{gruops}\mathrm{can}\mathrm{be}\mathrm{interchanged}$$$$\mathrm{among}\mathrm{themselves}\mathrm{in}3!\mathrm{ways}.$$$$\mathrm{so}\mathrm{required}\mathrm{number}\mathrm{of}\mathrm{ways}$$$$\mathrm{is}\frac{{12}_{{\mathrm{C}}_4}.8_{{\mathrm{C}}_4}.4_{{\mathrm{C}}_4}}{3!}=\frac{12!}{{(4!)}^3.3!}$$

Commented by mr W last updated on 06/Jun/21

$$thanksforexplanation!butthe$$$$answerisnotcorrect{i}^{\prime }mafraid.$$$$sincetheorderofthegroups{doesn}^{\prime }t$$$$matter,thenumberofwaysisjust$$$$\frac{12!}{{(4!)}^3}.$$$$eveniftheorderofthegroups$$$$matters,thenumberofwaysis$$$$\frac{12!}{{(4!)}^3}\times 3!,not\frac{12!}{{(4!)}^3\times 3!}.$$

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