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# 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-

Question Number 142765 by mr W last updated on 05/Jun/21
$${A}\:{class}\:{has}\:\mathrm{13}\:{children}.\:{To}\:{play}\:{a} \\$$$${game}\:{one}\:{child}\:{is}\:{the}\:{referee}\:{and}\:{the} \\$$$${other}\:{children}\:{are}\:{divided}\:{in}\:{three}\: \\$$$${teams}\:{with}\:{four}\:{children}\:{in}\:{each}\:{team}. \\$$$${In}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{class}\:{play} \\$$$${the}\:{game}? \\$$
Commented by mr W last updated on 06/Jun/21
$${generally}\:{if}\:{n}\:{distinct}\:{objects}\:{are} \\$$$${divided}\:{in}\:{several}\:{identical}\:{boxes} \\$$$${with}\:{n}_{\mathrm{1}} ,{n}_{\mathrm{2}} ,…,{n}_{{m}} \:{objects}\:{respectively}. \\$$$${n}_{\mathrm{1}} +{n}_{\mathrm{2}} +…+{n}_{{m}} ={n}.\:{the}\:{number}\:{of}\:{ways} \\$$$${is}\:\frac{{n}!}{{n}_{\mathrm{1}} !{n}_{\mathrm{2}} !…{n}_{{m}} !}. \\$$$${in}\:{current}\:{case}\:{it}\:{is}\:\frac{\mathrm{13}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{1}!}=\mathrm{450450}. \\$$$$\\$$$$\:{one}\:{can}\:{also}\:{get}: \\$$$${for}\:{the}\:{first}\:{group}\:{there}\:{are}\:{C}_{\mathrm{4}} ^{\mathrm{13}} \: \\$$$${ways}.\:{for}\:{the}\:{second}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{9}} \:{ways}\: \\$$$${and}\:{the}\:{third}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{5}} \:{ways}.\:{the}\: \\$$$${remaining}\:{one}\:{is}\:{automaticall}\:{the} \\$$$${referee}. \\$$$$\Rightarrow{C}_{\mathrm{4}} ^{\mathrm{13}} ×{C}_{\mathrm{4}} ^{\mathrm{9}} ×{C}_{\mathrm{4}} ^{\mathrm{5}} =\mathrm{450450}. \\$$$${or}\:{to}\:{select}\:{the}\:{referee}\:{there}\:{are} \\$$$${C}_{\mathrm{1}} ^{\mathrm{13}} \:{ways}.\:{for}\:{the}\:{first}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{12}} \:{ways} \\$$$${and}\:{the}\:{second}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{8}} \:{ways},\:{the} \\$$$${remaining}\:{children}\:{are}\:{automaticall} \\$$$${the}\:{third}\:{group}. \\$$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{13}} ×{C}_{\mathrm{4}} ^{\mathrm{12}} ×{C}_{\mathrm{4}} ^{\mathrm{8}} =\mathrm{450450} \\$$
Answered by gsk2684 last updated on 05/Jun/21
$$\mathrm{selecting}\:\mathrm{a}\:\mathrm{referee}\:\mathrm{in}\:\mathrm{13}\:\mathrm{ways}. \\$$$$\mathrm{remaining}\:\mathrm{12}\:\mathrm{can}\:\mathrm{be}\:\mathrm{grouped}\:\mathrm{into}\:\mathrm{three}\:\mathrm{each}\:\mathrm{contain}\: \\$$$$\mathrm{4}\:\mathrm{children}.\:\mathrm{It}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{3}!} \\$$$$\mathrm{simultaneously}\:\left(\mathrm{13}\right)\left(\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} .\mathrm{3}!}\right) \\$$
Commented by mr W last updated on 05/Jun/21
$${can}\:{you}\:{explain}\:{how}\:{you}\:{get}\:{the} \\$$$${number}\:{of}\:{ways}\:{to}\:{divide}\:\mathrm{12}\:{children} \\$$$${in}\:{three}\:{groups}\:{with}\:\mathrm{4}\:{children}\:{each}? \\$$
Commented by gsk2684 last updated on 06/Jun/21
$$\mathrm{selecing}\:\mathrm{4}\:\mathrm{from}\:\mathrm{12}\:\mathrm{in}\:\mathrm{12}_{\mathrm{C}_{\mathrm{4}} } \mathrm{ways}\:\mathrm{for}\:\mathrm{first}\:\mathrm{group} \\$$$$\mathrm{selecing}\:\mathrm{4}\:\mathrm{from}\:\mathrm{remain}\:\mathrm{8}\:\mathrm{in}\:\mathrm{8}_{\mathrm{C}_{\mathrm{4}} } \mathrm{ways}\:\mathrm{for}\:\mathrm{second}\:\mathrm{group} \\$$$$\mathrm{selecing}\:\mathrm{4}\:\mathrm{from}\:\mathrm{remain}\:\mathrm{4}\:\mathrm{in}\:\mathrm{4}_{\mathrm{C}_{\mathrm{4}} } \mathrm{ways}\:\mathrm{for}\:\mathrm{third}\:\mathrm{group} \\$$$$\mathrm{but}\:\mathrm{groups}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{members}\: \\$$$$\mathrm{so}\:\mathrm{they}\:\mathrm{can}\:\mathrm{be}\:\mathrm{interchange}\:\mathrm{entirely} \\$$$$\mathrm{3}\:\mathrm{gruops}\:\mathrm{can}\:\mathrm{be}\:\mathrm{interchanged}\: \\$$$$\mathrm{among}\:\mathrm{themselves}\:\mathrm{in}\:\mathrm{3}!\:\mathrm{ways}. \\$$$$\mathrm{so}\:\mathrm{required}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\: \\$$$$\mathrm{is}\:\frac{\mathrm{12}_{\mathrm{C}_{\mathrm{4}} } .\mathrm{8}_{\mathrm{C}_{\mathrm{4}} } .\mathrm{4}_{\mathrm{C}_{\mathrm{4}} } }{\mathrm{3}!}=\frac{\mathrm{12}!\:}{\left(\mathrm{4}!\right)^{\mathrm{3}} .\mathrm{3}!} \\$$
Commented by mr W last updated on 06/Jun/21
$${thanks}\:{for}\:{explanation}!\:{but}\:{the}\: \\$$$${answer}\:{is}\:{not}\:{correct}\:{i}'{m}\:{afraid}. \\$$$${since}\:{the}\:{order}\:{of}\:{the}\:{groups}\:{doesn}'{t} \\$$$${matter},\:{the}\:{number}\:{of}\:{ways}\:{is}\:{just} \\$$$$\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }.\: \\$$$${even}\:{if}\:{the}\:{order}\:{of}\:{the}\:{groups} \\$$$${matters},\:{the}\:{number}\:{of}\:{ways}\:{is} \\$$$$\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }×\mathrm{3}!,\:{not}\:\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} ×\mathrm{3}!}. \\$$