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Question Number 142765 by mr W last updated on 05/Jun/21
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?
$${A}\:{class}\:{has}\:\mathrm{13}\:{children}.\:{To}\:{play}\:{a} \\ $$$${game}\:{one}\:{child}\:{is}\:{the}\:{referee}\:{and}\:{the} \\ $$$${other}\:{children}\:{are}\:{divided}\:{in}\:{three}\: \\ $$$${teams}\:{with}\:{four}\:{children}\:{in}\:{each}\:{team}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{class}\:{play} \\ $$$${the}\:{game}? \\ $$
Commented by mr W last updated on 06/Jun/21
generally if n distinct objects are divided in several identical boxes with n_1 ,n_2 ,...,n_m objects respectively. n_1 +n_2 +...+n_m =n. the number of ways is ((n!)/(n_1 !n_2 !...n_m !)). in current case it is ((13!)/((4!)^3 1!))=450450. one can also get: for the first group there are C_4 ^(13) ways. for the second group C_4 ^9 ways and the third group C_4 ^5 ways. the remaining one is automaticall the referee. ⇒C_4 ^(13) ×C_4 ^9 ×C_4 ^5 =450450. or to select the referee there are C_1 ^(13) ways. for the first group C_4 ^(12) ways and the second group C_4 ^8 ways, the remaining children are automaticall the third group. ⇒C_1 ^(13) ×C_4 ^(12) ×C_4 ^8 =450450
$${generally}\:{if}\:{n}\:{distinct}\:{objects}\:{are} \\ $$$${divided}\:{in}\:{several}\:{identical}\:{boxes} \\ $$$${with}\:{n}_{\mathrm{1}} ,{n}_{\mathrm{2}} ,…,{n}_{{m}} \:{objects}\:{respectively}. \\ $$$${n}_{\mathrm{1}} +{n}_{\mathrm{2}} +…+{n}_{{m}} ={n}.\:{the}\:{number}\:{of}\:{ways} \\ $$$${is}\:\frac{{n}!}{{n}_{\mathrm{1}} !{n}_{\mathrm{2}} !…{n}_{{m}} !}. \\ $$$${in}\:{current}\:{case}\:{it}\:{is}\:\frac{\mathrm{13}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{1}!}=\mathrm{450450}. \\ $$$$ \\ $$$$\:{one}\:{can}\:{also}\:{get}: \\ $$$${for}\:{the}\:{first}\:{group}\:{there}\:{are}\:{C}_{\mathrm{4}} ^{\mathrm{13}} \: \\ $$$${ways}.\:{for}\:{the}\:{second}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{9}} \:{ways}\: \\ $$$${and}\:{the}\:{third}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{5}} \:{ways}.\:{the}\: \\ $$$${remaining}\:{one}\:{is}\:{automaticall}\:{the} \\ $$$${referee}. \\ $$$$\Rightarrow{C}_{\mathrm{4}} ^{\mathrm{13}} ×{C}_{\mathrm{4}} ^{\mathrm{9}} ×{C}_{\mathrm{4}} ^{\mathrm{5}} =\mathrm{450450}. \\ $$$${or}\:{to}\:{select}\:{the}\:{referee}\:{there}\:{are} \\ $$$${C}_{\mathrm{1}} ^{\mathrm{13}} \:{ways}.\:{for}\:{the}\:{first}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{12}} \:{ways} \\ $$$${and}\:{the}\:{second}\:{group}\:{C}_{\mathrm{4}} ^{\mathrm{8}} \:{ways},\:{the} \\ $$$${remaining}\:{children}\:{are}\:{automaticall} \\ $$$${the}\:{third}\:{group}. \\ $$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{13}} ×{C}_{\mathrm{4}} ^{\mathrm{12}} ×{C}_{\mathrm{4}} ^{\mathrm{8}} =\mathrm{450450} \\ $$
Answered by gsk2684 last updated on 05/Jun/21
selecting a referee in 13 ways. remaining 12 can be grouped into three each contain 4 children. It can be done in ((12!)/((4!)^3 3!)) simultaneously (13)(((12!)/((4!)^3 .3!)))
$$\mathrm{selecting}\:\mathrm{a}\:\mathrm{referee}\:\mathrm{in}\:\mathrm{13}\:\mathrm{ways}. \\ $$$$\mathrm{remaining}\:\mathrm{12}\:\mathrm{can}\:\mathrm{be}\:\mathrm{grouped}\:\mathrm{into}\:\mathrm{three}\:\mathrm{each}\:\mathrm{contain}\: \\ $$$$\mathrm{4}\:\mathrm{children}.\:\mathrm{It}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{3}!} \\ $$$$\mathrm{simultaneously}\:\left(\mathrm{13}\right)\left(\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} .\mathrm{3}!}\right) \\ $$
Commented by mr W last updated on 05/Jun/21
can you explain how you get the number of ways to divide 12 children in three groups with 4 children each?
$${can}\:{you}\:{explain}\:{how}\:{you}\:{get}\:{the} \\ $$$${number}\:{of}\:{ways}\:{to}\:{divide}\:\mathrm{12}\:{children} \\ $$$${in}\:{three}\:{groups}\:{with}\:\mathrm{4}\:{children}\:{each}? \\ $$
Commented by gsk2684 last updated on 06/Jun/21
selecing 4 from 12 in 12_C_4 ways for first group selecing 4 from remain 8 in 8_C_4 ways for second group selecing 4 from remain 4 in 4_C_4 ways for third group but groups have equal members so they can be interchange entirely 3 gruops can be interchanged among themselves in 3! ways. so required number of ways is ((12_C_4 .8_C_4 .4_C_4 )/(3!))=((12! )/((4!)^3 .3!))
$$\mathrm{selecing}\:\mathrm{4}\:\mathrm{from}\:\mathrm{12}\:\mathrm{in}\:\mathrm{12}_{\mathrm{C}_{\mathrm{4}} } \mathrm{ways}\:\mathrm{for}\:\mathrm{first}\:\mathrm{group} \\ $$$$\mathrm{selecing}\:\mathrm{4}\:\mathrm{from}\:\mathrm{remain}\:\mathrm{8}\:\mathrm{in}\:\mathrm{8}_{\mathrm{C}_{\mathrm{4}} } \mathrm{ways}\:\mathrm{for}\:\mathrm{second}\:\mathrm{group} \\ $$$$\mathrm{selecing}\:\mathrm{4}\:\mathrm{from}\:\mathrm{remain}\:\mathrm{4}\:\mathrm{in}\:\mathrm{4}_{\mathrm{C}_{\mathrm{4}} } \mathrm{ways}\:\mathrm{for}\:\mathrm{third}\:\mathrm{group} \\ $$$$\mathrm{but}\:\mathrm{groups}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{members}\: \\ $$$$\mathrm{so}\:\mathrm{they}\:\mathrm{can}\:\mathrm{be}\:\mathrm{interchange}\:\mathrm{entirely} \\ $$$$\mathrm{3}\:\mathrm{gruops}\:\mathrm{can}\:\mathrm{be}\:\mathrm{interchanged}\: \\ $$$$\mathrm{among}\:\mathrm{themselves}\:\mathrm{in}\:\mathrm{3}!\:\mathrm{ways}. \\ $$$$\mathrm{so}\:\mathrm{required}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\: \\ $$$$\mathrm{is}\:\frac{\mathrm{12}_{\mathrm{C}_{\mathrm{4}} } .\mathrm{8}_{\mathrm{C}_{\mathrm{4}} } .\mathrm{4}_{\mathrm{C}_{\mathrm{4}} } }{\mathrm{3}!}=\frac{\mathrm{12}!\:}{\left(\mathrm{4}!\right)^{\mathrm{3}} .\mathrm{3}!} \\ $$
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 ((12!)/((4!)^3 )). even if the order of the groups matters, the number of ways is ((12!)/((4!)^3 ))×3!, not ((12!)/((4!)^3 ×3!)).
$${thanks}\:{for}\:{explanation}!\:{but}\:{the}\: \\ $$$${answer}\:{is}\:{not}\:{correct}\:{i}'{m}\:{afraid}. \\ $$$${since}\:{the}\:{order}\:{of}\:{the}\:{groups}\:{doesn}'{t} \\ $$$${matter},\:{the}\:{number}\:{of}\:{ways}\:{is}\:{just} \\ $$$$\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }.\: \\ $$$${even}\:{if}\:{the}\:{order}\:{of}\:{the}\:{groups} \\ $$$${matters},\:{the}\:{number}\:{of}\:{ways}\:{is} \\ $$$$\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }×\mathrm{3}!,\:{not}\:\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} ×\mathrm{3}!}. \\ $$

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