Menu Close

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 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!)).
$${thanks}\:{for}\:{explanation}!\:{but}\:{the}\: \\ $$$${answer}\:{is}\:{not}\:{correct}\:{i}'{m}\:{afraid}. \\ $$$${since}\:{the}\:{order}\:{of}\:{the}\:{groups}\:{doesn}'{t} \\ $$$${matter},\:{the}\:{number}\:{of}\:{ways}\:{is}\:{just} \\ $$$$\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }.\: \\ $$$${even}\:{if}\:{the}\:{order}\:{of}\:{the}\:{groups} \\ $$$${matters},\:{the}\:{number}\:{of}\:{ways}\:{is} \\ $$$$\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }×\mathrm{3}!,\:{not}\:\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} ×\mathrm{3}!}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *