About average grouping...... (1 Viewer)

[Hikari Clover]

16 people to be divided into 4 teams of 4
how many ways can they be divided into teams if there r no restrictions......

isnt the answer (16c4*12c4*8c4)/4! ?

where is the 4! on the bottom come from?
because of 4 people in each team or just because there r 4 teams?


how about if we have 24 people, divided into 8 teams ,
4 teams contain 2 people each,
2 teams contain 5 people each,
2 teams contain 3 people each,
how many ways we could have if there r no restrictions......


thanks:wave:

 

[roadrage75]

regarding the first question, here's how i'd do it... it does get the same answer as yours....

abcdefghijlkmnop

imagine that these are the 16 players in a line.

ways of rearranging the line: 16!

now let's divide the teams.

abcd/efgh/ijkl/mnop

when i claimed there are 16! combinations, I made a lot of repetitions.

we must firstly divide by the number of ways the players can be re-arranged in their own, individual teams, because, for example, abcd and acdb are the same team.

each team consists of 4 people, therefore there are 4! possible ways of arranging each team. therefore we must divide by 4!x4!x4!x4!

now the number of combinations = 16!/(4!x4!x4!x4!)

we also must take into account the order of these four groups...
that is to say efgh/abcd/ijkl/mnop is the same as abcd/efgh/ijkl/mnop
the number of ways in which these 4 groups can be arranged is also 4!

for this reason, we must divide again by 4! (and i think this is the answer to your question, you divide by 4! because there are 4 teams)

thus numebr of combinations = 16!(4!x4!x4!x4!x4!) = 2,627,625 --and this is what you thought it was

 

[elseany]

hey, with your first question the thing to remember is that although we are dealing with people, the four teams are indistinguishable, like there are no ways of differentiating between which team is which, so if we counted the individual cases we would see we over counted by creating the same team a few times. like this is kinda hard to explain on the net but say from the 16 people we have people 1, 2, 3 and 4. If we counted all the cases we would see that 1, 2, 3 and 4 would have been in the same team more than once, in fact since we are creating 4 teams, they would have been in the same team 4! times, so to eliminate this error we have to divide by 4!

for your other questions its a similair method

so obviously you understand this part: 24C2*22C2*20C2*18C2*13C2*8C2*6C3*3C3 but again we've overcounted, so we have to consider the identical cases and eliminate them, now these identical cases occur where we have indistinguishable entity's; we have FOUR teams of 2, TWO teams of 5, and TWO teams of 3, so we have to divide by 4!*2!*2!

so the answer is; (24C2*22C2*20C2*18C2*13C2*8C2*6C3*3C3)/(4!*2!*2!)

... i think

 

[roadrage75]

as far as your second questions is concerned, i'd do it the same way....

put 24 people in a line. thus there 24! possible ways of arranging them. but i made a llot of repetitions....

let's devide the teams:

ab/cd/ef/gh/ijklm/nopqr/stu/vwx

first divide by the number of number of ways the players can be divided in their own teams, (as ba/cd.... is same as ab/cd.....)

hence divide by 2!x2!x2!x2!x5!x5!x3!x3!

now we must take into acount the order of these 8 groups, as ab/cd.... is same as cd/ab....thus divide by 8!

hence total = 24!/(2!x2!x2!x2!x5!x5!x3!x3!x8!) = 1.855x10^12. i hope i'm right!