arrange 12 people into 4 teams of 3 people

would u choose the team first then the players: 4C1 x 12C3 first team

3C1 x 9C3 second team and carry on for third and fourth then multiply all of ur results together?

Printable View

- May 23rd 2011, 11:01 AMqwerty10counting
arrange 12 people into 4 teams of 3 people

would u choose the team first then the players: 4C1 x 12C3 first team

3C1 x 9C3 second team and carry on for third and fourth then multiply all of ur results together? - May 23rd 2011, 11:52 AMPlato
- May 26th 2011, 03:21 AMqwerty10
But how would u then extend that to allow for the restriction that 2 particular people are not allowed in the same team?

- May 26th 2011, 04:12 AMPlato
- May 26th 2011, 06:30 AMSoroban
Hello, qwerty10!

Plato is correct . . . Here is an explanation.

Quote:

Arrange 12 people into 4 teams of 3 people.

Suppose we label the teams A, B, C and D.

Choose 3 of the 12 people for team A.

. . There are: .$\displaystyle _{12}C_3$ ways.

Choose 3 of the remaining 9 people for team B.

. . There are: .$\displaystyle _9C_3$ ways.

Choose 3 of the remaining 6 people for team C.

. . There are: .$\displaystyle _6C_3$ ways.

Choose 3 of the remaining 3 people for team D.

. . There are: .$\displaystyle _3C_3$ ways.

To arrange 12 people into 4*distinguishable*teams of 3 players each,

. . there are: .$\displaystyle \left(_{12}C_3\right)\left(_9C_3\right)\left(_6C_3 \right)\left(_3C_3\right)$

. . $\displaystyle =\; \frac{12!}{3!\,9!}\cdot\frac{9!}{3!\,6!}\cdot\frac {6!}{3!\,3!}\cdot\frac{3!}{3!\,0!} \;=\;\frac{12!}{(3!)^4} $ ways.

Since the four teams aredistinguishable, we must divide by $\displaystyle 4!$*not*

. . $\displaystyle \text{There are: }\:\frac{12!}{(3!)^44!} \:=\:15,\!400 $ ways.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle \text{Suppose two players, }X\text{ and }Y\text{, are }not\text{ allowed on the same team.}$

$\displaystyle \text{Count the ways that }X\text{ and }Y\;are\text{ on the same team.}$

$\displaystyle \text{Place }X\text{ and }Y\text{ on team A.}$

$\displaystyle \text{There are 10 choices for the third member of the team.}$

$\displaystyle \text{Then the other 9 people can be partitioned in }\,\frac{9!}{(3!)^3}\text{ ways.}$

Since the teams aredistinguishable we divide by 4!*not*

. . $\displaystyle \text{There are: }\:\frac{10\cdot 9!}{(3!)^34!} \:=\;700\text{ ways.}$

Therefore, there are: .$\displaystyle 15,\!400 - 700 \:=\:14,\!700\text{ ways}$

. . $\displaystyle \text{in which }X\text{ and }Y\text{ are }not\text{ teammates.}$

Darn . . . This doesn't agree with Plato's second answer.

Is one of us wrong? . . . maybe both?

. - May 26th 2011, 07:02 AMPlato
Your 15400 is correct. We both agree there.

But disagree on the 700.

There are ten ways to put**A & B**on a team together with one other person. That leaves nine to make up the other three teams.

That can be done in $\displaystyle \frac{9!}{(3!)^3(3!)}=280$. So there are 2800 ways that**A & B**on a team together. So the difference is 12600.

Here is a second front-door approach.

$\displaystyle \binom{10}{2}\binom{8}{2}\binom{5}{2}=12600$.

First pick**A's**team mates, then pick**B's**team mates then select the first person in alphabetical order from those not selected then pick**his**team mates, and we has a team of three left over. - May 28th 2011, 04:45 AMqwerty10
Hello,

Thank you for the help.

I see where I have gone wrong-ive just been doing the standard distinguishable approach and not taking into account that the teams are not distinguishable.

For part 1 for the total number of teams, I don't understand why in the distinguishable case( say we have teams 1,2,3) why we dont choose the team first then choose the members of the team? - May 28th 2011, 05:04 AMPlato
There is no need to select teams that are already given.

Say we have a red team, blue team, and green team, distinguishable teams.

We have a roster of the twelve people is alphabetical order.

If we rearrange the string "BBBBGGGGRRRR" in any order and put is next to that roster have one possible division into teams.

That can be done in $\displaystyle \frac{12!}{(4!)^3}$ ways.

BUT also notice that

$\displaystyle \frac{12!}{(4!)^3}=\underbrace {\binom{12}{4}}_{blue}\underbrace {\binom{8}{4}}_{green}\underbrace {\binom{4}{4}}_{red}.$

In other words, we could pick the blue team, then the green and finally the red. - May 28th 2011, 05:34 AMqwerty10
Thanks.

So then when we proceed to part 2 where 2 people A and B can't be on the same team and we want number of ways teams can be selected, do we then have to choose the team first then the members because its possible that couple could go into either of the three teams, assuming their distinguishable teams again to keep it simple