Question

in how many ways can 9 students be exactly divided into 3 teams .

Answer 1

Presumably order does not matter, e.g., we are just "grouping" people (so ABC is the same team as CBA). So it would be a combination, just use the formula:

9C3=9!/(3!(9-3)!)

Answer 2

\[\large _nC_r=\frac{ n! }{ r!(n-r)! }\]

Answer 3

can't we use permutation ?

Answer 4

in how many ways can 9 students be evenly divided into 3 teams

Answer 5

Oh wait.... evenly divided into 3 teams, that's different than what I was doing with a combination, which is just choosing 3. that's not quite right.....

But a permutation would give you how many ways to arrange the 9 people, which is not quite right either. E.g., it would count ABC and BCA twice, even though they are the same team.

Answer 6

my mistake i mixed up two questions

Answer 7

and wrote exactly in place of evenly

Answer 8

so it would be 9P3 ?

Answer 9

No, that was my mistake.... I was just doing the wrong thing. lol. I'm thinking.......

9! gives you the number of ways to "line up" the 9 people. The problem with that is that the same groups are counted multiple times, in different order. I'm thinking......

Answer 10

No, not 9P3. That gives you the number of ways to arrange 3 from 9. Again, order is not important.

Answer 11

i think order will be important ........

Answer 12

I'm thinking maybe 9!/6? Since each group of 3 can be arranged in 6 different orders?

Answer 13

But I'm not really confident about that... lol.

Answer 14

dang!! neither am i :/ i think it should be 9P3 just need someones confirmation

Answer 15

because i think the word evenly is specifying that order does matter

Answer 16

I am fairly confident that it is NOT 9P3.

Order does that matter - don't get too hung up on the word "evenly". It just means, group these 9 people into 3 teams of 3 people each. So order within each team does not matter - e.g., Sue, Sally and Joe is the SAME team as Joe, Sue and Sally.

Answer 17

9P3 would be the number of ways to pick 3 people from the 9, if order DOES matter. E.g., if I pick 3 names at random from 9, and the first will be the catcher, second will be the pitcher, and third will be the 1st baseman, then how may ways can I pick these 3 positions from these 9 people..... THAT would be 9P3.

Answer 18

9! (or 9P9) would give the number of ways to order the 9 people. E.g., how many ways can we line them up. Then we take the first 3 for one team, the second 3 for another team, and the third 3 for the third team.

The problem is, that counts the same groups multiple times, since:

ABC DEF GHI and
BAC FDE IGH

are counted as 2 distinct arrangements, but they are the same 3 teams.

Answer 19

@rational , i was like this question. any idea ?