Combinations

[noney08]

Q
Nine students are split into three equal teams to develop reports on one of three problems:
shortage of skilled labor, violence in schools, and low standardized test scores. How many different teams of students are possible?
(A) 5040
(B) 1680
(C) 1512
(D) 504
(E) 168

[GMAT-HELP]

Q
Nine students are split into three equal teams to develop reports on one of three problems:
shortage of skilled labor, violence in schools, and low standardized test scores. How many different teams of students are possible?
(A) 5040
(B) 1680
(C) 1512
(D) 504
(E) 168

Is it 9!/(3!*3!*3!)? = 1680 => B?

Thanks,
GMAT-HELP

[transyt7]

9C3*6C3*3 = 5040 or A...

[GMAT-HELP]

9C3*6C3*3 = 5040 or A...

Transyt, it should be 9c3*6c3*3c3 = 1680 or B.. Please check and let me know.

Thanks,
GMAT-HELP

[transyt7]

Transyt, it should be 9c3*6c3*3c3 = 1680 or B.. Please check and let me know.

Thanks,
GMAT-HELP

U can select 3 out of 9 in 9C3 ways
Again, u can select 3 out of remaining 6 in 6C3 ways.
Since u got thy groups figured out, u can assign to each group the 3 tasks in 3 ways.

Total num of ways = (9C3*6C3) * 3 = 5040

Am I missing something ? Please correct me ...

[noney08]

The answer is B or 1680

[noney08]

Is it 9!/(3!*3!*3!)? = 1680 => B?

Thanks,
GMAT-HELP

How did you get this ?? could you please elaborate. THanks

[shaq]

U can select 3 out of 9 in 9C3 ways
Again, u can select 3 out of remaining 6 in 6C3 ways.
Since u got thy groups figured out, u can assign to each group the 3 tasks in 3 ways.

Total num of ways = (9C3*6C3) * 3 = 5040

Am I missing something ? Please correct me ...

The first team you can select in in 9C3 ways
The second team you can select in 6C3 ways
The remaining 3 people can be selected in only 1 way ( 3C3)
so total combinations= 9C3 * 6C3 * 3C3 = 1680

HTH ...

[thebullfighter]

U can select 3 out of 9 in 9C3 ways
Again, u can select 3 out of remaining 6 in 6C3 ways.
Since u got thy groups figured out, u can assign to each group the 3 tasks in 3 ways.

Total num of ways = (9C3*6C3) * 3 = 5040

Am I missing something ? Please correct me ...

u took the third combination as 1 only.
but transyt the red part was not required to be considered in the Q. not asked.
U were not missing anything, rather u had something X-tra..

[transyt7]

u took the third combination as 1 only.
but transyt the red part was not required to be considered in the Q. not asked.
U were not missing anything, rather u had something X-tra..

Thanx, Oh well, in haste I interpreted the question as how many different ways these groups can be assigned to given 3 tasks...OK, I get it now...