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Combinations

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

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Re: Combinations

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

Thanks,
GMAT-HELP

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Re: Combinations

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

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Re: Combinations

Originally Posted by transyt7
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

5. Good post? |

Re: Combinations

Originally Posted by GMAT-HELP
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 ...

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Re: Combinations

The answer is B or 1680

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Re: Combinations

Originally Posted by GMAT-HELP
Is it 9!/(3!*3!*3!)? = 1680 => B?

Thanks,
GMAT-HELP
How did you get this ?? could you please elaborate. THanks

8. Good post? |

Re: Combinations

Originally Posted by transyt7
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 ...

9. Good post? |

Re: Combinations

Originally Posted by transyt7
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..

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Re: Combinations

Originally Posted by thebullfighter
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...

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