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Nine students are split into three equal teams to develop

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Nine students are split into three equal teams to develop [#permalink] New post 27 Jun 2006, 11:56
3. 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. If each team will report on a different problem, then how many different assignments of students to problems are possible?

(A) 5040
(B) 1680
(C) 1512
(D) 504
(E) 168


Can you please let me know how much time you took to solve this problem ?
I will post the answer later..
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Re: permutation.. [#permalink] New post 27 Jun 2006, 12:12
sharadGmat wrote:
3. 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. If each team will report on a different problem, then how many different assignments of students to problems are possible?

(A) 5040
(B) 1680
(C) 1512
(D) 504
(E) 168


Can you please let me know how much time you took to solve this problem ?
I will post the answer later..


Took me 1 min 45 secs approx.
Is the answer 1680?

9C3 * 6C3 * 3C3 ?
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Re: permutation.. [#permalink] New post 27 Jun 2006, 12:18
Futuristic wrote:
sharadGmat wrote:
3. 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. If each team will report on a different problem, then how many different assignments of students to problems are possible?

(A) 5040
(B) 1680
(C) 1512
(D) 504
(E) 168


Can you please let me know how much time you took to solve this problem ?
I will post the answer later..


Took me 1 min 45 secs approx.
Is the answer 1680?

9C3 * 6C3 * 3C3 ?



I will post the OA later.. But try googling for phrase "Nine students are split into three equal teams to develop" and check the answer.. The answer you see there is different from the one you got.. :)

Why aren't you considering the problems-team assignments?
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 [#permalink] New post 27 Jun 2006, 12:18
My answer is (B): 1680 different combinations of dividing 9 students in 3 groups of 3. I got to this number in about 90 seconds. Though I am still debating why I should not multiply 1680 again by 6(=3!) becase 6 is the number of different ways 3 assignements can be rearranged between 3 groups.
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 [#permalink] New post 27 Jun 2006, 12:19
B is the answer.


Total number of combinations of team: (9C3 * 6C3 * 3C3)/3! = 280

Total team to report combinations = 280 * 3! = 1680

It took me about 1 min 30 secs. Half of it in reading the question carefully.
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Last edited by ps_dahiya on 27 Jun 2006, 12:19, edited 1 time in total.
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 [#permalink] New post 27 Jun 2006, 12:19
v1rok wrote:
My answer is (B): 1680 different combinations of dividing 9 students in 3 groups of 3. I got to this number in about 90 seconds. Though I am still debating why I should not multiply 1680 again by 6(=3!) becase 6 is the number of different ways 3 assignements can be rearranged between 3 groups.


Why not ? Good question.. :)
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 [#permalink] New post 27 Jun 2006, 12:27
On the exam day, I would have marked (B) in 1.5 mins..

Now, let me go back and see, if (A) is really an alternative..
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 [#permalink] New post 27 Jun 2006, 12:28
sharadGmat wrote:
v1rok wrote:
My answer is (B): 1680 different combinations of dividing 9 students in 3 groups of 3. I got to this number in about 90 seconds. Though I am still debating why I should not multiply 1680 again by 6(=3!) becase 6 is the number of different ways 3 assignements can be rearranged between 3 groups.


Why not ? Good question.. :)


Bold part above is incorrect. This will be 1680/3! = 280

Then multiply by 3! and you get the answer. If you directly do 9C3 * 6C3 * 3C3 then you are duplicating the combinations. You need to divide it by factorial of the number of groups i.e. 3!.

I made the same mistake when I encountered a similar question " How many different combinations of 4 team of 2 people each are possible out of total 8 people."

Answer for this question is (8C2 * 6C2 * 4C2 * 2C2)/4!

Lets try the above question with two teams of 2 people out of 4 people then the combinations are
If we employ 4C2*2C2 (i.e
12, 34..................1
13, 24..................2
14, 23..................3
23, 14..................4...Same as 3
24, 13..................5...Same as 2
34, 12..................6...Same as 1

So you need to divide by 2! now we get the correct combinations.

Hope this helps.
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 [#permalink] New post 27 Jun 2006, 12:34
I think ps_dahiya explanation is correct and we do need to multiply by 3! at the end. the problem was that I overestimated number of student combinations. Once a student is assigned to a group, the order in that group does not matter, so the real number of different 3-student groups is 1680/3!=280. Then you multiply by 3! to get to te final answer of 1680.

I just checked this with a simplified example of 4 students (A,B,C,D) in 2 groups 1 and 2:

Group1 Group2
AB CD
AC BD
AD BC

Only 3 possible combinations, not 6
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 [#permalink] New post 27 Jun 2006, 12:43
ps_dahiya wrote:
sharadGmat wrote:
v1rok wrote:
My answer is (B): 1680 different combinations of dividing 9 students in 3 groups of 3. I got to this number in about 90 seconds. Though I am still debating why I should not multiply 1680 again by 6(=3!) becase 6 is the number of different ways 3 assignements can be rearranged between 3 groups.


Why not ? Good question.. :)


Bold part above is incorrect. This will be 1680/3! = 280

Then multiply by 3! and you get the answer. If you directly do 9C3 * 6C3 * 3C3 then you are duplicating the combinations. You need to divide it by factorial of the number of groups i.e. 3!.

I made the same mistake when I encountered a similar question " How many different combinations of 4 team of 2 people each are possible out of total 8 people."

Answer for this question is (8C2 * 6C2 * 4C2 * 2C2)/4!

Lets try the above question with two teams of 2 people out of 4 people then the combinations are
If we employ 4C2*2C2 (i.e
12, 34..................1
13, 24..................2
14, 23..................3
23, 14..................4...Same as 3
24, 13..................5...Same as 2
34, 12..................6...Same as 1

So you need to divide by 2! now we get the correct combinations.

Hope this helps.


Thanks for the explanation.. that makes it clear why OA is B..:)
This forum has the wrong answer.. :)
http://www.pagalguy.com/forum/quantitat ... tions.html
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 [#permalink] New post 28 Jun 2006, 06:11
sharadGmat wrote:
ps_dahiya wrote:
sharadGmat wrote:
v1rok wrote:
My answer is (B): 1680 different combinations of dividing 9 students in 3 groups of 3. I got to this number in about 90 seconds. Though I am still debating why I should not multiply 1680 again by 6(=3!) becase 6 is the number of different ways 3 assignements can be rearranged between 3 groups.


Why not ? Good question.. :)


Bold part above is incorrect. This will be 1680/3! = 280

Then multiply by 3! and you get the answer. If you directly do 9C3 * 6C3 * 3C3 then you are duplicating the combinations. You need to divide it by factorial of the number of groups i.e. 3!.

I made the same mistake when I encountered a similar question " How many different combinations of 4 team of 2 people each are possible out of total 8 people."

Answer for this question is (8C2 * 6C2 * 4C2 * 2C2)/4!

Lets try the above question with two teams of 2 people out of 4 people then the combinations are
If we employ 4C2*2C2 (i.e
12, 34..................1
13, 24..................2
14, 23..................3
23, 14..................4...Same as 3
24, 13..................5...Same as 2
34, 12..................6...Same as 1

So you need to divide by 2! now we get the correct combinations.

Hope this helps.


Thanks for the explanation.. that makes it clear why OA is B..:)
This forum has the wrong answer.. :)
http://www.pagalguy.com/forum/quantitat ... tions.html


Can someone explain me the bold part in detial above? I suck at probability questions. thanks
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 [#permalink] New post 28 Jun 2006, 06:48
shampoo wrote:
sharadGmat wrote:
ps_dahiya wrote:
sharadGmat wrote:
v1rok wrote:
My answer is (B): 1680 different combinations of dividing 9 students in 3 groups of 3. I got to this number in about 90 seconds. Though I am still debating why I should not multiply 1680 again by 6(=3!) becase 6 is the number of different ways 3 assignements can be rearranged between 3 groups.


Why not ? Good question.. :)


Bold part above is incorrect. This will be 1680/3! = 280

Then multiply by 3! and you get the answer. If you directly do 9C3 * 6C3 * 3C3 then you are duplicating the combinations. You need to divide it by factorial of the number of groups i.e. 3!.

I made the same mistake when I encountered a similar question " How many different combinations of 4 team of 2 people each are possible out of total 8 people."

Answer for this question is (8C2 * 6C2 * 4C2 * 2C2)/4!

Lets try the above question with two teams of 2 people out of 4 people then the combinations are
If we employ 4C2*2C2 (i.e
12, 34..................1
13, 24..................2
14, 23..................3
23, 14..................4...Same as 3
24, 13..................5...Same as 2
34, 12..................6...Same as 1

So you need to divide by 2! now we get the correct combinations.

Hope this helps.


Thanks for the explanation.. that makes it clear why OA is B..:)
This forum has the wrong answer.. :)
http://www.pagalguy.com/forum/quantitat ... tions.html


Can someone explain me the bold part in detial above? I suck at probability questions. thanks


There are 8 people. How many ways to select first team of two people = 8C2.
Now six people are remaining. How many ways to select second team of two people = 6C2.
Now four people are remaining. How many ways to select third team of two people = 4C2.
Now two people are remaining. How many ways to select fourth team of two people = 2C2.

Total ways = 8C2 * 6C2 * 4C2 * 2C2. But these contain duplicates as described in my earlier post. To remove those duplicate divide by th efactorial of total number of teams. i.e. 4!

Final result = (8C2 * 6C2 * 4C2 * 2C2)/4!

Hope this helps.
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 [#permalink] New post 28 Jun 2006, 07:21
Just had a small idea. If you order the topic then you could simply consider the three groups as ordered (that means abc, def, ghi is different from abd, ghi, def) and do not need to trouble about getting rid of the repeats and then assign the groups to topics. In other words you get the answer directly from 3C9*3C6. Might save a bit of time.
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  [#permalink] 28 Jun 2006, 07:21
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