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unique teams

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Manager
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Status: Still Struggling
Joined: 02 Nov 2010
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Location: India
GMAT Date: 10-15-2011
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unique teams [#permalink] New post 16 Dec 2010, 00:40
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There are 12 co-workers who work on projects in teams each month. Each team is comprised of 3 people and every team is together only once before the rotation begins again. How many unique teams can be created?

My effort:
From the above scenario, it is clear that there are 4 teams and each comprises of 3 people.
Hence, i took m=4 and n=3 in the formula (mn)! / (n!)^m * m!

but the answer coming is wrong. Can somebody please tell me whats wrong in here?

OA : 220
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Re: unique teams [#permalink] New post 16 Dec 2010, 01:10
krishnasty wrote:
There are 12 co-workers who work on projects in teams each month. Each team is comprised of 3 people and every team is together only once before the rotation begins again. How many unique teams can be created?

My effort:
From the above scenario, it is clear that there are 4 teams and each comprises of 3 people.
Hence, i took m=4 and n=3 in the formula (mn)! / (n!)^m * m!

but the answer coming is wrong. Can somebody please tell me whats wrong in here?

OA : 220


Hi!

You've vastly overcomplicated the problem.

The simplest interpretation of the question is "how many unique groups of 3 people can be made out of a total pool of 12?"

The answer is then 12C3. Plugging into the combinations formula:

12C3 = 12!/3!9! = 12*11*10/3*2*1 = 2*11*10 = 220

To be honest, I'm not even sure where they formula you used comes from - I've never seen it before.
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Manager
Manager
User avatar
Status: Still Struggling
Joined: 02 Nov 2010
Posts: 139
Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 15 [0], given: 8

Re: unique teams [#permalink] New post 16 Dec 2010, 03:16
well, that seemed a very easy approach to do so..

The formula that i have used was posted by Bunuel in some of his post. And to tell you the truth, i have often used this formula to derive the correct answer..

@Bunuel, can you please point out if i used the formula incorrectly or where exactly i made a mistake.

Thanks Guys!!
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Appreciation in KUDOS please!
Knewton Free Test 10/03 - 710 (49/37)
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Re: unique teams [#permalink] New post 16 Dec 2010, 03:32
I agree with skovinsky that the solution is simply 12!/(9!*3!) = 220.

I've studied quite a lot of combinatorics for my GMAT and I've never come across the formula you listed above. It looks too fancy to be relevant for the GMAT :) . For the GMAT it's enough to remember just the basic formulas for combinations and permutations. If you understand these formulas and know how to apply them you should be able to answer pretty much all the questions.
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Re: unique teams [#permalink] New post 16 Dec 2010, 05:34
Expert's post
krishnasty wrote:
well, that seemed a very easy approach to do so..

The formula that i have used was posted by Bunuel in some of his post. And to tell you the truth, i have often used this formula to derive the correct answer..

@Bunuel, can you please point out if i used the formula incorrectly or where exactly i made a mistake.

Thanks Guys!!


As OA given to be 220 then skovinsky's interpretation of the question is correct and the answer is simply 12C3=220.

You are applying the formula of dividing a group of people (items) equally into smaller groups when order matters, but it's not the case here, as we don't have 4 projects to make 4 groups of 3 to assign to them. Note that even in this case you can solve the question without this formula, for more on this check:

probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups
combination-and-selection-into-team-106277.html

Hope it's clear.
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Re: unique teams   [#permalink] 16 Dec 2010, 05:34
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