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# Hi guys here's another combination type question

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Senior Manager
Joined: 14 Jul 2006
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Hi guys here's another combination type question [#permalink]

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24 Jul 2006, 07:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I got this from the 1000 math question bank downloaded from this site. Isnt the answer suppose to be 6C2?

6 people are to be divided to 3 groups with the same numbers of people. How many such groups are possible? 90

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SVP
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25 Jul 2006, 01:17
Ok we can safely say that there will be 2 persons in each group.

Now number of ways 2 people can be chosen for the 1st group = 6C2
Number of ways 2 people can be chosen for the second group = 4C2
Number of ways 2 people can be chosen for the 3 group = 2C2

Remember for the second group you are chossing from 4 people not 6 people.

Total = 6C2 * 4C2 *2C2 = 90

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GMAT Instructor
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25 Jul 2006, 01:31
Are the groups themselves distingushable? In other words,

Group 1..........Group 2...........Group 3

....AB.................CD..................EF
....EF..................AB..................CD

Are these assignments distinguishable? If not, answer should be 6C2 * 4C2/3!= 6C2= 15

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Senior Manager
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25 Jul 2006, 03:51
Hi guys,

Thanks for being so helpful! I'm really not sure whether the groups are distinguishable though.

In the event that they are not, can you explain to me the logic of dividing the second term by 3!?

Thanks

The answer given in the question bank is 90.

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Manager
Joined: 21 Jul 2006
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25 Jul 2006, 09:48
kevincan wrote:
Are the groups themselves distingushable? In other words,

Group 1..........Group 2...........Group 3

....AB.................CD..................EF
....EF..................AB..................CD

Are these assignments distinguishable? If not, answer should be 6C2 * 4C2/3!= 6C2= 15

Kevin.
Can you pls explaing why the 3!.
Thanks

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VP
Joined: 02 Jun 2006
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25 Jul 2006, 10:42
We can look at this problems as :

1. Form 1 group of 2 people from a group of 6 people

2. Form 1 group of 2 people from a group of 4 people

3. Form 1 group of 2 people from a group of 2 people

(1) = 6x5/2! = 15
(2) = 4x3/2! = 6
(3) = 1

Total number of groups = 15x6x1 = 90

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CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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25 Jul 2006, 11:12
ghantark wrote:
kevincan wrote:
Are the groups themselves distingushable? In other words,

Group 1..........Group 2...........Group 3

....AB.................CD..................EF
....EF..................AB..................CD

Are these assignments distinguishable? If not, answer should be 6C2 * 4C2/3!= 6C2= 15

Kevin.
Can you pls explaing why the 3!.
Thanks

To remove duplicates.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Intern
Joined: 07 Nov 2003
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Location: USA

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25 Jul 2006, 11:33
apollo168 wrote:
Hi guys,

Thanks for being so helpful! I'm really not sure whether the groups are distinguishable though.

In the event that they are not, can you explain to me the logic of dividing the second term by 3!?

Thanks

The answer given in the question bank is 90.

which math question bank are you referring to? is it the ps 1000 (a zipped file)? I downloaded that but it does not seem to have answers in the documents. could you please post the file that you are referring to, here again?

Thanks

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Retired Moderator
Joined: 05 Jul 2006
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06 Aug 2006, 04:38
I believe you can think of the problem this way:

a) number of possibilities of forming 3 groups of two is 6C2.

B) each new group has 2! ways of arrangment>

c) they are 3 GROUPS

multiply 3*2! *6C2 = 90

I hope i am right.

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Senior Manager
Joined: 09 Aug 2005
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06 Aug 2006, 09:19
kevincan wrote:
Are the groups themselves distingushable? In other words,

Group 1..........Group 2...........Group 3

....AB.................CD..................EF
....EF..................AB..................CD

Are these assignments distinguishable? If not, answer should be 6C2 * 4C2/3!= 6C2= 15

another way of solving is

6 people

pick 1 you have 5 choices for the first team (1x5)

for team 2 pick 1 you have 3 choices (1x3)

team three pick 1 you have 1 choice (1x1)

total teams (with no one serving on more than 1 team) = 1x3x5=15

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Manager
Joined: 20 Mar 2006
Posts: 200

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Re: Hi guys here's another combination type question [#permalink]

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06 Aug 2006, 11:10
apollo168 wrote:
I got this from the 1000 math question bank downloaded from this site. Isnt the answer suppose to be 6C2?

6 people are to be divided to 3 groups with the same numbers of people. How many such groups are possible? 90

=6c2.4c2.2c2
=15*6*1
= 90

Heman

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Re: Hi guys here's another combination type question   [#permalink] 06 Aug 2006, 11:10
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# Hi guys here's another combination type question

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