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# If 6 people are to be divided to 3 different groups, each of

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Senior Manager
Joined: 08 Aug 2005
Posts: 251
If 6 people are to be divided to 3 different groups, each of [#permalink]

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05 May 2006, 23:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If 6 people are to be divided to 3 different groups, each of which has 2 people. How many such groups are possible?
VP
Joined: 29 Dec 2005
Posts: 1341

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16 May 2006, 12:34
getzgetzu wrote:
If 6 people are to be divided to 3 different groups, each of which has 2 people. How many such groups are possible?

does it make any difference to say 3 groups pf 2 prople?

no of group with 2 people = 6c2 = 15.

so no of group = 15/3 = 5
Senior Manager
Joined: 09 Mar 2006
Posts: 444

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16 May 2006, 13:00
6! / (2!*2!*2!) = 90
Senior Manager
Joined: 15 Mar 2005
Posts: 418
Location: Phoenix

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16 May 2006, 14:38
getzgetzu wrote:
If 6 people are to be divided to 3 different groups, each of which has 2 people. How many such groups are possible?

Forming first group C(6,2) = 15
Forming second group C(4,2) = 6
Forming third group C(2,2) = 1.

Number of ways of forming 3 groups = 90.
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The answer changes if the groups are named groups. Say, we have 3 groups, Reps, Dems and Inds. Since any group of 2 can be assigned to any named group, we have 6 more ways of doing that.

Then number of ways = 540.

Btw I think the question "How many such groups are possible" has a simple answer : 3. You already mentioned that we divide 6 people into 3 groups
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Director
Joined: 16 Aug 2005
Posts: 938
Location: France

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16 May 2006, 15:34
deowl wrote:
6! / (2!*2!*2!) = 90

Wont this include groups of same people, just in different orders.
I am not sure if that is what the question is asking. Can you please elaborate?
VP
Joined: 28 Mar 2006
Posts: 1369

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16 May 2006, 16:46
deowl wrote:
6! / (2!*2!*2!) = 90

I concur with this solution

It is like having 6 elements of which every 2 elements are similar and we are trying to arrange them.
Manager
Joined: 27 Jan 2006
Posts: 156
Location: Europe

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19 May 2006, 03:04
IMO, Just 15

6C2 = 15
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Manager
Joined: 27 Jan 2006
Posts: 156
Location: Europe

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20 May 2006, 12:18
plz provide OA
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Senior Manager
Joined: 09 Aug 2005
Posts: 283

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20 May 2006, 12:26
getzgetzu wrote:
If 6 people are to be divided to 3 different groups, each of which has 2 people. How many such groups are possible?

1 x 5c1 + 1x3c1 + 1x1c1 = 15
Re: Group of people   [#permalink] 20 May 2006, 12:26
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