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

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Senior Manager
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If 6 people are to be divided to 3 different groups, each of [#permalink]  05 May 2006, 22:06
If 6 people are to be divided to 3 different groups, each of which has 2 people. How many such groups are possible?
VP
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Re: Group of people [#permalink]  16 May 2006, 11:34
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
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6! / (2!*2!*2!) = 90
Senior Manager
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Re: Group of people [#permalink]  16 May 2006, 13:38
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
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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
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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
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IMO, Just 15

6C2 = 15
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Manager
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plz provide OA
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Re: Group of people [#permalink]  20 May 2006, 11:26
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, 11:26
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