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# Combinations - Groups

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CEO
Joined: 15 Aug 2003
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Combinations - Groups [#permalink]  11 Sep 2003, 02:52
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In how many ways can 4 groups of 2 each be selected from a group of 8 Students?

Is it 8C2 * 6C2 * 4C2 * 2C2 ?

Thanks
Praetorian
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Joined: 03 Feb 2003
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I think no. There is the doubling effect: the set of some combinations defines the set of others.

For example, we have a group of 2 people, in how many ways we can divide them? The answer is clear -- there is the only way to do so. However, if we employ combinatorics, we count twice: 2C1=2.

In your case the doubling effect is much more complicated.
I think it is [8C2*6C2*4C2*2C2]/[4!*3!*2!*1!]
Senior Manager
Joined: 22 May 2003
Posts: 334
Location: Uruguay
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Kudos [?]: 41 [0], given: 0

stolyar wrote:
I think no. There is the doubling effect: the set of some combinations defines the set of others.

For example, we have a group of 2 people, in how many ways we can divide them? The answer is clear -- there is the only way to do so. However, if we employ combinatorics, we count twice: 2C1=2.

In your case the doubling effect is much more complicated.
I think it is [8C2*6C2*4C2*2C2]/[4!*3!*2!*1!]

Maybe I didn't get precisely what you mean, but the number of ways you can arrange a group of 2 people is 2C2=1

Therefore I agree with Praetorian's solution. (not that I don't have any doubts)
GMAT Instructor
Joined: 07 Jul 2003
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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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Re: Combinations - Groups [#permalink]  14 Sep 2003, 02:50
praetorian123 wrote:
In how many ways can 4 groups of 2 each be selected from a group of 8 Students?

Is it 8C2 * 6C2 * 4C2 * 2C2 ?

Thanks
Praetorian

Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

CEO
Joined: 15 Aug 2003
Posts: 3467
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Kudos [?]: 718 [0], given: 781

Re: Combinations - Groups [#permalink]  14 Sep 2003, 03:41
AkamaiBrah wrote:
Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.

That was awesome!
Thanks..My solution gives the same answer.
Re: Combinations - Groups   [#permalink] 14 Sep 2003, 03:41
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