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# m24 q16

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Intern
Joined: 23 Mar 2010
Posts: 23
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Kudos [?]: 1 [0], given: 0

m24 q16 [#permalink]  16 Apr 2010, 18:26
Hi All,

I couldnt find this problem in the forum...so if I missed it...I apologize!

How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

8
10
16
20
24
[Reveal] Spoiler:
10

I am not quite sure about the answer...shouldnt it be as simple as 6C3...do we really have to consider leaving groups behind etc...I have never encountered this kind before...any further explanation would help.
Manager
Joined: 21 Jan 2010
Posts: 232
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Re: m24 q16 [#permalink]  16 Apr 2010, 21:27
Should be 6C3 / 2

Because half of the combination would be repeated in the other group.
For example, if group 1 has "A B C", group 2 must be "D E F"
Hence, the combination "D E F" in group 1 should be double counted.
Senior Manager
Joined: 29 Sep 2009
Posts: 396
GMAT 1: 690 Q47 V38
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Re: m24 q16 [#permalink]  22 Oct 2010, 04:26
calvinhobbes wrote:
Should be 6C3 / 2

Because half of the combination would be repeated in the other group.
For example, if group 1 has "A B C", group 2 must be "D E F"
Hence, the combination "D E F" in group 1 should be double counted.

6C3 simply means selecting 3 out of 6 and the order is not important.
Why divide the original expression by 2?

Say the set is: {A,B,C,D,E,F}
Using 6C3 select {A,B,D} and form 1 group - the other group would be formed automatically{C,E,F} - I dont see any "doubling".
Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 53359 [3] , given: 8155

Re: m24 q16 [#permalink]  22 Oct 2010, 06:18
3
KUDOS
Expert's post
vicksikand wrote:
calvinhobbes wrote:
Should be 6C3 / 2

Because half of the combination would be repeated in the other group.
For example, if group 1 has "A B C", group 2 must be "D E F"
Hence, the combination "D E F" in group 1 should be double counted.

6C3 simply means selecting 3 out of 6 and the order is not important.
Why divide the original expression by 2?

Say the set is: {A,B,C,D,E,F}
Using 6C3 select {A,B,D} and form 1 group - the other group would be formed automatically{C,E,F} - I dont see any "doubling".

Not so.

For example if we choose with $$C^3_6$$ the group {ABC} then the group {DEF} is left and we would have two groups {ABC} and {DEF} but then we could choose also {DEF}, so in this case second group would be {ABC}, so we would have the same two groups: {ABC} and {DEF}. So to get rid of such duplications we should divide $$C^3_6*C^3_3$$ by factorial of number of groups - 2!.

Check similar questions:
combinations-problems-95344.html?hilit=dividing%20objects%20order#p734396
split-the-group-101813.html?hilit=split
9-people-and-combinatorics-101722.html?hilit=divided%20equally%20into#p788744
ways-to-divide-99053.html?hilit=divided%20equally%20into#p763471

Hope it helps.
_________________
Re: m24 q16   [#permalink] 22 Oct 2010, 06:18
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# m24 q16

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