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How many ways are there to split a group of 6 boys into two

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Manager
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How many ways are there to split a group of 6 boys into two [#permalink] New post 02 Jul 2007, 14:16
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)

Explanations too, if you will :)
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 [#permalink] New post 02 Jul 2007, 14:26
combination is my weak spot but this looks like:

6 total people

2 groups

6C2 = 6!/4!2! = (6*5*4!)/(4!)(2) = 30/2 = 15
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 [#permalink] New post 02 Jul 2007, 16:43
dahcrap wrote:
I think it is 6C3


That's what i thought too...
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 [#permalink] New post 03 Jul 2007, 05:15
OA is (6C3)/2 = 10.

Does anyone know why we divide by two?
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 [#permalink] New post 02 Oct 2007, 16:35
I'd like to revive this question....I too would like to know why we divide by 2.

Thanks!
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 [#permalink] New post 02 Oct 2007, 18:11
6 boys. Pick one (doesn't matter). After you pick one, you have 5 boys left from which you have to pick 2 more to make a team of 3. So C(5,2) = 10. Once the first team of 3 is picked, you're left 3 boys and only one way to pick C(3,3). So 10*1= 10.
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 [#permalink] New post 02 Oct 2007, 19:13
Let there be mn objects split into m groups of n objects.

# of ways = (mn) ! / (m! * (n!)^m)

Hence 6! / (2! * (3!) ^2) = 10
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 [#permalink] New post 02 Oct 2007, 23:39
Bluebird wrote:
I'd like to revive this question....I too would like to know why we divide by 2.

Thanks!


Last note of the question, part in the parathesis is the answer.

6C3, gives number of ways 6 people can be organized into group of 3 people.
f.e. group A and group B OR group B and group A, which is 20 in total, but since the order of groups does not matter group A and group B is the same as group B and group A, thus we have to devide by 2.
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 [#permalink] New post 03 Oct 2007, 01:26
Consider the case with 6 boys:

Joe, John, James, Mike, Mark, Mitch

* Pick 3 boys for group A: Joe, John, James
--> This leaves 3 boys for group B: Mike, Mark, Mitch

* Pick 3 boys for group B: Mike, Mark, Mitch
--> This leaves 3 boys for group A: Joe, John, James

We see the two choices results in the same groups - this occurs because selection of the first group decided what boys would be in the second group. So half of the 6C3 results will be mirror images of each other --> divide the result by 2.
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Re: Quick Combinations question... [#permalink] New post 03 Oct 2007, 11:33
plaguerabbit wrote:
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)

Explanations too, if you will :)


here the caveat is that the order of the groups do not matter. Therefore it is half of the total, since we only need to figure out how many groups of 3 we can create out of the six, but since there is no A or B group, there is only A and A group. so it should look like this 6!/3!2!3!
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Re: Quick Combinations question... [#permalink] New post 03 Oct 2007, 11:46
plaguerabbit wrote:
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)

Explanations too, if you will :)


here the caveat is that the order of the groups do not matter. Therefore it is half of the total, since we only need to figure out how many groups of 3 we can create out of the six, but since there is no A or B group, there is only A and A group. so it should look like this 6!/3!2!3!
Re: Quick Combinations question...   [#permalink] 03 Oct 2007, 11:46
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