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In how many different ways can a group of 8 be divided into

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VP
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In how many different ways can a group of 8 be divided into [#permalink] New post 09 Nov 2007, 10:49
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In how many different ways can a group of 8 be divided into 4 teams of 2 people each?
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 [#permalink] New post 09 Nov 2007, 11:09
8C2 = 8!/2!6! = 28

:)
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 [#permalink] New post 09 Nov 2007, 11:29
KillerSquirrel wrote:
8C2 = 8!/2!6! = 28

:)


Wouldn't 8C2 be the number of ways you can pick 2 elements from a group of 8, where order doesn't matter? We are looking for how many ways can you arrange 8 elements in groups of 2.

I get 8C2 * 6C2 * 4C2 * 2C2 = 10080.

Correct me if I'm wrong please
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Re: Permutation/Combo problem -Tricky one [#permalink] New post 09 Nov 2007, 12:37
Ozmba2006 wrote:
In how many different ways can a group of 8 be divided into 4 teams of 2 people each?


105. Start with the question, how many ways can you divide a group of 4 into 2 teams. The answer is 3. Suppose our group is A B C D. Then the possible teams are:

AB CD
AC BD
AD BC

(It's 4C2/2, because determining one of the teams automatically determines the other one.)

Now consider the case of a group of 6 divided into 3 teams. Our group: A B C D E F. A has to be on some team, and there are 5 possibilities: AB, AC, AD, AE, or AF. For each of these possibilities, the rest of the members form a group of 4 divided into 2 teams. So the overall result is

5*(4C2/2) = 15.

Now we're up to our case. Group = A B C D E F G H. Again, A must be on a team--there are 7 possibilities. For each of those possibilities, there are 15 ways of dividing the remaining 6 members into 3 teams. So total possibilities = 7 * 15 = 105.
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 [#permalink] New post 09 Nov 2007, 14:12
Hmmm ... is it 8C2 divided by 4.

If 8C2 is all the combinations of 2 from 8 elements.

Would you take that number and divide by 4 since the question wants to know how many different 4 team combos there are?

My guess is 7.
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 [#permalink] New post 09 Nov 2007, 18:17
Im goin w/ 7 as well.
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Re: Permutation/Combo problem -Tricky one [#permalink] New post 10 Nov 2007, 08:24
johnrb wrote:
Ozmba2006 wrote:
In how many different ways can a group of 8 be divided into 4 teams of 2 people each?


105. Start with the question, how many ways can you divide a group of 4 into 2 teams. The answer is 3. Suppose our group is A B C D. Then the possible teams are:

AB CD
AC BD
AD BC

(It's 4C2/2, because determining one of the teams automatically determines the other one.)

Now consider the case of a group of 6 divided into 3 teams. Our group: A B C D E F. A has to be on some team, and there are 5 possibilities: AB, AC, AD, AE, or AF. For each of these possibilities, the rest of the members form a group of 4 divided into 2 teams. So the overall result is

5*(4C2/2) = 15.

Now we're up to our case. Group = A B C D E F G H. Again, A must be on a team--there are 7 possibilities. For each of those possibilities, there are 15 ways of dividing the remaining 6 members into 3 teams. So total possibilities = 7 * 15 = 105.


After spending too much time on this problem, I agree with this approach.
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 [#permalink] New post 10 Nov 2007, 15:39
My approach was different. Not sure if it's correct.

8C2 + 6C2 + 4C2 + 2C2 = 50

28 + 15 + 6 + 1 = 50
  [#permalink] 10 Nov 2007, 15:39
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