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  #1  
Old 05-01-2006, 04:46 PM
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Math Dunce needs help from Math Wizards

OK - I'm the worst in the world when it comes to this type of thinking, so I thought maybe someone here could help...

Here's the scenario:
I have 4 workshop sessions to run at work and 35 folks attending. Each session is to have 4 teams (3 teams of 9, 1 team of 8). I need to mix up the teams as much as possible so that everyone has the opportunity to work with as many different folks as possible.

So, that means I'll need a total of 16 eight or nine-person teams and I need to have the same people working together as little as possible.

I'm sure this is some type of simple math problem, but as I mentioned, it's way beyond my ability!
TIA!
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  #2  
Old 05-01-2006, 05:03 PM
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Not sure what you're asking Jay. There will be four teams per session, 4 sessions so 16 teams. Are you looking to do multiple teams per session or see how to mix up 35 people within the teams?
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Last edited by cmyX6go; 05-01-2006 at 05:12 PM.
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  #3  
Old 05-01-2006, 05:12 PM
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Quote:
Originally Posted by cmyX5go
Not sure what you're asking Jay. There will be four teams per session, 4 sessions so 16 teams. Are you looking to see how to mix up 35 people within the teams?
Yes - I need to keep the same people working with each other on a team to a minimum. Given that there are 4 sessions with 4 teams for each session, I have 4 opportunities to mix them up. Sorry if I'm not explaining clearly. We're doing the 4 sessions sequentially...

35 people total to mix & match...
Session 1: 35 folks to split into 4 teams
Team A = 9
Team B = 9
Team C = 9
Team D = 8

Session 2: Same 35 folks, 4 new teams...
Team E = 9
Team F = 9
Team G = 9
Team H = 8

Session 3: Same 35 folks, 4 new teams...
Team I
Team J
Etc...
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Old 05-01-2006, 06:38 PM
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Quick and dirty method:

1) Get 35 cards, set aside 3.
2) Divide the 32 cards into 2 stacks of 16.
3) Number each stack with this binary sequence:
0000
0001
0010
0011
0100
...
1111
4) Number the remaining 3 cards 0000, 0001, 0010 (this is somewhat arbitrary).
5) Hand out the cards.
6) On the first day the teams will be comprised of the people with numbers ending with the same two digits. There will be four teams: 00, 01, 10 and 11.
7) On the second day people with the first digit of 1 will move to the adjacent team.
8) On the third day people with the second digit of 1 will move to the adjacent team.
9) On the fourth day people with the third digit of 1 will move to the adjacent team.
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Old 05-01-2006, 10:02 PM
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There are only 10 types of people in this world:those who understand binary and those who don't.

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Old 05-01-2006, 10:26 PM
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Hey C3PO, can you come over and talk to my evaporator?

*that whizzing sound is that joke going over everyone's heads...
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Old 05-02-2006, 07:25 AM
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Quote:
Originally Posted by gresch
There are only 10 types of people in this world:those who understand binary and those who don't.

LOL Ron! I think it took me 1010 minutes to count from 0 - 15 in binary, but I finally got it! I have such a non-math mind...

Looks like this method will work! Thanks rayxi - very much appreciated!
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  #8  
Old 05-02-2006, 11:56 AM
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The solution isn't necessarily binary in nature. It was just convenient for explaining the method. My command of words isn't as good as my command of numbers.
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Old 05-02-2006, 01:17 PM
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Quote:
Originally Posted by X5Jay
LOL Ron! I think it took me 1010 minutes to count from 0 - 15 in binary, but I finally got it! I have such a non-math mind...

Looks like this method will work! Thanks rayxi - very much appreciated!
I love that joke... it's been a long time since I've been able to use it
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