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January 8th, 2017, 07:01 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 464 Thanks: 70  How Many Ways Can 9 People Be Made Into 3 Groups Of 3?
How many ways can 9 people be made into 3 groups of 3 without any two people being in the same group more than once?

January 8th, 2017, 08:22 PM  #2 
Senior Member Joined: Feb 2010 Posts: 583 Thanks: 80 
I'll take a stab and let others correct me if I'm wrong. I think it is $\displaystyle \dfrac{\binom{9}{3}\binom{6}{3}}{3!}=280$ 
January 9th, 2017, 05:19 AM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 464 Thanks: 70  That's not what I'm looking for. Two people cannot be in the same group in multiple arrangements. Each person will be grouped with 2 of the other 8 people, so that limits it to 4 arrangements, but I don't know if the answer is 4 or less than 4.

January 9th, 2017, 03:46 PM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 595 Thanks: 319  Quote:
$\{(1,2,3), (4,5,6), (7,8,9)\}$ $\{(1,4,7), (2,5,8 ), (3,6,9)\}$ $\{(1,5,9), (2,6,7), (3,4,8 )\}$ $\{(1,6,8 ), (2,4,9), (3,5,7)\}$ since each member shares exactly 1 group with every other member I don't think it's possible that there are any additional arrangements that can be added. it's possible that there is another set of 4 arrangements but 4 is going to be the maximum size of any set of valid arrangements.  
January 11th, 2017, 04:44 PM  #5 
Newbie Joined: Jan 2017 From: California Posts: 14 Thanks: 0 
Is there a formula for this?

January 11th, 2017, 05:36 PM  #6 
Senior Member Joined: Sep 2015 From: CA Posts: 595 Thanks: 319  
January 11th, 2017, 06:56 PM  #7 
Senior Member Joined: Sep 2015 From: CA Posts: 595 Thanks: 319  
January 11th, 2017, 06:59 PM  #8 
Newbie Joined: Jan 2017 From: California Posts: 14 Thanks: 0  

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