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 January 8th, 2017, 06:01 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79 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, 07:22 PM #2 Senior Member     Joined: Feb 2010 Posts: 630 Thanks: 102 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, 04:19 AM   #3
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Quote:
 Originally Posted by mrtwhs I'll take a stab and let others correct me if I'm wrong. I think it is $\displaystyle \dfrac{\binom{9}{3}\binom{6}{3}}{3!}=280$
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, 02:46 PM   #4
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Quote:
 Originally Posted by EvanJ 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?
I can make 4

$\{(1,2,3), (4,5,6), (7,8,9)\}$

$\{(1,4,7), (2,5,8 ), (3,6,9)\}$

$\{(1,5,9), (2,6,7), (3,4,8 )\}$

$\{(1,6,8 ), (2,4,9), (3,5,7)\}$

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, 03:44 PM #5 Member   Joined: Jan 2017 From: California Posts: 80 Thanks: 8 Is there a formula for this?
January 11th, 2017, 04:36 PM   #6
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 Originally Posted by dthiaw Is there a formula for this?
let $n = \text{ # of people total}$

let $m = \text{ # of people per group}$

and note that $n \pmod{m} = 0$

I suspect the formula is

$\text{# arrangements } = \dfrac{n-1}{m-1}$

January 11th, 2017, 05:56 PM   #7
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 Originally Posted by romsek let $n = \text{ # of people total}$ let $m = \text{ # of people per group}$ and note that $n \pmod{m} = 0$ I suspect the formula is $\text{# arrangements } = \dfrac{n-1}{m-1}$
nah, this can't be right. I'll keep looking

January 11th, 2017, 05:59 PM   #8
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 Originally Posted by romsek nah, this can't be right. I'll keep looking
yeah I bet it involves some computations without replacement

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