My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Thanks Tree2Thanks
  • 1 Post By romsek
  • 1 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
January 8th, 2017, 07:01 PM   #1
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 502
Thanks: 74

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?
EvanJ is offline  
 
January 8th, 2017, 08:22 PM   #2
Senior Member
 
mrtwhs's Avatar
 
Joined: Feb 2010

Posts: 600
Thanks: 87

I'll take a stab and let others correct me if I'm wrong. I think it is

$\displaystyle \dfrac{\binom{9}{3}\binom{6}{3}}{3!}=280$
mrtwhs is offline  
January 9th, 2017, 05:19 AM   #3
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 502
Thanks: 74

Quote:
Originally Posted by mrtwhs View Post
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.
EvanJ is offline  
January 9th, 2017, 03:46 PM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,111
Thanks: 580

Quote:
Originally Posted by EvanJ View Post
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.
Thanks from EvanJ
romsek is offline  
January 11th, 2017, 04:44 PM   #5
Member
 
Joined: Jan 2017
From: California

Posts: 31
Thanks: 2

Is there a formula for this?
dthiaw is offline  
January 11th, 2017, 05:36 PM   #6
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,111
Thanks: 580

Quote:
Originally Posted by dthiaw View Post
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

$\text{# arrangements } = \dfrac{n-1}{m-1}$
Thanks from dthiaw
romsek is offline  
January 11th, 2017, 06:56 PM   #7
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,111
Thanks: 580

Quote:
Originally Posted by romsek View Post
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
romsek is offline  
January 11th, 2017, 06:59 PM   #8
Member
 
Joined: Jan 2017
From: California

Posts: 31
Thanks: 2

Quote:
Originally Posted by romsek View Post
nah, this can't be right. I'll keep looking
yeah I bet it involves some computations without replacement
dthiaw is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
groups, made, people, ways



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How many ways can nine people line up? RedBarchetta Probability and Statistics 8 June 29th, 2014 09:15 PM
About minimal normal groups and subnormal groups Sheila496 Abstract Algebra 0 October 20th, 2011 09:45 AM
Homomorphism of Lie groups as groups but not as manifolds lime Abstract Algebra 3 October 25th, 2010 07:30 AM
Algorithm for Pairing 2 Groups of People math92037 Applied Math 3 October 7th, 2009 04:38 PM





Copyright © 2017 My Math Forum. All rights reserved.