My Math Forum > Math How many ways...
 User Name Remember Me? Password

 Math General Math Forum - For general math related discussion and news

 November 25th, 2017, 09:33 AM #1 Senior Member   Joined: Nov 2011 Posts: 209 Thanks: 2 How many ways... How many ways I can seprate 7 disks to 2 groups? There is a formula for that?! Thank for answering me!
 November 25th, 2017, 12:41 PM #2 Global Moderator   Joined: May 2007 Posts: 6,541 Thanks: 592 Essentially you use the binomial distribution. Net result is $2^7$ ways.
 November 25th, 2017, 10:33 PM #3 Senior Member   Joined: Nov 2011 Posts: 209 Thanks: 2 I don't understand why that formula is used for it? How was it discovered?
November 26th, 2017, 12:49 PM   #4
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 12,771
Thanks: 862

Quote:
 Originally Posted by shaharhada How many ways I can seprate 7 disks to 2 groups?
Are the disks alike or different from each other?

November 26th, 2017, 01:23 PM   #5
Global Moderator

Joined: May 2007

Posts: 6,541
Thanks: 592

Quote:
 Originally Posted by shaharhada I don't understand why that formula is used for it? How was it discovered?
https://en.wikipedia.org/wiki/Binomial_distribution

Above will give you a complete picture.

 November 27th, 2017, 06:28 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,248 Thanks: 887 For each disk there are two possibilities- it may be put into group 1 or group 2. Now use the "fundamental theorem of counting": If A can happen in m ways and B can happen, independently of A, in n ways they can happen together in mn ways. The first disk can be put into either of two groups- that can happen in 2 ways and, whatever group the first disk in put into, the second disk can be put into either of the two groups. The two disks can be put into the two groups in any of $\displaystyle 2(2)= 4= 2^2$ ways. If we write "A" for one group and "B" for the other those 4 ways are "AA", "AB", "BA", and "BB". Now the third disk can be put in either of those groups so now there are 4(2)= 8 ways to do it: "AAA", "ABA", "BAA", "BBA", "AAB", "ABB", "BAB". and "BBB". Notice that those are just the 4 ways above with "A" appended then the 4 ways above with "B" appended so $\displaystyle 4+ 4= 2(4)= 8= 2^3$ ways to put 3 objects in two groups. You can continue that way to see that there are $\displaystyle 2^7$ ways to put 7 things into 2 groups.

 Tags ways

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post shunya Elementary Math 10 December 10th, 2015 06:41 AM LuLu99 Probability and Statistics 2 April 30th, 2014 05:13 PM need2knw Number Theory 5 August 18th, 2013 09:06 PM need2knw Applied Math 0 August 17th, 2013 02:51 PM sivela Abstract Algebra 1 January 25th, 2010 09:25 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2018 My Math Forum. All rights reserved.