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November 25th, 2017, 10:33 AM  #1 
Senior Member Joined: Nov 2011 Posts: 170 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, 01:41 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,398 Thanks: 546 
Essentially you use the binomial distribution. Net result is $2^7$ ways.

November 25th, 2017, 11:33 PM  #3 
Senior Member Joined: Nov 2011 Posts: 170 Thanks: 2 
I don't understand why that formula is used for it? How was it discovered? 
November 26th, 2017, 01:49 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,074 Thanks: 722  
November 26th, 2017, 02:23 PM  #5  
Global Moderator Joined: May 2007 Posts: 6,398 Thanks: 546  Quote:
Above will give you a complete picture.  
November 27th, 2017, 07:28 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,875 Thanks: 766 
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. 

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