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 November 25th, 2017, 09:33 AM #1 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 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: 6,755 Thanks: 695 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: 250 Thanks: 3 I don't understand why that formula is used for it? How was it discovered?
November 26th, 2017, 12:49 PM   #4
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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
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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: 3,264 Thanks: 902 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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