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April 6th, 2013, 01:05 AM  #1 
Joined: Apr 2013 Posts: 4 Thanks: 0  Question in combinatorics
Hello all I had an exercise in maths that said: 4 Greeks, 6 Russians and 3 Albanians are candidates of a 4 member group, but in that group all these 3 nationalities must be present. The teacher said that we must separate the problem in 3 cases: 1st: we take 2 Greeks and 1 of each of the others 2nd: we take 2 Russians and 1 of each of the others 3rd: we take 2 Albanians and 1 of each of the others I tried to solve it by separating it in 2 cases: We take 1 of each nationality We take 1 of the 10 people that remain (subtract 1 of each and add them together But the answer was wrong, actually was 2 times the correct one. I didn't quite understand my teacher's explanation. Can someone explain why my method is wrong? Thank you 
April 6th, 2013, 02:21 AM  #2 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,946 Thanks: 114 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Question in combinatorics
1st. 2nd 3rd [color=#000000] What is the result you get?[/color] 
April 6th, 2013, 03:01 AM  #3 
Joined: Apr 2013 Posts: 65 Thanks: 0  Re: Question in combinatorics
At the second case you actually do a permutation. So you have to divide it by 2!

April 6th, 2013, 03:05 AM  #4 
Joined: Apr 2013 Posts: 4 Thanks: 0  Re: Question in combinatorics
That's my teacher's method, however at the 3rd part it's 4 x 3 x 6=72 so the total of the 3 parts is 360. That's the correct solution However, with my method I get a total of 720. Why? 
April 6th, 2013, 03:10 AM  #5 
Joined: Apr 2013 Posts: 4 Thanks: 0  Re: Question in combinatorics
@Drake: why am I doing a permutation? I am using the 10C1 formula. Can you explain please? Thanks 
April 6th, 2013, 04:13 AM  #6 
Joined: Apr 2013 Posts: 65 Thanks: 0  Re: Question in combinatorics
Well, the fourth person will be from the same nation with one of the 3. Let me give an example: and are same group, but you count them twice. 
April 6th, 2013, 11:07 AM  #7 
Joined: Apr 2013 Posts: 4 Thanks: 0  Re: Question in combinatorics
Hmm now I see... So I had to devide by 2! Thank you for the help! 

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