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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: 1,974 Thanks: 129 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Question in combinatorics 1st. $\binom{4}{2}\cdot\binom{6}{1}\cdot\binom{3}{1}=6\c dot 6 \cdot 3=108$ 2nd $\binom{6}{2}\cdot\binom{4}{1}\cdot\binom{3}{1}=15\ cdot 4\cdot 3=180$ 3rd $\binom{3}{2}\cdot\binom{6}{1}\cdot\binom{4}{1}=3\c dot 6\cdot 4=72$ [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: $G_{1}, R_{1}, A_{1}, G_{2}$ and $G_{2}, R_{1}, A_{1}, G_{1}$ 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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