
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 1,675 Thanks: 8  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! 

Tags 
combinatorics, question 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Combinatorics question  rummi  Advanced Statistics  0  January 16th, 2013 02:58 PM 
Combinatorics QUESTION 13  maximus101  Applied Math  1  October 22nd, 2012 03:58 AM 
Combinatorics QUESTION  maximus101  Number Theory  1  October 21st, 2012 04:28 PM 
Combinatorics QUESTION 33  maximus101  Applied Math  1  October 20th, 2012 01:05 PM 
Combinatorics question  proglote  Algebra  3  May 1st, 2011 06:06 AM 