## Question in combinatorics

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### 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
michaelg9
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### Re: Question in combinatorics

1st. $\binom{4}{2}\cdot\binom{6}{1}\cdot\binom{3}{1}=6\cdot 6 \cdot 3=108$
2nd $\binom{6}{2}\cdot\binom{4}{1}\cdot\binom{3}{1}=15\cdot 4\cdot 3=180$
3rd $\binom{3}{2}\cdot\binom{6}{1}\cdot\binom{4}{1}=3\cdot 6\cdot 4=72$

What is the result you get?
Last edited by ZardoZ on Sat Apr 06, 2013 6:11 am, edited 1 time in total.

ZardoZ
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Location: Greece, Thessaloniki

### Re: Question in combinatorics

At the second case you actually do a permutation. So you have to divide it by 2!
Drake
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Joined: Mon Apr 01, 2013 5:00 pm

### 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?
michaelg9
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Joined: Sat Apr 06, 2013 3:51 am

### Re: Question in combinatorics

@Drake: why am I doing a permutation? I am using the 10C1 formula. Can you explain please?
Thanks
michaelg9
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Joined: Sat Apr 06, 2013 3:51 am

### 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.
Drake
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Joined: Mon Apr 01, 2013 5:00 pm

### Re: Question in combinatorics

Hmm now I see... So I had to devide by 2!
Thank you for the help!
michaelg9
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Posts: 4
Joined: Sat Apr 06, 2013 3:51 am