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Combinatorics #6

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Combinatorics #6

by papgust » Sun Oct 18, 2009 5:18 am
Several teams take part in a competition, each of which must play one game with all the other teams. How many teams took part in the competition if they played 45 games in all?

A. 5
B. 10
C. 15
D. 20

Solution and explanation pls?
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by NikolayZ » Sun Oct 18, 2009 8:00 am
Hi there !
is the answer is B?

I did it without combinatorics. Try solving backwards from the answers.
So, if there were 5 teams.

1st must play with - 2 3 4 5
2nd -3 4 5
3d -4 5
4th - 5
5 (already played with all possible teams)
clearly - there are less games played than 45.

Try 10 the same way. You ll get 45 games total.
9+8+7+6+5+4+3+2+1=45
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by papgust » Sun Oct 18, 2009 8:13 am
Hi NikolayZ,

Thanks for the solution. The answer is indeed 10. But i'm looking to solve this problem by combinatorics theory. Even i solved this by backsolving. I'm looking to build my fundamentals in combinatorics as i'm very weak in this subject.

Can you solve this prob using combinations and help me understand how the theory is applied to this kind of a prob?
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