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8 teams in conference

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by mberkowitz » Fri Sep 26, 2008 2:49 pm
there are 8 teams in a conference. how many games are possible if each team plays eachother twice?

OA 56

if each team played the other only once, the answer would be 8c2 or 28. in this case, bc they play eachother twice, the total number of games will be 28 times 2 or 56.

can somebody please explain the theory behind this counting method? i understand the ratoinal behind the factorial in this case, but have trouble distiguishing between cases in which team only play once, twice, and how to avoid counting a v b and b v a as seperate games...

thanks
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by Morgoth » Fri Sep 26, 2008 3:00 pm
A B C D E F G H

Games played

AB AC AD AE AF AG AH----7 GAMES
BC BD BE BF BG BH--------6 GAMES
CD CE CF CG CH-----------5 GAMES
DE DF DG DH----------------4 GAMES
EF EG EH---------------------3 GAMES
FG FH-------------------------2 GAMES
GH-----------------------------1 GAME

1+2+3+4+5+6+7 = 28 GAMES

Each team plays with the other twice

28*2 = 56 Games

Hope its clear.
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by mberkowitz » Fri Sep 26, 2008 3:05 pm
appreciate that, i guess the best method to learn these ones is to picture it like that, i just have trouble going from there to the factorial. i.e. thats a great method if you have 4 mins, but clearly not the fastest option.
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by Morgoth » Fri Sep 26, 2008 3:17 pm
mberkowitz wrote:appreciate that, i guess the best method to learn these ones is to picture it like that, i just have trouble going from there to the factorial. i.e. thats a great method if you have 4 mins, but clearly not the fastest option.
I thought you were looking for clarity rather than which method to follow.

If you are looking for the method its hard to beat 8C2, which is the most effective on such type of questions.
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by mberkowitz » Fri Sep 26, 2008 4:05 pm
i was looking for an explanation of the method in this problem. in other words this problem stumped me even though i know the method and can usually execute it well, as i find the method arbitrary at times.

thanks for the help.
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by stop@800 » Sat Sep 27, 2008 3:24 am
8c2 is clear
when a person is playing once with everyother

so 2 * 8c2 means each person played twice.

Perhaps I could not understand your problem well.....
please help me understanding your doubt.

Thanks
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by mberkowitz » Sat Sep 27, 2008 4:31 am
i was having trouble getting 8c2 quickly bc i was tending to worry about counting teams more than once (counting a v b and b v a as two seperate games). id imagine learning that property as is is the best way for me to proceed. thanks.
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