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October 26, 2014

October 26, 2014

Posted by **BWB** on Friday, May 7, 2010 at 12:35am.

if there are eight teams and all teams play each other twice?

(8*7)*2= 56*2=112 games

- math: Am I right? -
**Reiny**, Friday, May 7, 2010 at 1:01amno,

number of games in one round = C(8,2) = 28

so for two rounds that would be 56

When you take 8*7 that would include cases such as

teamA vs Team B and Team B vs Team A, but that is the same game and in your first 8*7 or 56 games you have 28 such pairs.

So you must divide by 2.

In other words, you have taken the order into play, but this is not a permutation, but rather a combination.

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