Posted by BWB on Friday, May 7, 2010 at 12:35am.
no,
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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