Subject: Puzzles & Brain Teasers : Football Matches

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alternative solution procedures, or questions about this puzzles, feel free
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Problem seems to be a combination. Using the nCr(n!/(r!(n-r)!) notation,
and knowing that we are choosing 2 teams at a time, the formula for 78 games
would reduce to:
78 = n!/(2!(n-2)!)
if imagining a fraction bar would look like:
78 = n!
2!(n-2)!
which equals 78*2! = n!/(n-2)!
which equals 156 = n!/(n-2)!
which equals 156 = n(n-1) =====> imagine 5!/(5-2)! = 120/6
which is 5*4
then you have n^2 - n - 156 = 0
which can be factored to (n-13)(n+12) = 0
so n = 13
(n = -12 is ignored because you cannot have negative teams)