If I had a bracket style tournament, how many possibilities are there for a 8 team tournament. Once a team gets eliminated, they do not play a game for the 3rd place and below.
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$\begingroup$Let me just take the nice ASCII-Art @Kaligule provided:
1 2 3 4 5 6 7 8
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+-O-+ +-O-+ +-O-+ +-O-+ | | | | +---O---+ +---O---+ | | +-------O-------+ | Winner!We'll start by establishing a numbered order of the eight teams as seen in the graphic. For this, there are $8!$ possibilities. We don't really care for whether a team is "on the left side" or "on the right side" of a match, so we'll divide by $2!$ for every match. So for the initial setup of the entry matches, that gives us$$ n = \frac{8!}{2!2!2!2!} $$possibilities.
Now we determine the winner of each match. Clearly, it's either the left or the right side that wins, so that gives us $2$ possibilities each. So for the (ordered) outcomes of the entry matches, we have in total$$ n = 2^4 \cdot \frac{8!}{2!2!2!2!} = 8! $$possibilities.
Again, we don't really care what side each team is on for the second-level matches. So we divide by $2!$ for each. At the same time, there'll be two possible outcomes for each match again. It just so happens that we end up multiplying by $\frac{2^2}{2!2!}$ possibilities, which of course is one. So the number of (ordered) outcomes of the second level matches is$$ n = 8! \cdot 1 = 8! $$
Finally, we go for the final match. Divide by $2!$, multiply by $2$.. again it comes out to one. So, if I'm not entirely mistaken, our final amount of possibilities for the matches should be:$$ n = 8! $$
$\endgroup$ $\begingroup$We can uniquely encode the outcome in the tournament in an $8$-character string as follows. The code "abcdefgh" means
- a was the winner
- b came second
- c lost to a in the semifinals
- d lost to b in the semifinals
- e lost to a in the first round
- f lost to b in the first round
- g lost to c in the first round
- f lost to d in the first round
Since there are $8!$ such strings, there are $8!$ possible outcomes.
$\endgroup$ $\begingroup$You are not really specific about what you want to count, but lets assume the starting positions are fix already.
There will be $7$ games (just count the Os in the diagram below) and each game can have $2$ outcomes (left wins or right wins). The outcomes of the single games are all that matters for the tournament, so we only have to look at them.
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+-O-+ +-O-+ +-O-+ +-O-+ | | | | +---O---+ +---O---+ | | +-------O-------+ | Winner!Since there are 2 possibilities for every game you get $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ possibilities how the tournament might go.
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