Bracket of 8 Tournament

Let me just take the nice ASCII-Art @Kaligule provided:

1   2   3   4   5   6   7   8
|   |   |   |   |   |   |   |
+-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: $$ n = 8! $$

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.

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.

1   2   3   4   5   6   7   8
|   |   |   |   |   |   |   |
+-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.