This is almost the definition of a "$1$-factorization of $K_{2k}$", except that a $1$-factorization has an unordered set of matchings instead of a sequence of rounds. Since there are $2k-1$ rounds, this means that there are $(2k-1)!$ times as many tournaments, according to the definition above, as there are $1$-factorizations.
Counting $1$-factorizations of $K_{2k}$ seems to be a nontrival problem; see the Encyclopedia of Mathematics entry. The number of $1$-factorizations of $K_{2k}$ is OEIS sequence A000438. Also, see this paper (also here) for a count in the $k=7$ case.
We have $2^n$ players, so basically there are $\frac{2^n}{2} = 2^{n-1}$ games in the first round. The combination of possible players going to the next round is given by $2^{2^{n-1}}$ because in each game we have two possible winners.
In the second round, we will have $2^{n-1}$ players and $2^{n-2}$ games. Following the same reasoning, the combination of winners is given by $2^{2^{n-2}}$.
And so on, until we reach the final, for which we have $2^1$ possible winners.
Therefore, the number of possibilities we have is given by the summation of all possible outcomes in different rounds:
$\displaystyle \sum_{i=1}^{n} 2^{2^{n-i}}$.
If we take only the power, we can see that it follows a geometric series form, so we can calculate the sum of all powers $= 2^n - 1$
Therefore, the final number of possibilities is $2^{2^n - 1}$.
The explanation of Tony K is much simpler, however I couldn't figure out why the number of games was $2^n - 1$ until I developed this reasoning.
Hope it helps!
Best Answer
you have to chose the first and the second teams. for the first team you have $\binom{P}T$ and for the second team you have $\binom{P-T}T$ however you are counting each match twice in this case so the total number of possible matches is $$\frac{\binom{P}T\binom{P-T}T}{2}$$
The number of possible combinations of matches per round (assuming you get as many teams to play as possible in each round) is $$\prod_{n=0}^{n=2\lfloor{T/2P}\rfloor-1}\binom{P-(Tn)}T/(\lfloor{T/2P}\rfloor!(2^{\lfloor{T/2P}\rfloor})$$