Counting all possible combinations of a playoff tournament I have a question about a chess tournament, and I need to count all possible different tournaments. It's a playoff so that after the first round there are $2^n$ players, after the second round there are $2^{(n-1)}$ and so on....
So, how do I count it? Thanks!
 A: For a tournament starting with $2^n$ players, let $A_n$ be  the number off all possible match results. That is, $A_n$ is the number of different player pools for the next round.
So, I am calling two tournaments different if there is a round where the player pools are different. I am ignoring the different ways for which a particular player pool in round $i$
can be obtained from the games played in the previous round. 
Then $A_{n+1} = B_{n+1}\cdot A_n$, where $B_{n+1}$ is the number of different match results in the first round of a tournament with $2^{n+1}$ players.
Now,
$B_n= { 2^n\choose 2^{n-1}}$.
$A_1$=2
$A_2= B_2 \cdot A_1=  { { 4\choose 2}}\cdot 2 $
$A_3= B_3\cdot A_2= {8\choose 4} \cdot  { { 4\choose 2}}\cdot2 $
$A_4= B_4\cdot A_3={16\choose 8}\cdot {8\choose 4}\cdot { { 4\choose 2}}\cdot2 $ 
$\ \ \ \vdots$
$A_n={2^n\choose 2^{n-1}}\cdot{2^{n-1}\choose 2^{n-2}}\cdot\cdots\cdot{8\choose 4} \cdot  { { 4\choose 2}}\cdot2 $.
Of course, the above can be simplified considerably:
Note that  $2^n-2^{n-1}=2^{n-1}$. 
So ${ 2^n\choose 2^{n-1}}={(2^n)!\over (2^{n-1})!(2^{n-1})!}$, and we have:
$$
\eqalign{
A_n&=
{(2^n)!\over (2^{n-1})! (2^{n-1})!}\cdot{(2^{n-1})!\over(2^{n-2})!(2^{n-2})!}\cdot\cdots\cdot{8!\over 4!\cdot4! } \cdot  { { 4!\over 2!\cdot2!}}\cdot2 \cr
&=
{(2^n)!\over (2^{n-1})!\cdot 1}\cdot{1\over(2^{n-2})!\cdot1}\cdot\cdots\cdot{1\over 4!\cdot1 } \cdot  { {1\over 2!\cdot1}}\cdot1 \cr 
&={(2^n)!\over (2^{n-1})!(2^{n-2})!\cdots4!\cdot 2!}.
}
$$
