Outcome of a Tournament (Combinatorics) I'd like a tip or a hint in this elementary problem:
Three Tennis Players A, B and C will compete in a tournament with 10 rounds (one play per round). Two players face each every round and the winner faces the third player (that did not participate the round) in the next round. The winner of the last play (the tenth round) is the tournament's champion. In the initial round A and B face each other. Therefore, a possible outcome is the sequence ACCAAABBBC, with C declared champion. Determine the number of possible outcomes to the sequence of games in which B is champion.
 A: Hint.
Work backwards from $B$ as the final winner (round 10). The previous winner (round 9) could have been any of $A,B,C$. Suppose it was $A$. That means $A$ played $B$ in round 10. In this case there are now two possibilities for previous winner (round 8). But if the round 9 winner was $B$, there are three possibilities for the previous (round 8) winner.
A: Using @almagest hint I was able to solve the problem, with some generalization:
The generalized problem is: How many ways can $B$ win the game in the nth round?
For $n=1$ and $n=2$ the answer is $1$
($B$ and $BB$)
Suppose that we know how many ways can $B$ win the game in the nth round and lets call that number $W(n)$, so that $W(1)=W(2)=1$
To calculate $W(n)$ it is sufficient to note that, if $B$ wins the nth game, then we have three possibilities: $A, B$ or $C$ won the (n-1)th game. 
If $A$ won the (n-1)th game, it was playing against $C$ (otherwise $B$ wouldnt play the n-th game). So that in the (n-2)th game, either $A$ played against $B$  (and $A$ won) or $C$ played against $B$ (and $C$ won), but the different ways it could have happened is precisely $W(n-2)$
If $C$ won the (n-1)th game, it was playing against $A$, so that in the (n-2)th game either $C$ played against $B$ or $A$ played against $B$ and, as before, there are $W(n-1)$ ways of doing that.
If $B$ won the (n-1)th game there are $W(n-1)$ ways of doing that, so we have
$W(n) = W(n-1) + 2W(n-2)$
With $W(1)=W(2)=1$
Solving the linear recurrence we get
$W(n) = \frac{1}{3} \left ( 2^n + (-1)^{n+1} \right)$
And then $W(10) = 341$
