Can you use combinatorics rather than a tree for a best of five match? 
The Chicago Cubs are playing a best-of-five-game series (the first team to win 3 games win the series and no other games are played) against the St. Louis Cardinals.  Let X denotes the total number of games played in the series. Assume that the Cubs win 59% of their games versus their arch rival Cardinals and that the probability of winning game is independent of other games. 
(a)  calculate the mean and standard deviation for X.

In this problem I think I understand how to calculate the mean and standard deviation, but first you have to calculate the total combinations of plays.  Is there a way using combinations or permutations to derive all possibilities rather than drawing a tree?
The tricky thing with this using combinatorics is that its a best of five.  So I don't see how to use permutations directly because as soon as you have 3 wins the series is over--meaning you'll over count if you don't take this into consideration.
I'd appreciate any feedback.  Thank you!

k=3
$$
\begin{array}{c|c|c}
\text{Loss} & \text{Win} & \text{Combinations}\\
\hline
3 & 0 & 1\\
2 & 1 & 3\\
1 & 2 & 3\\
0 & 3 & 1\\
\end{array}
$$
Total Combinations = 8
k=4
$$
\begin{array}{c|c|c}
\text{Loss} & \text{Win} & \text{Combinations}\\
\hline
1 & 3 & 4\\
2 & 2 & 6\\
3 & 1 & 4\\
4 & 0 & 1\\
\end{array}
$$
**Here have to subtract 1 from total combinations because cannot have WWWL.
Total combinations = 15-1 = 14
k=5
$$
\begin{array}{c|c|c}
\text{Loss} & \text{Win} & \text{Combinations}\\
\hline
2 & 3 & 10\\
3 & 2 & 10\\
4 & 1 & 5\\
5 & 0 & 1\\
\end{array}
$$
**Here have to subtract from total combinations because cannot have WWWLL and LWWWL.
Total combinations = 26-2 = 24
 A: Let us solve a more general problem. Imagine we have a series between two teams, A and B. Team A wins a game with probability $p$, and we make the independence assumption of the OP. Let us suppose that the first team to win $w$ games wins the series.  Let $X$ be the number of games. Then $w\le X\le 2w-1$. Yours is the case $w=3$.
For any $k$ in this interval, we find the probability that $X=k$. In order for the series to have length exactly $k$, one of the teams has to have won $w-1$ of the first $k-1$ games, and has to win the last game.
The probability that A wins the series in $k$ games is therefore 
$$\binom{k-1}{w-1}p^{w-1}(1-p)^{w-k}p=\binom{k-1}{w-1}p^w(1-p)^{k-w},$$
and the probability that B wins the series  in $k$ games is 
$$\binom{k-1}{w-1}(1-p)^wp^{k-w}.$$
The events A wins the series in $k$ games and B wins the series in $k$ games are disjoint, so to find $\Pr(X=k)$ we add the two expressions above.
Remark: A distribution closely related to the above calculations is the negative binomial. For details, Wikipedia has a discussion, as do many introductory probability books. 
For the case $w=3$, we don't need all this machinery, and can simply draw a tree.  
Let $p=0.59$. We have $\Pr(X=3)=p^3+(1-p)^3$. For $k=4$, we get $3p^3(1-p)+3(1-p)^3p$. For $k=5$ we get $6p^3(1-p)^2+6(1-p)^3p^2$. All these expressions can be simplified somewhat.
