Find the total number of ways in which A can win this series of games Two players A and B plays a series of $2n$ games. Each game can result in either a loss or win for A. Find the total number of ways in which A can win this series of games. (All games are to be played)
I'm feeling a bit confused with this one. How to solve?
 A: Since $2n$ games are going to be played each string of $2n$ games can be described by the games which A wins. Now A wins $k$ out of $2n$ games in $\dbinom{2n}{k}$ ways (order matters here when we count like this), so the total number of outcomes is equal to $$\sum_{k=0}^{2n}\dbinom{2n}{k}=(1+1)^{2n}=2^{2n}=4^n$$ by the Binomial Theorem. If A wins exactly $n$ games, then we have a tie. So there are $\dbinom{2n}{n}$ ties. Now due to symmetry, A wins exactly half of the rest of the strings, so the answer is $$\frac{\text{#total outcomes$-\#$ties}}{2}=\frac{4^n-\dbinom{2n}{n}}{2}=2^{2n-1}-\dbinom{2n-1}{n}$$

Plug in some values for $n$ to verify that this works. For example:


*

*$n=1$. Then there are $4^1$ outcomes: $ww, wl, lw, ll$ (where $w$ stands for a win of A and $l$ for a loss). In $1$ of those A wins.

*$n=2$. Then there are $4^2=16$ outcomes: $wwww, wwwl, wwlw, \dots$ and A wins in $$\frac{4^2-\dbinom{4}{2}}{2}=\frac{16-6}{2}=5$$ of them. B wins in other $5$ (by symmetry) and in $6$ outcomes we have a tie. You can check this by writing out all $16$ strings.

A: If the series runs until all $2n$ games are played, then the answer is:
$$\sum\limits_{k=n+1}^{2n}\binom{2n}{k}$$

You simply need to sum up the number of ways in which A can win:


*

*$n+1$ games out of $2n$ games

*$n+2$ games out of $2n$ games

*$\dots$

*$2n$ games out of $2n$ games

A: Since all games are to be played you want to count how many combinations of $0100 \ldots 0011$ there are (where $0 \to A$ loses and $1 \to A$ wins).
The formal result is
$$\sum\limits_{k=n+1}^{2n} (\text{combination where there are $k$ ones}) = \sum\limits_{k=n+1}^{2n}\binom{2n}{k}$$
A: if you seek the number of combinations to arrange A and B (for which players who wins), the number of ways if all games are played is (2n)!
