Counting votes, as long as one has more votes all the way through. 
*

*Two competitors won $n$ votes each.
How many ways are there to count the $2n$ votes, in a way that one competitor is always ahead of the other?

*One competitor won $a$ votes, and the other won $b$ votes. $a>b$.
How many ways are there to count the votes, in a way that the first competitor is always ahead of the other?
(They can have the same amount of votes along the way)
I know that the first question is the same as the number of different legal strings built of brackets, which is equal to the Catalan number; $\frac{1}{n+1}{2n\choose n}$, by the proof with the grid.
I am unsure about how to go about solving the second problem.

We can use a similar idea to the solution to #1:
The total number of ways the votes can be counted is $a+b \choose a$, so we have to subtract the number of ways competitor B can get ahead of competitor A: 
In any count where B gets ahead of A, change all the votes up to and including the first vote where B takes the lead; this will give a count where A gets $a+1$ votes and B gets $b-1$ votes.
Conversely, in any count where A gets $a+1$ votes and B gets $b-1$ votes, changing all the votes up to and including the first vote where A gets the lead gives a vote sequence of the first type.
Thus the answer is $\binom{a+b}{a}-\binom{a+b}{a+1}$.
After referring to the link provided by Brian Scott, I realized that I am answering a slightly different question, which is the number of ways the votes can be counted so that A never gets behind B in the count, not the number of ways that A always remains ahead of B.  (In #1, the answer is $C_n$ to this question, not to the question in which A always remains ahead of B.)  
For the version of the problem as stated, use the fact that A must get the first vote and then apply the same argument to the remaining $a+b-1$ votes.
