Counting Parentheses

There are ${20 \choose 10}$ ways in which a score of $10:10$ can be achieved in a football game between teams $A$ , $B$ . By the Reflection Principle, there are ${20 \choose 9}$ ways in which a score of $10:10$ can be achieved for which $A$ is behind in the score at some time. Thus there are ${20 \choose 10} - {20 \choose 9} = 16796$ ways in which a score of $10:10$ , with team $A$ never falling behind in the score.

Regarding a left brace as a goal for team $A$ , and a right brace as a goal for team $B$ , the answer to the question is again $\boxed{16796}$ .

Let $f(n)$ be the number of ways we can arrange $n$ pairs of parentheses.

$f(0)=1$ because $0$ pairs of parentheses aren't rearrengeable.

Suppose we have $n$ pairs of parentheses, take the first pair $p$ : then there can be $k$ pairs of parentheses into $p$ and $n-1-k$ pairs out of $p$ for all possible $k$ .

So, $f(n)=\displaystyle \sum_{k=0}^{n-1} f(k)f(n-1-k)$

$f(10)= \boxed{16796}$

From the question, we know that we are basically finding the 10th Catalan number, which is 16796.