I was trying a question that required me to find the number of ways of arranging $y$ opening brackets and $x$ closing brackets, such that there is no prefix where the number of closing brackets exceed the number of opening brackets. $y\ge x$
Obviously, the number of arrangements is $\binom{x+y}{y}$.
Now to find the number of incorrect sequences, I noticed that there must be a minimal(possibly empty) perfectly balanced sequence, followed by a closing bracket. The length of this sequence cannot exceed $2x-2$ otherwise it would require more closing brackets than available. The rest can be arranged randomly, as this sequence is already wrong. So the equation I ended up with is $$\sum_{i=0}^{x-1} C_i \times \binom{x+y-2i-1}{y-i}$$ How do I simplify this? I know almost certainly that it evaluates to $\binom{x+y}{y+1}$
If there is a better way to solve it, I would like to know that too.
I find it easiest to think in terms of up-down paths [PDF]. You want to count the paths consisting of $y$ up-steps and $x$ down-steps that drop below the baseline. The first step dropping to height $-1$ must be an odd-numbered step, say $2k+1$. At that point the path has $k$ up-steps and $k+1$ down-steps and ends in a down-step. The remainder of the path must therefore have $y-k$ up-steps and $x-k-1$ down-steps.
Now reflect that remainder in the height $-1$ line; that changes each of its up-steps into a down-step and vice versa, so the reflection has $x-k-1$ up-steps and $y-k$ down-steps, so the resulting path has altogether $x-1$ up-steps and $y+1$ down-steps and therefore ends at a height of $$(x-1)-(y+1)=x-y-2=-(y-x+2)\;.$$ (The original path, before the reflection, ended at a height of $y-x$.) Every path that ends at height $-(y-x+2)$ after after $x+y$ steps is uniquely generated in this way, since the reflection procedure is reversible, and there are $\binom{x+y}{y+1}$ such paths, so there are, as you thought, $\binom{x+y}{y+1}$ bad bracket strings with $y$ opening and $x$ closing brackets.
This is just a slight generalization of the proof that $C_n=\frac1{n+1}\binom{2n}n$ using André’s reflection method.
Added: Since there are $\binom{x+y}y$ bracket strings altogether,there are
$$\binom{x+y}y-\binom{x+y}{y+1}=\left(1-\frac{x}{y+1}\right)\binom{x+y}y=\frac{y-x+1}{y+1}\binom{x+y}y$$
good bracket strings.