Counting by critical parenthesis method Suppose we have to arrange $n$ pairs of parentheses in such a way that every arrangements is matched i.e., number of left parentheses are always greater than or equal to number of right parentheses in any length of the chain from start.
I found the answer to be $ \frac{(2n)!}{n!(n+1)!} $.
This is how the proof goes 
S is the number of ways of arranging $n$ right and $n$ left parentheses in a row = $ \frac{(2n)!}{n!n!} $. Let T be the arrangement of $(n+1)$ right and $(n-1)$ left parentheses = $ \frac{(2n)!}{(n-1)!(n+1)!} $.
Now this is where I do no not understand. After that proof says that It can be shown that set of mismatched arrangements of parentheses in 'S' has bijective relation with the set of 'T'
After that the final matched arrangement = S-T= $ \frac{(2n)!}{n!(n+1)!} $
Please show me the bijection. I can not see it. And do not use catalan-number.
 A: This is the reflection principle in disguise.
You want to go from $(0,0)$ to $(n,n)$ on a grid in $n$ steps up and $n$ steps to the right without ever going below the diagonal, that is, without touching the line $L$ given by $y=x-1$. There are $\binom{2n}n$ such paths if we ignore the constraint. For paths that do touch $L$, reflect the part of the path up to the point where it first touches $L$ in $L$. This maps $(0,0)$ to $(1,-1)$. Each path from $(1,-1)$ to $(n,n)$ in $n+1$ steps up and $n-1$ steps to the right arises in exactly one way by this operation (since such a path necessarily at some point touches $L$). Thus there is a bijection between the paths from $(0,0)$ to $(n,n)$ that violate the constraint and the paths from $(1,-1)$ to $(n,n)$; and there are $\binom{2n}{n-1}$ of these paths.
To transfer this to the parenthesis paradigm, find the first closing parenthesis that makes the string unbalanced and toggle it and all parentheses before it (both opening and closing) to the other kind. This produces a string with $n+1$ opening and $n-1$ closing parentheses, and each such string is produced once. (I don’t know why the proof you have has it the other way around; you can instead toggle the parentheses after the first one that violates the constraint in order to get a bijection to strings with $n+1$ closing and $n-1$ opening parentheses.)
