${2n\choose n}$ counts the total number of collections of $n$ left and $n$ right parentheses. So if we can show that ${2n\choose n+1}$ is the number of ways to write those $n$ left and $n$ right parentheses in a way that is not legal, then we are done.
Note that a sequence is legal if when we read from left to right, we have always encountered at least as many left parentheses as right parentheses. Suppose a sequence $L$ is not legal. Then there is a least $k$ where there is a right parenthesis at position $k$ and equally many left and right parentheses before $k$ ( necessarily $\frac{k-1}{2}$ ). Now swap all left parentheses for right and all right for left in the first $k$ positions of $L$. This gives us a collection of $n+1$ left parentheses and $n-1$ right parentheses.
Conversely, let us say we are given a sequence $M$ of $n+1$ left parentheses and $n-1$ right parentheses, and let $k$ be the first position where there are more left parentheses up to that point than right. Flipping those parentheses gives us back a sequence of $n$ left parenthese and $n$ right parentheses that is not legal, because there are more right parentheses up to $k$ than left.
It should be clear that the second map and the first are inverses. Therefore, the number of illegal sequences of $n$ left and $n$ right parentheses is equal to the number of sequences, legal or not, of $n+1$ left and $n-1$ right parentheses, which is ${2n\choose n+1}$.
Therefore the number of legal sequences of parentheses is ${2n\choose n} - {2n\choose n+1}$.
But shouldn't the number be just same for matched arrangements? If there are more left parentheses then wouldn't it be unmatched arrangement?
They say "number of left parentheses are always greater than or equal to number of right parentheses in any length of the chain from start".
(Of course, the total number of left parenthesis and the total number of right parenthesis has to be the same.)
Also, for unmatched arrangement, they are just taking $n+1$ and $nā1$ right and left parentheses. But can't it be $n+2$ and $nā2$?
In another post, they are talking about flipping the parentheses. I don't understand why.
In joriki's answer, $\uparrow$ represents a left parenthesis, and $\rightarrow$ represents a right parenthesis.
The answer says "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 $y=xā1$".
The reason why we take the line $y=x-1$ is that we touch the line $y=x-1$ if and only if the sequence is not legal.
For example, $\uparrow\uparrow\rightarrow\rightarrow\rightarrow\uparrow\uparrow\uparrow\rightarrow\rightarrow$ represents $(()))((())$ which is not legal. It touches the line $y=x-1$ at $(x,y)=(3,2)$.
Next, Callus - Reinstate Monica's answer says
"Suppose a sequence $L$ is not legal. Then there is a least $k$ where there is a right parenthesis at position $k$ and equally many left and right parentheses before $k$ ( necessarily $\frac{k-1}{2}$ ). Now swap all left parentheses for right and all right for left in the first $k$ positions of $L$. This gives us a collection of $n+1$ left parentheses and $n-1$ right parentheses".
To understand these sentences, it might be helpful to consider an example. Take $(())\color{red})((())$ as an example of illegal $L$ where $n=5$. Then, $k$ is equal to $5$. Then, swapping all left parentheses for right and all right for left in the first $k=5$ positions of $L$ gives $\color{red}{))(((}((())$ where there are $5+1$ left parentheses and $5-1$ right parentheses.
In yet another post, they are talking about two separate terms: legal and balanced. I don't know how they are different.
As this question says, a legal sequence of parentheses is one in which the parentheses can be properly matched (each opening parenthesis should be matched to a closing one that lies further to its right).
As this question says, a sequence of length $2n$ that contains only $1,ā1$ is said to be balanced if all the $2n$ partial sums are non-negative, and the total sum is $0$.
Best Answer
Associate to a sequence of parantheses a sequence of $1$, $-1$, by $(\mapsto 1$, $)\mapsto -1$.
Balanced sequence $a_1 a_2 \ldots a_{2n}$ means
$ \sum_1^k a_i\ge 0$ for all $1\le k \le 2n$
$ \sum_1^{2n} a_i = 0$
Let's show first: a sequence of $1$, $-1$, is legal if and only if it is balanced.
Now, legal sequences are sequences obtained as follows: start from the empty sequence, perform several steps of form
$$ A,B \mapsto A, 1,-1, B$$ that is, we insert into the word $A,B$ at some position the piece $1, -1$ ($A$, $B$ are words in $1$, $-1$, possibly empty).
We can see that if $AB$ is balanced, so is the new sequence. Hence, by induction, all legal sequences are balanced.
Conversely: say we have a balanced sequence of $1$, $-1$. Consider the first appearance of $-1$ from the left. Before it, there must be $1$. So our sequence is of the form $A, 1, -1, B$. It is easy to see that $A,B$ is again balanced.
Therefore: legal is equivalent to balanced.
Consider now a sequence with $\sum_1^{2n} a_i = 0$ that is not legal. Then there must exist $k$ such that $\sum_1^k a_i < 0$. Consider $k$ smallest with such property. Then $k=1$, or $\sum_1^{k-1} a_i \ge 0$. The cannot be $> 0$, since then $a_k\le -2$. Therefore we have $\sum_1^{k-1}a_i = 0$, and $a_k=-1$ $\square$