Let's call a sequence with $n$ occurrences of $L$ and $n$ occurrences of $R$ a balanced sequence if on counting from left to right the number of $L$'s is always greater than or equal to the number of $R$'s. How many such balanced sequences are there?
The answer given in my textbook is $$ {{2n}\choose{n}} \cdot \frac{1}{n+1}$$
I could devise no counting arguments for this.
EDIT: The number of balanced and unbalanced sequences is $$ {{2n}\choose{n}}$$
For finding the unbalanced sequences,I've taken hint that interchanging $R$ and $L$ from the place where the number of occurrences of $R$ exceeds the number of occurrences of $L$. This gives the number of such sequences to be$$ {{2n}\choose{n+1}} $$
I infer that this interchanging produces a bijection between unbalanced sequences with $n$ occurrences of $L$ and $n$ occurrences of $R$ and $n-1$ occurrences of $L$ and $n+1$ occurrences of $R$.
But I am not sure if my reason for this bijection is correct.After the interchange the number of occurrences of $R$ is always greater than that of $L$.So all the sequences will be unbalanced.But how could I prove that this will contain exactly the same number of unbalanced sequences as the original sequence?