A legal sequence of parentheses is one in which the parentheses can be properly matched,like ()(()). I should calculate the number of legal sequences of length $2n$, the answer is $C_n = {2n \choose n}  {2n \choose n + 1}$, how can it be proved without recurrence and induction?

I edited the question please verify that is what you meant. For future reference, use LaTeX notation for math – NazimJ Nov 09 '18 at 13:23

Thanks will keep it in mind – stackoverload Nov 09 '18 at 13:24
1 Answers
${2n\choose n}$ 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 $\frac{k1}{2}$ ). 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 $n1$ right parentheses.
Conversely, let us say we are given a sequence $M$ of $n+1$ left parentheses and $n1$ 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 $n1$ right parentheses, which is ${2n\choose n+1}$.
Therefore the number of legal sequences of parentheses is ${2n\choose n}  {2n\choose n+1}$.
 5,192
 15
 24