To solve this problem, we recognize it as a classic application of Catalan numbers, which count the number of valid parentheses sequences (among other combinatorial structures like binary trees).

Key Insight:

The number of valid parentheses sequences with (n) pairs is given by the (n)-th Catalan number:
[C(n) = \frac{1}{n+1} \binom{2n}{n}]

Example for (n=3) (common problem case):

For 3 pairs of parentheses, the Catalan number is:
[C(3) = \frac{1}{4} \binom{6}{3} = \frac{1}{4} \times 20 = 5]

Valid Sequences for (n=3):

The 5 valid sequences are:
(((())), (()()), (())(), ()(()), ()()())

Answer: 5 (assuming the problem asks for the number of valid parentheses sequences with 3 pairs).

If the problem had a different (n), adjust using the Catalan formula, but 5 is a typical result here.

(\boxed{5})

作者声明:本文包含人工智能生成内容。