Example 2-60 $P=a_{1} a_{2} \cdots a_{n}$ is the product of $n$ numbers. According to the associative law of multiplication, without changing their order, only by adding parentheses to indicate the order of multiplication, how many ways of multiplication are there?
Let $p_{n}$ denote the number of different ways to insert $n-1$ pairs of parentheses into the product of $n$ numbers.
$$
\begin{array}{c}
p_{n}=p_{1} p_{n-1}+p_{2} p_{n-2}+\cdots+p_{n-1} p_{1} \\
p_{1}=p_{2}=1
\end{array}
$$

Obviously, $p_{k}=C_{k+1}, k=1,2, \cdots, n$. For example, when $k=4$:
$$
p_{4}=C_{5}=5
$$

Figure 2-22 shows the one-to-one correspondence between different multiplication orders and the triangulation of a pentagon. The product $a \cdot b$ is represented by a binary tree with the operator * as the branching node.
The 5 multiplication orders are
$$
\begin{array}{l}
\left(\left(a_{1}\left(a_{2} a_{3}\right)\right) a_{4}\right),\left(\left(\left(a_{1} a_{2}\right) a_{3}\right) a_{4}\right),\left(\left(a_{1} a_{2}\right)\left(a_{3} a_{4}\right)\right) \\
\left(a_{1}\left(a_{2}\left(a_{3} a_{4}\right)\right)\right),\left(a_{1}\left(\left(a_{2} a_{3}\right) a_{4}\right)\right) .
\end{array}
$$