Example 21 (2001 British Mathematical Olympiad (Round 1) Question) 12 people sit around a table in a circle. How many ways are there for 6 pairs of people to shake hands so that no arms cross?
Given the problem, handshakes only occur on the table and not behind others.
Let $N_{p}$ denote the total number of ways $2p$ people can sit around a circular table and shake hands without crossing arms, with each person shaking hands only once. Consider each person as a point. Without loss of generality, let the point where Wang is located be denoted as A. Since handshakes occur in pairs, the number of people on either side of Wang and the person he shakes hands with must be 0 or even.
If there are 2 people, there is only one way to shake hands, so $N_{1}=1$.
If there are 4 people, Wang can shake hands with his adjacent person (such as B or D in Figure 5-21), but not with the opposite person C, as this would cause a crossing. Therefore, $N_{2}=2$.

If there are 6 people, Wang can shake hands with his adjacent person (such as B or F in Figure 5-22), and the remaining 4 people have $N_{2}$ ways to shake hands. Wang can also shake hands with the opposite person D, with 2 people on each side of AD, and BC and FE each have $N_{1}$ ways to shake hands. Therefore, $N_{3}=2 N_{2}+N_{1}^{2}=5$.
Using this method, we can gradually find $N_{6}$.
When there are 8 people, Wang can shake hands with his adjacent person (such as B or H in Figure 5-23). The remaining 6 people on one side have $N_{3}$ ways to shake hands. Wang can also shake hands with D or F, with 2 people on one side having $N_{1}$ ways to shake hands and 4 people on the other side having $N_{2}$ ways to shake hands. Therefore, $N_{4}=2 N_{3}+2 N_{1} \times N_{2}=14$.
Similarly, we get
$$
\begin{array}{l}
N_{5}=2 N_{4}+2 N_{3}+N_{1}+N_{2}^{2}=42, \\
N_{6}=2 N_{5}+2 N_{4} \times N_{1}+2 N_{3} \times N_{2}=132 .
\end{array}
$$

Therefore, the solution to this problem is 132.
Note: The sequence $1,2,5,14,42,132 \cdots$ is the famous Catalan number. If we define $N_{0}=1$, then the general term of this sequence is
$$
N_{k}=\sum_{r=0}^{k-1}\left(N_{r} \times N_{k-r-1}\right) .
$$