12.12 People are sitting around a table in a circle. How many ways are there for 6 pairs of people to shake hands so that no arms cross?
(2001, British Mathematical Olympiad (Round 1))
Solution: According to the problem, handshakes only occur on the table, not behind other people.

Let $N_{p}$ denote the total number of ways for $2p$ people sitting around a circular table to shake hands without crossing arms, and each person shakes hands only once. Consider each person as a point. Without loss of generality, let the point where Arthur is located be denoted as A. Since handshakes occur in pairs, the number of people on either side of Arthur 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, Arthur can shake hands with his adjacent person (such as B or D in Figure 4), but not with the opposite person C, otherwise, it would cause a crossing. Therefore, $N_{2}=2$.

If there are 6 people, Arthur can shake hands with his adjacent person (such as B or F in Figure 5), and the remaining 4 people have $N_{2}$ ways to shake hands. Arthur can also shake hands with the opposite person D, and on both sides of AD, there are 2 people, 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, Arthur can shake hands with his adjacent person (such as B or H in Figure 6). The remaining 6 people on one side have $N_{3}$ ways to shake hands. Arthur can also shake hands with D or F, then the 2 people on one side have $N_{1}$ ways to shake hands, and the 4 people on the other side have $N_{2}$ ways to shake hands. Therefore, $N_{4}=2 N_{3}+2 N_{1} \times N_{2}=14$.
Similarly, we get
$$
\begin{array}{l}
N_{5}=2 N_{4}+2 N_{3} \times N_{1}+N_{2}^{2}=42, \\
N_{6}=2 N_{5}+2 N_{4} \times N_{1}+2 N_{3} \times N_{2}=132 .
\end{array}
$$

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
$$
N_{k}=\sum_{r=0}^{k-1}\left(N, \times N_{k-r-1}\right) .
$$