13. 6 different points are given on the plane, no three of which are collinear. Each pair of points is to be joined by a red line or a blue line subject to the following restriction: if the lines joining $A B$ and $A C$ (where $A, B, C$ denote the given points) are both red, then the line joining $B C$ is also red. How many different colourings of the lines are possible?
(2 marks)
在平面上給定 6 個不同的點, 當中沒有三點共線。現要把任意兩點均以一條紅線或監線連起, 並須符合以下規定: 若 $A B$ 和 $A C$ (這裡 $A 、 B 、 C$ 代表給定的點）均以紅線連起, 則 $B C$ 亦必須以紅線連起。那麼, 連線的顏色有多少個不同的組合？
13. 203
13. We only have to consider which points are to be connected by red lines, and the rest of the lines must be blue. Note that the restriction means that the 6 points must be divided into some number of groups, such that two points are joined by a red line if and only if they are in the same group. So the question becomes counting the number of such groupings, and we count using the pattern of grouping, as follows:
\begin{tabular}{|c|l|}
\hline Pattern & \\
\hline 6 & 1 \\
\hline $5-1$ & $C_{1}^{6}=6$ \\
\hline $4-2$ & $C_{4}^{6}=15$ \\
\hline $4-1-1$ & $C_{4}^{6}=15$ \\
\hline $3-3$ & $C_{3}^{6} \div 2=10$ \\
\hline $3-2-1$ & $C_{3}^{6} \times C_{2}^{3}=60$ \\
\hline $3-1-1-1$ & $C_{3}^{6}=20$ \\
\hline $2-2-2$ & $C_{2}^{6} \times C_{2}^{4} \div 3!=15$ \\
\hline $2-2-1-1$ & $C_{2}^{6} \times C_{2}^{4} \div 2=45$ \\
\hline $2-1-1-1-1$ & $C_{2}^{6}=15$ \\
\hline $1-1-1-1-1-1$ & 1 \\
\hline
\end{tabular}

Hence the answer is $1+6+15+15+10+60+20+15+45+15+1=203$.
Remark. This question essentially asks for the number of partitions of the set $\{1,2,3,4,5$, 6\}. The answer is in fact the 6th Bell mumber.