Given ten points on the boundary line of a half-plane. In how many ways can the points be paired so that the paired points can be connected by lines lying within the half-plane and not intersecting each other?
Let $a_{i}$ denote the number of connection possibilities when $i$ points are given on the boundary line of a half-plane. Our task is then to determine $a_{10}$. Let's number the points from 1 to 10 in a certain direction along the line. There cannot be an odd number of points between two connected points, because then we could not pair them up in a way that satisfies the conditions. Therefore, $a_{10}$ can be expressed by distinguishing the cases where the first point is connected to the second, fourth, sixth, eighth, or tenth point.

If the first point is connected to the second, the remaining eight points can be paired in $a_{8}$ ways. If the first point is connected to the fourth, the number of ways to connect the points between the first and fourth points - which can only be connected to each other - must be multiplied by the number of ways to connect the remaining six points: $a_{2} \cdot a_{6}$. Similarly, for the other three cases, the number of possibilities is $a_{4} \cdot a_{4}, a_{6} \cdot a_{2}$, and $a_{8}$. Thus:

$$
a_{10}=a_{8}+a_{2} \cdot a_{6}+a_{4} \cdot a_{4}+a_{6} \cdot a_{2}+a_{8}
$$

Similarly, we get that

$$
a_{8}=a_{6}+a_{2} \cdot a_{4}+a_{4} \cdot a_{2}+a_{6}, a_{6}=a_{4}+a_{2} \cdot a_{2}+a_{4}, a_{4}=a_{2}+a_{2}
$$

It is obvious that two points can be paired in one way, so $a_{2}=1$. Substituting this, we get $a_{4}=2, a_{6}=5$, $a_{8}=14$, and finally $a_{10}=42$.

Thus, the number of pairing possibilities is: 42.

Remark. It is generally true that $a_{2 n}=a_{2 n-2}+a_{2 n-4} \cdot a_{2}+\cdots+a_{2 n-2}$. Using this, it can also be proven that $a_{2 n}=\frac{1}{n+1}\binom{2 n}{n}$.

Zsolt Katona (Fazekas M. Főv. Gyak. Gimn., I. o.t.)

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