noncrossing partitions of two elements (catalan numbers)

Question

I was reading the solution to a problem which goes like this:

Problem: There is a round table with n chairs. n is even. In each chair there is a person. How many ways are there that each person can handshake with somebody else without two pairs of hands crossing each other. For example if n = 6(six chairs) the answer is 5: the permutations.

Solution:The answer is the n/2th Catalan number.

I know that Catalan numbers are used to represent the number of noncrossing partitions in a set, but I am not sure why the answer is the n/2th Catalan number. Does that mean that the number of noncrossing partitions of size 2 in a set of n points is equal to all the noncrossing partitions in a set of n/2 points? Can someone explain why?

Answer 1

Hint:

It is easy to see that the Catalan numbers satisfy the recurrence relation $$\tag{1} C_n = \sum_{k=0}^{n-1} C_kC_{n-k-1},$$ Where $C_k$ is the $k$-th Catalan number. Now if the chairs are numbered from $1$ to $2n$, if $1$ shakes hands with $k$, $k$ must be even. So by induction there are $C_{\frac{k}{2}-1}$ ways for $2$ to $k-1$ to shake hands and $C_{\frac{n}{2}-\frac{k}{2}}$ ways for $k+1$ to $2n$ to shake hands.

Summation over $k$ gives the $RHS$ of $(1)$ for $\frac{n}{2}$.