TITLE: British Mathematical Olympiad - December 2001 - Round 1 - Question 4
QUESTION [10 upvotes]: Twelve people are seated around a circular table. In how many ways
can six pairs of people engage in handshakes so that no arms cross?
(Nobody is allowed to shake hands with more than one person at once.) 
I don't even understand what this question is asking. My best guess is that people can shake hands over the table (i.e. with someone on the opposite to them on the table) but that seems like something they'd mention in the brackets at the end.

REPLY [9 votes]: Let $S_{2n}$ = ways for for 2n people to shake n pairs, without crossing.  
Example, for 4 people ABCD, we have 2 ways: (AB,CD),(AD,BC)  
$S_2 = 1$
$S_4 = 2S_2 = 2$
$S_6 = 2S_4 + S_2 S_2 = 4+1 = 5$
$S_8 = 2S_6 + 2S_2 S_4 = 10 + 4 = 14$
$S_{10} = 2S_8 + 2S_2 S_6 + S_4 S_4= 28+10+4 = 42$
$S_{12} = 2S_{10} + 2S_2 S_8 + 2S_4 S_6= 84+28+20 = 132$