TITLE: Catalan numbers and triangulations
QUESTION [11 upvotes]: The number of ways to parenthesize an $n$ fold product is a Catalan number in the list $1,1,2,5,14,\cdots$ where these are in order of the number of terms in the product. The $n$th such number is also the numbr of ways to triangulate an $n+1$-gon.
I'm wondering whether there is a simple translation between a specific parenthesized $n$ fold product to a specific triangulation of an $n+1$-gon.
For example one parenthesized 5-fold product is $(12)(3(45))$ This then would (hopefully) be translatable to one of the $14$ triangulations of a $6$-gon.
I have tried various ways to label the vertices of the $n+1$ gon to see which parenthesized $n$ fold product a given triangulation goes with, but no luck.

REPLY [4 votes]: These are cool (as much of has to do with Catalan Numbers), but trees are not necessary in this case. Just number the edges of the hexagon clockwise (and not including) the base. Each diagonal is now the product of the other two sides of the triangle, and you'll get your product at the base. In your example, it looks like this.
          3
      -–––––---
     /|\      |\
   2/ | \     | \4      
   /  |  \    |  \
  < 12| 3(45) |45 >
   \  |    \  |  /
   1\ |     \ | /5
     \|      \|/
      ---–––––-
     (12)(3(45))