# Why does the term ${\frac{1}{n-1}} {2n-4\choose n-2}$ counts the number of possible triangulations in a polygon?

In the given picture bellow, it counts the number of different triangloations in a polygon, how do the get to this expression, why is it:

$${2n-4\choose n-2}$$

and why do we multiply it by $${\frac{1}{n-1}}$$

In this book they don't explain this issue.

I took the counting expression from here:

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Every triangulation of an $n$-sided polygon can be associated with a binary tree with $(n-2)$ nodes, since there are $(n-2)$ triangles in the triangulation. Fix a side of the polygon, then consider the triangle on that side. It can have one or two neighbours, left or right. The triangles with only one neighbour are the ones that share a side (or two, in the case of the starting triangle) with the perimeter of the polygon, and the triangles with no neighbours are the ones that share two sides with the perimeter of the polygon. On the other hand, every rooted binary tree with $(n-2)$ nodes (where a right child differs from a left child) gives a different triangulation. Moreover, every rooted binary tree with $(n-2)$ nodes can be encoded as a string of $(2n-4)$ balanced parenthesis: assign the encoding "()" to every leaf, then recursively assign the encoding "(encoding-of-the-left-child) encoding-of-the-right-child" to every parent, until the root is reached. A couple of parenthesis is added every time an edge of the tree is traversed. Moreover, every string of $(2n-4)$ balanced parenthesis can be uniquely "decoded" into a rooted binary left/right tree with $(n-2)$ nodes. The last bijection is between the string of $(2n-4)$ balanced parenthesis and the up-or-right paths in a $(n-2)\times (n-2)$ square grid: starting from the lower left corner, we move up every time we encounter a "(" and to the right every time we encounter a ")": the balancing condition is equivalent to the fact that we never cross the diagonal of the grid. Now see the second proof on Wikipedia about Catalan numbers and everything is done.
Draw a regular $n$-gon, and pick one of the sides. Once you triangulate it, this side will be on one of the triangles. If we remove this triangle, we are left with two triagulated polygons. The triangulations we find on them are arbitrary, so this gives us a recurrence.