Question

By joining the vertices of a 'n' sided polygon, pentagons are formed. Find the number of pentagons hence formed such that none of their side is common with the sides of the polygon.

Answer 1

I have solved this using x -sided polygonsa re fomrd from the vertices of n-sided polygon.
I will assume that the polygons have to be non-self-intersecting.

Number the vertices of the original polygon from 1 to N. Now, there are two possibilities. Either vertex 1 is part of the solution polygon or not.

If vertex 1 is part of the solution polygon, vertex 2 and vertex N cannot be. This means that x-1 vertices have to be chosen out of the remaining N-3 vertices such that no two are adjacent. The number of ways of choosing this is equal to the number of integer solutions of y1+y2+....yx=(N-3)-(x-1) with the additional constraints that y1,yx >= 0; y2,y3...yx-1 >=1. This is equal to the number of whole number solutions of y1+y2+..yx=(N-3)-(x-1)-(x-2). This is just (N-x-1)C(x-1).

If vertex 1 is not part of the solution polygon, we need to choose x vertices out of the remaining N-1 vertices under the same constraints. Following the same logic (or just a change of variables), we get the number of such polygons to be equal to (N-x)C(x).

But here given that x=5.
So, substitute x=5 in the equation.
(N-4)C4 + (N-5)C5 = (N-4)(N-5)(N-6)(N-7)[1/24 +(N-8)(N-9)/120 ]