Question : If no 3 diagonals of a convex decagon meet at the same point, inside the decagon. Into how many line segments are diagonals divided by their intersections?
Answer : The no. of ways of selecting 2 vertices out of 10 vertices is : $C(10,2) = 45$.
10 sides, so the number of vertices is 45-10=35 (why?-1)
(Why?-2) The number of intersections $C(10,4) = 210$.
(Why?-3) The required number is = $35+420 = 455$
My misunderstandings are labelled as Why-1,2,3.
Please help me understanding this problem. There is a similar question already discussed here but that doesn't solve my issue.
They have a typo here; they meant to say that the number of diagonals is $45-10=35$. There are $45$ line segments in the picture, but $10$ of these are the sides of the decagon, so there are $35$ diagonals left over.
An intersection point is uniquely determined by choosing $4$ vertices out of the ten decagon vertices. These four vertices define a quadrilateral, and the two diagonals of that quadrilateral are also diagonals of the decagon. The intersection of the two diagonals are an intersection point.
This is the tricky part. Think about this; if you have a string, and you cut it at $k$ places, then you have $k+1$ pieces of string. That is, each cut adds a piece. Therefore, to count the number of pieces the diagonals are cut into, you take the number of diagonals, $35$, and you add the number of cuts, which is the number of intersection points, $420$.