Number of points of intersection of diagonals of polygon with $2009$ sides which are situated inside the polygon.

^{2009} C_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{2009} C_{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 \times^{2009} C_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{2009} C_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $^{2009} C_{4}$
As we know that no two diagonals are parallel in the polygon with odd number of sides.
If we take any $4$ vertices then we get one intersection point.
So, we need to select $4$ vertices out of $2009$ vertices.
Hence, required number of ways $=^{2009} C_{4}$

Suggest Corrections

Similar questions

Q. Number of points of intersection of diagonals of polygon with

2009

sides which are situated inside the polygon.

Q. The number of intersection points of diagonals of

2009

sides polygon, which lie inside the polygon.

Q. If a convex polygon has

35

diagonals, then the number of points of intersection of diagonals which lies inside the polygon is

Q. If a convex polygon has

35

diagonals, then the number of points of intersection of diagonals which lies inside the polygon is

In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70, then the number of diagonals of the polygon is

Join BYJU'S Learning Program

Number of points of intersection of diagonals of polygon with 2009 sides which are situated inside the polygon.

The correct option is B 2009C4As we know that no two diagonals are parallel in the polygon with odd number of sides. If we take any 4 vertices then we get one intersection point. So, we need to select 4 vertices out of 2009 vertices. Hence, required number of ways = 2009C4

Number of points of intersection of diagonals of polygon with $2009$ sides which are situated inside the polygon.

The correct option is B $^{2009} C_{4}$
As we know that no two diagonals are parallel in the polygon with odd number of sides.
If we take any $4$ vertices then we get one intersection point.
So, we need to select $4$ vertices out of $2009$ vertices.
Hence, required number of ways $=^{2009} C_{4}$