There are ‘n’ points in a plane, out of which ‘m’ points are co-linear. Find the number of triangles formed by the points as vertices ?
Input : n = 5, m = 4 Output : 6 Out of five points, four points are collinear, we can make 6 triangles. We can choose any 2 points from 4 collinear points and use the single point as 3rd point. So total count is 4C2 = 6 Input : n = 10, m = 4 Output : 116
Number of triangles = nC3 – mC3
How does this formula work?
Consider the second example above. There are 10 points, out of which 4 collinear. A triangle will be formed by any three of these ten points. Thus forming a triangle amounts to selecting any three of the 10 points. Three points can be selected out of the 10 points in nC3 ways.
Number of triangles formed by 10 points when no 3 of them are co-linear = 10C3……(i)
Similarly, the number of triangles formed by 4 points when no 3 of them are co-linear = 4C3……..(ii)
Since triangle formed by these 4 points are not valid, required number of triangles formed = 10C3 – 4C3 = 120 – 4 = 116
- Count of different straight lines with total n points with m collinear
- Number of triangles that can be formed with given N points
- Number of triangles formed from a set of points on three lines
- Count the number of possible triangles
- Count number of unique Triangles using STL | Set 1 (Using set)
- Count of sub-sets of size n with total element sum divisible by 3
- Count Integral points inside a Triangle
- Number of triangles after N moves
- Find all possible triangles with XOR of sides zero
- Program to check similarity of given two triangles
- Program to check congruency of two triangles
- Area of the circumcircle of any triangles with sides given
- Number of Triangles in an Undirected Graph
- Counting Triangles in a Rectangular space using BIT
- Number of triangles possible with given lengths of sticks which are powers of 2
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : Ita_c