Count of triangles with total n points with m collinear

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 ?

Examples :

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Number of triangles = nC3mC3

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 = 10C34C3 = 120 – 4 = 116

C++

 // CPP program to count number of triangles // with n total points, out of which m are // collinear. #include using namespace std;    // Returns value of binomial coefficient // Code taken from https://goo.gl/vhy4jp int nCk(int n, int k) {     int C[k+1];     memset(C, 0, sizeof(C));        C = 1;  // nC0 is 1        for (int i = 1; i <= n; i++)     {         // Compute next row of pascal triangle         // using the previous row         for (int j = min(i, k); j > 0; j--)             C[j] = C[j] + C[j-1];     }     return C[k]; }    /* function to calculate number of triangle    can be formed */ int counTriangles(int n,int m) {     return (nCk(n, 3) - nCk(m, 3)); }    /* driver function*/ int main() {     int n = 5, m = 4;     cout << counTriangles(n, m);     return 0; }

Java

 //Java program to count number of triangles // with n total points, out of which m are // collinear. import java.io.*; import java.util.*;    class GFG {    // Returns value of binomial coefficient // Code taken from https://goo.gl/vhy4jp static int nCk(int n, int k) {     int[] C=new int[k+1];     for (int i=0;i<=k;i++)     C[i]=0;            C = 1; // nC0 is 1        for (int i = 1; i <= n; i++)     {         // Compute next row of pascal triangle         // using the previous row         for (int j = Math.min(i, k); j > 0; j--)             C[j] = C[j] + C[j-1];     }     return C[k]; }    /* function to calculate number of triangle can be formed */ static int counTriangles(int n,int m) {     return (nCk(n, 3) - nCk(m, 3)); }        public static void main (String[] args) {       int n = 5, m = 4;       System.out.println(counTriangles(n, m));        } }    //This code is contributed by Gitanjali.

Python3

 # python program to count number of triangles # with n total points, out of which m are # collinear. import math     # Returns value of binomial coefficient # Code taken from https://goo.gl / vhy4jp def nCk(n, k):     C = [0 for i in range(0, k + 2)]         C = 1; # nC0 is 1     for i in range(0, n + 1):                 # Compute next row of pascal triangle         # using the previous row         for j in range(min(i, k), 0, -1):             C[j] = C[j] + C[j-1]             return C[k]     # function to calculate number of triangle # can be formed  def counTriangles(n, m):     return (nCk(n, 3) - nCk(m, 3))     # driver code n = 5 m = 4 print (counTriangles(n, m))     # This code is contributed by Gitanjali

C#

 //C# program to count number of triangles // with n total points, out of which m are // collinear. using System;    class GFG {        // Returns value of binomial coefficient     // Code taken from https://goo.gl/vhy4jp     static int nCk(int n, int k)     {         int[] C=new int[k+1];         for (int i = 0; i <= k; i++)         C[i] = 0;                    // nC0 is 1         C = 1;                 for (int i = 1; i <= n; i++)         {             // Compute next row of pascal triangle             // using the previous row             for (int j = Math.Min(i, k); j > 0; j--)                 C[j] = C[j] + C[j - 1];         }         return C[k];     }            /* function to calculate number of triangle     can be formed */     static int counTriangles(int n,int m)     {         return (nCk(n, 3) - nCk(m, 3));     }            // Driver code      public static void Main ()      {         int n = 5, m = 4;         Console.WriteLine(counTriangles(n, m));        } }    // This code is contributed by vt_m.

PHP

 0; \$j--)             \$C[\$j] = \$C[\$j] + \$C[\$j - 1];     }     return \$C[\$k]; }    /* function to calculate number  of triangles that can be formed */ function counTriangles(\$n, \$m) {     return (nCk(\$n, 3) - nCk(\$m, 3)); }    // Driver Code \$n = 5; \$m = 4; echo counTriangles(\$n, \$m); return 0;    // This code is contributed by ChitraNayal ?>

Output :

6

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