Count of triangles with total n points with m collinear

20 May, 2022
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
 
 
Number of triangles = nC3 – mC3How 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 
 
C++
Java
Python3
C#
PHP
Javascript
// CPP program to count number of triangles// with n total points, out of which m are// collinear.#include <bits/stdc++.h>using namespace std; // Returns value of binomial coefficient// Code taken from https://goo.gl/vhy4jpint nCk(int n, int k){    int C[k+1];    memset(C, 0, sizeof(C));     C[0] = 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 countTriangles(int n,int m){    return (nCk(n, 3) - nCk(m, 3));} /* driver function*/int main(){    int n = 5, m = 4;    cout << countTriangles(n, m);    return 0;}
//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/vhy4jpstatic int nCk(int n, int k){    int[] C=new int[k+1];    for (int i=0;i<=k;i++)    C[i]=0;         C[0] = 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 trianglecan be formed */static int countTriangles(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(countTriangles(n, m));     }} //This code is contributed by Gitanjali.
# 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 / vhy4jpdef nCk(n, k):    C = [0 for i in range(0, k + 2)]      C[0] = 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 formeddef countTriangles(n, m):    return (nCk(n, 3) - nCk(m, 3))  # driver coden = 5m = 4print (countTriangles(n, m))  # This code is contributed by Gitanjali
//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[0] = 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 countTriangles(int n,int m)    {        return (nCk(n, 3) - nCk(m, 3));    }         // Driver code    public static void Main ()    {        int n = 5, m = 4;        Console.WriteLine(countTriangles(n, m));     }} // This code is contributed by vt_m.
<?php// PHP program to count number// of triangles with n total// points, out of which m are collinear. // Returns value of binomial coefficient// Code taken from https://goo.gl/vhy4jpfunction nCk($n, $k){    for ($i = 0; $i <= $k; $i++)    $C[$i] = 0;     $C[0] = 1; // nC0 is 1     for ($i = 1; $i <= $n; $i++)    {        // Compute next row of pascal        // triangle using the previous row        for ($j = min($i, $k); $j > 0; $j--)            $C[$j] = $C[$j] + $C[$j - 1];    }    return $C[$k];} /* function to calculate numberof triangles that can be formed */function countTriangles($n, $m){    return (nCk($n, 3) - nCk($m, 3));} // Driver Code$n = 5;$m = 4;echo countTriangles($n, $m);return 0; // This code is contributed by ChitraNayal?>
<script>    // Javascript program to count number of triangles    // with n total points, out of which m are    // collinear.         // Returns value of binomial coefficient    // Code taken from https://goo.gl/vhy4jp    function nCk(n, k)    {        let C = new Array(k+1);        C.fill(0);         C[0] = 1;  // nC0 is 1         for (let i = 1; i <= n; i++)        {            // Compute next row of pascal triangle            // using the previous row            for (let 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 */    function countTriangles(n, m)    {        return (nCk(n, 3) - nCk(m, 3));    }         let n = 5, m = 4;    document.write(countTriangles(n, m));         // This code is contributed by divyesh072019.</script>
Output :  
6
 
ukasp
divyesh072019
simmytarika5
triangle
Combinatorial
Combinatorial
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