Count of different straight lines with total n points with m collinear

There are ‘n’ points in a plane out of which ‘m points are collinear. How many different straight lines can form?

Examples:

Input : n = 3, m = 3
Output : 1
We can form only 1 distinct straight line
using 3 collinear points

Input : n = 10, m = 4
Output : 40



Number of distinct Straight lines = nC2mC2 + 1

How does this formula work?
Consider the second example above. There are 10 points, out of which 4 collinear. A straight line will be formed by any two of these ten points. Thus forming a straight line amounts to selecting any two of the 10 points. Two points can be selected out of the 10 points in nC2 ways.

Number of straight line formed by 10 points when no 2 of them are co-linear = 10C2…..…(i)
Similarly, the number of straight lines formed by 4 points when no 2 of them are co-linear = 4C2….(ii)

Since straight lines formed by these 4 points are sane, straight lines formed by them will reduce to only one.
Required number of straight lines formed = 10C24C2 + 1 = 45 – 6 + 1 = 40

Implementation of the approach is given as:

C++

// CPP program to count number of straight lines
// 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/vhy4jp
int 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 straight lines
   can be formed */
int count_Straightlines(int n,int m)
{
  
    return (nCk(n, 2) - nCk(m, 2)+1);
  
}
   
  
/* driver function*/
int main()
{
  
    int n = 4, m = 3 ;
    cout << count_Straightlines(n, m);
    return 0;
  
}

Java

// Java program to count number of straight lines
// with n total points, out of which m are
// collinear.
import java.util.*;
import java.lang.*;
  
public class GfG  {
  
    // Returns value of binomial coefficient
    // Code taken from https://goo.gl/vhy4jp
    public static int nCk(int n, int k)
    {
        int[] C = new int[k + 1];
  
        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 straight lines
    can be formed */
    public static int count_Straightlines(int n, int m)
    {
        return (nCk(n, 2) - nCk(m, 2) + 1);
    }
  
  
    // Driver function
    public static void main(String argc[])
    {
        int n = 4, m = 3;
        System.out.println(count_Straightlines(n, m));
    }
  
    // This code is contributed by Sagar Shukla
}

Python

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

C#

// C# program to count number of straight
// lines with n total points, out of 
// which m are collinear.
using System;
  
public class GfG {
  
    // Returns value of binomial coefficient
    // Code taken from https://goo.gl/vhy4jp
    public static int nCk(int n, int k)
    {
        int[] C = new int[k + 1];
  
        // 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 
    // straight lines can be formed
    public static int count_Straightlines(int n, int m)
    {
        return (nCk(n, 2) - nCk(m, 2) + 1);
    }
  
  
    // Driver Code
    public static void Main(String []args)
    {
        int n = 4, m = 3;
        Console.WriteLine(count_Straightlines(n, m));
    }
  
  
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP program to count number of straight lines
// with n total points, out of which m are
// collinear.
  
// Returns value of binomial coefficient
function nCk($n, $k)
{
    $C = array_fill(0, $k + 1, NULL);
  
    $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
// number of straight lines
// can be formed 
function count_Straightlines($n, $m)
{
  
    return (nCk($n, 2) - nCk($m, 2) + 1);
  
}
  
// Driver Code
$n = 4;
$m = 3;
echo(count_Straightlines($n, $m));
  
// This code is contributed 
// by Prasad Kshirsagar
?>


Output:

4


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