Maximum points of intersection n lines

You are given n straight lines. You have to find maximum number of point of intersection with these n lines.

Examples:

Input : n = 4 
Output : 6

Input : n = 2
Output :1

Approach :

As we have n number of line, and we have to find maximum point of intersection using these n line. So this can be done using combination. This problem can be think as number of ways to select any two line among n line. As every line intersect with other that is selected.
So, total number of points = nC2

Below is the implementation of above approach:

C++

// CPP program to find maximum intersecting 
// points 
#include <bits/stdc++.h> 
using namespace std; 
#define ll long int 
  
  
// nC2 = (n)*(n-1)/2; 
ll countMaxIntersect(ll n) 
{ 
   return (n) * (n - 1) / 2; 
} 
  
// Driver code 
int main() 
{ 
    // n is number of line 
    ll n = 8; 
    cout << countMaxIntersect(n) << endl; 
    return 0; 
} 

Java

// Java program to find maximum intersecting 
// points 
  
public class GFG { 
      
    // nC2 = (n)*(n-1)/2; 
    static long countMaxIntersect(long n) 
    { 
       return (n) * (n - 1) / 2; 
    } 
  
      
    // Driver code 
    public static void main(String args[]) 
    { 
        // n is number of line 
        long n = 8; 
        System.out.println(countMaxIntersect(n)); 
  
  
    } 
    // This code is contributed by ANKITRAI1 
} 

Python3

# Python3 program to find maximum  
# intersecting points 
  
#nC2 = (n)*(n-1)/2 
def countMaxIntersect(n): 
    return int(n*(n - 1)/2) 
  
#Driver code 
if __name__=='__main__': 
      
# n is number of line 
    n = 8
    print(countMaxIntersect(n)) 
  
# this code is contributed by  
# Shashank_Sharma 

C#

// C# program to find maximum intersecting 
// points 
using System; 
  
class GFG  
{ 
      
    // nC2 = (n)*(n-1)/2; 
    public static long countMaxIntersect(long n) 
    { 
    return (n) * (n - 1) / 2; 
    } 
  
      
    // Driver code 
    public static void Main() 
    { 
        // n is number of line 
        long n = 8; 
        Console.WriteLine(countMaxIntersect(n)); 
    } 
} 
// This code is contributed by Soumik 

PHP

<?PHP 
// PHP program to find maximum intersecting  
// points 
  
// nC2 = (n)*(n-1)/2;  
function countMaxIntersect($n)  
{  
    return ($n) * ($n - 1) / 2;  
}  
  
// Driver code  
  
// n is number of line  
$n = 8; 
echo countMaxIntersect($n) . "\n";  
  
// This code is contributed by ChitraNayal 
?> 

Output:

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