Forming triangles using points on a square

Given a square with N points on each side of the square and none of these points co-incide with the corners of the square. The task is to calculate the total number of triangles that can be formed using these 4 * N points (N points on each side of the square) as vertices of the triangle.
Examples:

Input: N = 1 
Output: 4 
There is one point on each side. So we can make three rectangles by leaving one point and picking other three points out of the four. 
Input: N = 2 
Output: 56 

Approach: The number of ways of choosing 3 points among 4 * N points is ^{(4 * N)}C₃. However, some of them do not form a triangle. This happens when all the three chosen points are on the same side of the square. The count of these triplets is ^NC₃ for each of the side i.e. 4 * ^NC₃ in total. Therefore, the required count of triangles will be (^{(4 * N)}C₃) – (4 * ^NC₃).
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of possible triangles
int noOfTriangles(int n)
{
    int y = 4 * n;
    return ((y * (y - 2) * (y - 1))
            - (4 * n * (n - 2) * (n - 1)))
           / 6;
}

// Driver code
int main()
{
    int n = 1;

    cout << noOfTriangles(n);

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

    // Function to return the count
    // of possible triangles
    static int noOfTriangles(int n)
    {
        int y = 4 * n;
        return ((y * (y - 2) * (y - 1)) -
                (4 * n * (n - 2) * (n - 1))) / 6;
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 1;

        System.out.println(noOfTriangles(n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the count
# of possible triangles
def noOfTriangles(n):
    y = 4 * n
    return ((y * (y - 2) * (y - 1)) -
            (4 * n * (n - 2) * (n - 1)))// 6

# Driver code
n = 1

print(noOfTriangles(n))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the count
    // of possible triangles
    static int noOfTriangles(int n)
    {
        int y = 4 * n;
        return ((y * (y - 2) * (y - 1)) -
                (4 * n * (n - 2) * (n - 1))) / 6;
    }

    // Driver code
    public static void Main (String[] args)
    {
        int n = 1;

        Console.WriteLine(noOfTriangles(n));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// javascript implementation of the approach

// Function to return the count
// of possible triangles
function noOfTriangles(n)
{
    var y = 4 * n;
    return ((y * (y - 2) * (y - 1))
            - (4 * n * (n - 2) * (n - 1)))
           / 6;
}

// Driver code
var n = 1;
document.write(noOfTriangles(n));

</script>

Output:

4

Time Complexity: O(1)

Auxiliary Space: O(1)