# Number of triangles that can be formed

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

Input:N = 2

Output:56

**Approach:** The number of ways of choosing **3** points among **4 * N** points is ** ^{(4 * N)}C_{3}**. 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

**for each of the side i.e.**

^{N}C_{3}**4 ***in total. Therefore, the required count of triangles will be

^{N}C_{3}**(**.

^{(4 * N)}C_{3}) – (4 *^{N}C_{3})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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

4

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