Given three integers *m, n and k* that store the number of points on lines l1, l2 and l3 respectively that do not intersect. The task is to find the number of triangles that can possibly be formed from these set of points.

**Examples:**

Input:m = 3, n = 4, k = 5Output:205Input:m = 2, n = 2, k = 1Output:10

**Approach:**

- The total number of points are (m + n + k) which must give number of triangles.
- But ‘m’ points on ‘l1’ gives combinations which cannot form a triangle.
- Similarly, and number of triangles can not be formed.
- Therefore, Required Number of Triangles =

Below is the implementation of the above approach:

## C++

`// CPP program to find the possible number ` `// of triangles that can be formed from ` `// set of points on three lines ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Returns factorial of a number ` `int` `factorial(` `int` `n) ` `{ ` ` ` `int` `fact = 1; ` ` ` `for` `(` `int` `i = 2; i <= n; i++) ` ` ` `fact = fact * i; ` ` ` `return` `fact; ` `} ` ` ` `// calculate c(n, r) ` `int` `ncr(` `int` `n, ` `int` `r) ` `{ ` ` ` ` ` `return` `factorial(n) ` ` ` `/ (factorial(r) * factorial(n - r)); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `m = 3, n = 4, k = 5; ` ` ` `int` `totalTriangles ` ` ` `= ncr(m + n + k, 3) ` ` ` `- ncr(m, 3) - ncr(n, 3) - ncr(k, 3); ` ` ` `cout << totalTriangles << endl; ` `} ` |

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## Java

`//Java program to find the possible number ` `// of triangles that can be formed from ` `// set of points on three lines ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Returns factorial of a number ` `static` `int` `factorial(` `int` `n) ` `{ ` ` ` `int` `fact = ` `1` `; ` ` ` `for` `(` `int` `i = ` `2` `; i <= n; i++) ` ` ` `fact = fact * i; ` ` ` `return` `fact; ` `} ` ` ` `// calculate c(n, r) ` `static` `int` `ncr(` `int` `n, ` `int` `r) ` `{ ` ` ` ` ` `return` `factorial(n) ` ` ` `/ (factorial(r) * factorial(n - r)); ` `} ` ` ` `// Driver code ` ` ` ` ` `public` `static` `void` `main (String[] args) { ` ` ` ` ` `int` `m = ` `3` `, n = ` `4` `, k = ` `5` `; ` ` ` `int` `totalTriangles = ncr(m + n + k, ` `3` `) - ` ` ` `ncr(m, ` `3` `) - ncr(n, ` `3` `) - ncr(k, ` `3` `); ` ` ` `System.out.println (totalTriangles); ` ` ` ` ` ` ` `} ` `} ` |

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## Python 3

`# Python 3 program to find the ` `# possible number of triangles ` `# that can be formed from set of ` `# points on three lines ` ` ` ` ` `# Returns factorial of a number ` `def` `factorial(n): ` ` ` `fact ` `=` `1` ` ` `for` `i ` `in` `range` `(` `2` `, n ` `+` `1` `): ` ` ` `fact ` `=` `fact ` `*` `i ` ` ` `return` `fact ` ` ` `# calculate c(n, r) ` `def` `ncr(n, r): ` ` ` ` ` `return` `(factorial(n) ` `/` `/` `(factorial(r) ` `*` ` ` `factorial(n ` `-` `r))) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `m ` `=` `3` ` ` `n ` `=` `4` ` ` `k ` `=` `5` ` ` `totalTriangles ` `=` `(ncr(m ` `+` `n ` `+` `k, ` `3` `) ` `-` ` ` `ncr(m, ` `3` `) ` `-` `ncr(n, ` `3` `) ` `-` ` ` `ncr(k, ` `3` `)) ` ` ` `print` `(totalTriangles) ` ` ` `# This code is contributed ` `# by ChitraNayal ` |

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## C#

`// C# program to find the possible number ` `// of triangles that can be formed from ` `// set of points on three lines ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Returns factorial of a number ` `static` `int` `factorial(` `int` `n) ` `{ ` ` ` `int` `fact = 1; ` ` ` `for` `(` `int` `i = 2; i <= n; i++) ` ` ` `fact = fact * i; ` ` ` `return` `fact; ` `} ` ` ` `// calculate c(n, r) ` `static` `int` `ncr(` `int` `n, ` `int` `r) ` `{ ` ` ` ` ` `return` `factorial(n) / (factorial(r) * ` ` ` `factorial(n - r)); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main () ` `{ ` ` ` `int` `m = 3, n = 4, k = 5; ` ` ` ` ` `int` `totalTriangles = ncr(m + n + k, 3) - ` ` ` `ncr(m, 3) - ncr(n, 3) - ` ` ` `ncr(k, 3); ` ` ` ` ` `Console.WriteLine (totalTriangles); ` `} ` `} ` ` ` `// This code is contributed ` `// by anuj_67.. ` |

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## PHP

`<?php ` `// PHP program to find the possible ` `// number of triangles that can be ` `// formed from set of points on ` `// three lines ` ` ` `// Returns factorial of a number ` `function` `factorial(` `$n` `) ` `{ ` ` ` `$fact` `= 1; ` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$n` `; ` `$i` `++) ` ` ` `$fact` `= ` `$fact` `* ` `$i` `; ` ` ` `return` `$fact` `; ` `} ` ` ` `// calculate c(n, r) ` `function` `ncr(` `$n` `, ` `$r` `) ` `{ ` ` ` `return` `factorial(` `$n` `) / (factorial(` `$r` `) * ` ` ` `factorial(` `$n` `- ` `$r` `)); ` `} ` ` ` `// Driver code ` `$m` `= 3; ` `$n` `= 4; ` `$k` `= 5; ` `$totalTriangles` `= ncr(` `$m` `+ ` `$n` `+ ` `$k` `, 3) - ` ` ` `ncr(` `$m` `, 3) - ncr(` `$n` `, 3) - ` ` ` `ncr(` `$k` `, 3); ` `echo` `$totalTriangles` `. ` `"\n"` `; ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

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**Output:**

205

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