# 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 =

^{n}C_{2}–^{m}C_{2}+ 1How 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^{n}C_{2}ways.

Number of straight line formed by 10 points when no 2 of them are co-linear =â€¦..â€¦(i)^{10}C_{2}

Similarly, the number of straight lines formed by 4 points when no 2 of them are co-linear =….(ii)^{4}C_{2}

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 =^{10}C_{2}â€“^{4}C_{2}+ 1 = 45 â€“ 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` `?>` |

## Javascript

`<script>` ` ` `// Javascript 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` ` ` `function` `nCk(n, k)` ` ` `{` ` ` `let C = ` `new` `Array(k+1);` ` ` `C.fill(0);` ` ` `C[0] = 1; ` `// nC0 is 1` ` ` `for` `(let i = 1; i <= n; i++)` ` ` `{` ` ` `// Compute next row of pascal triangle` ` ` `// using the previous row` ` ` `for` `(let 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 */` ` ` `function` `count_Straightlines(n,m)` ` ` `{` ` ` `return` `(nCk(n, 2) - nCk(m, 2)+1);` ` ` `}` ` ` ` ` `let n = 4, m = 3 ;` ` ` `document.write(count_Straightlines(n, m));` ` ` ` ` `// This code is contributed by divyesh072019.` `</script>` |

Output:

4

**Time Complexity:** O(max(n,m)) **Auxiliary Space:** O(1)

This article is contributed by **Aarti_Rathi**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above