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 = 4, m = 3 Output : 40

Number of distinct Straight lines =

^{n}C_{2}–^{m}C_{2}+ 1

How 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 =

…..…(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}–^{4}C_{2}+ 1 = 45 – 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".

Output:

4

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