Skip to content
Related Articles

Related Articles

Improve Article
Number of quadrilateral formed with N distinct points on circumference of Circle
  • Last Updated : 15 Mar, 2021

Given an integer N which denotes the points on the circumference of a circle, the task is to find the number of quadrilaterals formed using these points.
Examples: 
 

Input: N = 5 
Output: 5
Input: N = 10 
Output: 210 
 

 

Approach: The idea is to use permutation and combination to find the number of possible quadrilaterals using the N points on the circumference of the circle. The number of possible quadrilaterals will be ^{N}C_4  .
Below is the implementation of the above approach:
 

C++




// C++ implementation to find the
// number of quadrilaterals formed
// with N distinct points
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the factorial
// of the given number N
int fact(int n)
{
    int res = 1;
 
    // Loop to find the factorial
    // of the given number
    for(int i = 2; i < n + 1; i++)
       res = res * i;
        
    return res;
}
 
// Function to find the number of
// combinations in the N
int nCr(int n, int r)
{
    return (fact(n) / (fact(r) *
                       fact(n - r)));
}
 
// Driver Code
int main()
{
    int n = 5;
 
    // Function Call
    cout << (nCr(n, 4));
}
 
// This code is contributed by rock_cool

Java




// Java implementation to find the
// number of quadrilaterals formed
// with N distinct points
class GFG{
     
// Function to find the number of
// combinations in the N
static int nCr(int n, int r)
{
    return (fact(n) / (fact(r) *
                       fact(n - r)));
}
 
// Function to find the factorial
// of the given number N
static int fact(int n)
{
    int res = 1;
 
    // Loop to find the factorial
    // of the given number
    for(int i = 2; i < n + 1; i++)
        res = res * i;
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 5;
 
    // Function Call
    System.out.println(nCr(n, 4));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation to find the
# number of quadrilaterals formed
# with N distinct points
 
# Function to find the number of
# combinations in the N
def nCr(n, r):
    return (fact(n) / (fact(r)
                * fact(n - r)))
 
# Function to find the factorial
# of the given number N
def fact(n):
    res = 1
     
    # Loop to find the factorial
    # of the given number
    for i in range(2, n + 1):
        res = res * i    
    return res
 
# Driver Code
if __name__ == "__main__":
    n = 5
     
    # Function Call
    print(int(nCr(n, 4)))

C#




// C# implementation to find the
// number of quadrilaterals formed
// with N distinct points
using System;
class GFG{
     
// Function to find the number of
// combinations in the N
static int nCr(int n, int r)
{
    return (fact(n) / (fact(r) *
                       fact(n - r)));
}
 
// Function to find the factorial
// of the given number N
static int fact(int n)
{
    int res = 1;
 
    // Loop to find the factorial
    // of the given number
    for(int i = 2; i < n + 1; i++)
        res = res * i;
    return res;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 5;
 
    // Function Call
    Console.Write(nCr(n, 4));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
 
// JavaScript implementation to find the
// number of quadrilaterals formed
// with N distinct points
 
// Function to find the factorial
// of the given number N
function fact(n)
{
    let res = 1;
 
    // Loop to find the factorial
    // of the given number
    for(let i = 2; i < n + 1; i++)
    res = res * i;
         
    return res;
}
 
// Function to find the number of
// combinations in the N
function nCr(n, r)
{
    return (fact(n) / (fact(r) *
                    fact(n - r)));
}
 
// Driver Code
 
    let n = 5;
 
    // Function Call
    document.write(nCr(n, 4));
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
5

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :