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Seating arrangement of N boys sitting around a round table such that two particular boys sit together

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  • Difficulty Level : Easy
  • Last Updated : 09 Jun, 2022

There are N boys which are to be seated around a round table. The task is to find the number of ways in which N boys can sit around a round table such that two particular boys sit together.
Examples: 
 

Input: N = 5 
Output: 12 
2 boy can be arranged in 2! ways and other boys 
can be arranged in (5 – 2)! (2 is subtracted because the 
previously selected two boys will be considered as a single boy now and No. of ways to arrange boys around a round table = (n-1)!) 
So, total ways are 2! * (n-2)!)  =  2! * 3! = 12
Input: N = 9 
Output: 10080
 

 

Approach: 
 

  • First, 2 boys can be arranged in 2! ways.
  • No. of ways to arrange remaining boys and the previous two boy pair is (n – 2)!.
  • So, Total ways = 2! * (n – 2)!.

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 total count of ways
int Total_Ways(int n)
{
 
    // Find (n - 2) factorial
    int fac = 1;
    for (int i = 2; i <= n - 2; i++) {
        fac = fac * i;
    }
 
    // Return (n - 2)! * 2!
    return (fac * 2);
}
 
// Driver code
int main()
{
    int n = 5;
 
    cout << Total_Ways(n);
 
    return 0;
}


Java




// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function to return the total count of ways
static int Total_Ways(int n)
{
 
    // Find (n - 2) factorial
    int fac = 1;
    for (int i = 2; i <= n - 2; i++)
    {
        fac = fac * i;
    }
 
    // Return (n - 2)! * 2!
    return (fac * 2);
}
 
// Driver code
public static void main (String[] args)
{
 
    int n = 5;
 
    System.out.println (Total_Ways(n));
}
}
 
// This code is contributed by Tushil.


Python3




# Python3 implementation of the approach
 
# Function to return the total count of ways
def Total_Ways(n) :
 
    # Find (n - 2) factorial
    fac = 1;
    for i in range(2, n-1) :
        fac = fac * i;
         
    # Return (n - 2)! * 2!
    return (fac * 2);
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 5;
 
    print(Total_Ways(n));
 
# This code is contributed by AnkitRai01


C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the total count of ways
static int Total_Ways(int n)
{
 
    // Find (n - 2) factorial
    int fac = 1;
    for (int i = 2; i <= n - 2; i++)
    {
        fac = fac * i;
    }
 
    // Return (n - 2)! * 2!
    return (fac * 2);
}
 
// Driver code
static public void Main ()
{
    int n = 5;
 
    Console.Write(Total_Ways(n));
}
}
 
// This code is contributed by ajit..


Javascript




<script>
// javascript implementation of the approach
 
    // Function to return the total count of ways
    function Total_Ways(n)
    {
 
        // Find (n - 2) factorial
        var fac = 1;
        for (i = 2; i <= n - 2; i++)
        {
            fac = fac * i;
        }
 
        // Return (n - 2)! * 2!
        return (fac * 2);
    }
 
    // Driver code
        var n = 5;
        document.write(Total_Ways(n));
 
// This code is contributed by aashish1995
</script>


Output

12

Time Complexity: O(n)
Auxiliary Space: O(1)


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