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Seating arrangement of N boys sitting around a round table such that two particular boys sit together
  • Difficulty Level : Easy
  • Last Updated : 03 Jul, 2019

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: 48
2 boy can be arranged in 2! ways and other boys
can be arranged in (5 – 1)! (1 is subtracted because the
previously selected two boys will be considered as a single boy now)
So, total ways are 2! * 4! = 48.

Input: N = 9
Output: 80640

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 – 1)!.
  • So, Total ways = 2! * (n – 1)!.

Below is the implementation of the above approach:

C++

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

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Java

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// 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 - 1) factorial
    int fac = 1;
    for (int i = 2; i <= n - 1; i++) 
    {
        fac = fac * i;
    }
  
    // Return (n - 1)! * 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. 

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Python3

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

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C#

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// 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 - 1) factorial
    int fac = 1;
    for (int i = 2; i <= n - 1; i++) 
    {
        fac = fac * i;
    }
  
    // Return (n - 1)! * 2!
    return (fac * 2);
}
  
// Driver code
static public void Main ()
{
    int n = 5;
  
    Console.Write(Total_Ways(n));
}
}
  
// This code is contributed by ajit.. 

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Output:

48

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