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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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; ` `} `

## 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 - 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.  `

## Python3

 `# 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 `

## 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 - 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..  `

Output:

```48
```

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Improved By : AnkitRai01, jit_t