# Number of ways to arrange 2*N persons on the two sides of a table with X and Y persons on opposite sides

Last Updated : 06 Mar, 2023

Given three integers N, X and Y. The task is to find the number of ways to arrange 2*N persons along two sides of a table with N number of chairs on each side such that X persons are on one side and Y persons are on the opposite side.
Note: Both X and Y are less than or equals to N.
Examples:

Input : N = 5, X = 4, Y = 2
Output : 57600
Explanation :
The total number of person 10. X men on one side and Y on other side, then 10 – 4 – 2 = 4 persons are left. We can choose 5 – 4 = 1 of them on one side in ways and the remaining persons will automatically sit on the other side. On each side arrangement is done in 5! ways. The number of ways is .5!5!

Input : N = 3, X = 1, Y = 2
Output : 108

Approach :
The total number of person 2*N. Let call both the sides as A and B. X men on side A and Y on side B, then 2*N – X – Y persons are left. We can choose N-X of them for side A in ways and the remaining persons will automatically sit on the other side B. On each side arrangement is done in N! ways. The number of ways to arrange 2*N persons along two sides of a table is .N!N!
Below is the implementation of the above approach :

## C++

 `#include ``using` `namespace` `std;` `// Function to find factorial of a number``int` `factorial(``int` `n)``{``    ``if` `(n <= 1)``        ``return` `1;``    ``return` `n * factorial(n - 1);``}` `// Function to find nCr``int` `nCr(``int` `n, ``int` `r)``{``    ``return` `factorial(n) / (factorial(n - r) * factorial(r));``}`  `// Function to find the number of ways to arrange 2*N persons``int` `NumberOfWays(``int` `n, ``int` `x, ``int` `y)``{``    ``return` `nCr(2*n-x-y, n-x) * factorial(n) * factorial(n); ``}`  `// Driver code``int` `main()``{``    ``int` `n = 5, x = 4, y = 2;``    ` `    ``// Function call``    ``cout << NumberOfWays(n, x, y);``    ` `    ``return` `0;``}`

## Java

 `// Java implementation for the above approach``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `GFG``{``    ` `    ``// Function to returns factorial of n``    ``static` `int` `factorial(``int` `n) ``    ``{ ``        ``if` `(n <= ``1``) ``            ``return` `1``; ``        ``return` `n * factorial(n - ``1``); ``    ``} ``    ` `    ``// Function to find nCr ``    ``static` `int` `nCr(``int` `n, ``int` `r) ``    ``{ ``        ``return` `factorial(n) / (factorial(n - r) *``                               ``factorial(r)); ``    ``} ``    ` `    ``// Function to find the number of ways``    ``// to arrange 2*N persons ``    ``static` `int` `NumberOfWays(``int` `n, ``int` `x, ``int` `y) ``    ``{ ``        ``return` `nCr(``2` `* n - x - y, n - x) * ``               ``factorial(n) * factorial(n); ``    ``} ``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``                  ``throws` `java.lang.Exception``    ``{``        ``int` `n = ``5``, x = ``4``, y = ``2``; ``        ` `        ``// Function call ``        ``System.out.println(NumberOfWays(n, x, y));         ``    ``}``}` `// This code is contributed by Nidhiva`

## Python3

 `# Python3 implementation for the above approach` `# Function to find factorial of a number``def` `factorial(n):` `    ``if` `(n <``=` `1``):``        ``return` `1``;``    ``return` `n ``*` `factorial(n ``-` `1``);` `# Function to find nCr``def` `nCr(n, r):` `    ``return` `(factorial(n) ``/``           ``(factorial(n ``-` `r) ``*` `factorial(r)));` `# Function to find the number of ways ``# to arrange 2*N persons``def` `NumberOfWays(n, x, y):` `    ``return` `(nCr(``2` `*` `n ``-` `x ``-` `y, n ``-` `x) ``*``            ``factorial(n) ``*` `factorial(n)); ` `# Driver code``n, x, y ``=` `5``, ``4``, ``2``;` `# Function call``print``(``int``(NumberOfWays(n, x, y)));` `# This code is contributed by PrinciRaj1992`

## C#

 `// C# implementation for the above approach``using` `System;``    ` `class` `GFG``{``    ` `    ``// Function to returns factorial of n``    ``static` `int` `factorial(``int` `n) ``    ``{ ``        ``if` `(n <= 1) ``            ``return` `1; ``        ``return` `n * factorial(n - 1); ``    ``} ``    ` `    ``// Function to find nCr ``    ``static` `int` `nCr(``int` `n, ``int` `r) ``    ``{ ``        ``return` `factorial(n) / (factorial(n - r) *``                               ``factorial(r)); ``    ``} ``    ` `    ``// Function to find the number of ways``    ``// to arrange 2*N persons ``    ``static` `int` `NumberOfWays(``int` `n, ``int` `x, ``int` `y) ``    ``{ ``        ``return` `nCr(2 * n - x - y, n - x) * ``            ``factorial(n) * factorial(n); ``    ``} ``    ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `n = 5, x = 4, y = 2; ``        ` `        ``// Function call ``        ``Console.WriteLine(NumberOfWays(n, x, y));         ``    ``}``}` `// This code is contributed by Princi Singh`

## PHP

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## Javascript

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Output:
`57600`

Time Complexity: O(N)
Auxiliary Space: O(N), for recursive stack space.

Method – 2 O(1) Space Solution

In the above solution, we implemented the factorial function in a recursive manner if we implement the factorial function in an iterative manner then we don’t need O(n) auxiliary extra stack space so this way solution becomes O(1) space complexity solution
Below is an implementation for the same   :

## C++

 `// C++ code for above mentioned solution` `#include ``using` `namespace` `std;` `// Iterative function to find factorial of a number``int` `factorial(``int` `n) {``    ``int` `fact = 1;``    ``for` `(``int` `i = 1; i <= n; i++) {``        ``fact = fact * i;``    ``}``    ``return` `fact;``}` `// Function to find nCr``int` `nCr(``int` `n, ``int` `r) {``    ``return` `factorial(n) / (factorial(n - r) * factorial(r));``}`  `// Function to find the number of ways to arrange 2*N persons``int` `NumberOfWays(``int` `n, ``int` `x, ``int` `y) {``    ``return` `nCr(2 * n - x - y, n - x) * factorial(n) * factorial(n);``}`  `// Driver code``int` `main() {``    ``int` `n = 5, x = 4, y = 2;` `    ``// Function call``    ``cout << NumberOfWays(n, x, y);` `    ``return` `0;``}`

## Java

 `// Java code for above mentioned solution``public` `class` `Main {` `  ``// Iterative function to find factorial of a number``  ``public` `static` `int` `factorial(``int` `n) {``    ``int` `fact = ``1``;``    ``for` `(``int` `i = ``1``; i <= n; i++) {``      ``fact = fact * i;``    ``}``    ``return` `fact;``  ``}` `  ``// Function to find nCr``  ``public` `static` `int` `nCr(``int` `n, ``int` `r) {``    ``return` `factorial(n) / (factorial(n - r) * factorial(r));``  ``}` `  ``// Function to find the number of ways to arrange 2*N persons``  ``public` `static` `int` `NumberOfWays(``int` `n, ``int` `x, ``int` `y) {``    ``return` `nCr(``2` `* n - x - y, n - x) * factorial(n) * factorial(n);``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args) {``    ``int` `n = ``5``, x = ``4``, y = ``2``;` `    ``// Function call``    ``System.out.println(NumberOfWays(n, x, y));``  ``}``}`

## Python3

 `import` `math` `# Function to find factorial of a number``def` `factorial(n):``    ``fact ``=` `1``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``fact ``=` `fact ``*` `i``    ``return` `fact` `# Function to find nCr``def` `nCr(n, r):``    ``return` `factorial(n) ``/``/` `(factorial(n ``-` `r) ``*` `factorial(r))` `# Function to find the number of ways to arrange 2*N persons``def` `NumberOfWays(n, x, y):``    ``return` `nCr(``2``*``n ``-` `x ``-` `y, n ``-` `x) ``*` `factorial(n) ``*` `factorial(n)` `# Driver code``n, x, y ``=` `5``, ``4``, ``2` `# Function call``print``(NumberOfWays(n, x, y))`

## C#

 `using` `System;` `public` `class` `Program {``  ` `  ``// Iterative function to find factorial of a number``  ``public` `static` `int` `Factorial(``int` `n) {``    ``int` `fact = 1;``    ``for` `(``int` `i = 1; i <= n; i++) {``      ``fact = fact * i;``    ``}``    ``return` `fact;``  ``}` `  ``// Function to find nCr``  ``public` `static` `int` `nCr(``int` `n, ``int` `r) {``    ``return` `Factorial(n) / (Factorial(n - r) * Factorial(r));``  ``}` `  ``// Function to find the number of ways to arrange 2*N persons``  ``public` `static` `int` `NumberOfWays(``int` `n, ``int` `x, ``int` `y) {``    ``return` `nCr(2 * n - x - y, n - x) * Factorial(n) * Factorial(n);``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main() {``    ``int` `n = 5, x = 4, y = 2;` `    ``// Function call``    ``Console.WriteLine(NumberOfWays(n, x, y));``  ``}``}` `// This code is contributed by divyansh2212`

## Javascript

 `//GFG``// Recursive function to find factorial of a number``function` `factorial(n) {``    ``let fact = 1;``    ``for` `(let i = 1; i <= n; i++) {``        ``fact = fact * i;``    ``}``    ``return` `fact;``}` `// Function to find nCr``function` `nCr(n, r) {``  ``return` `factorial(n) / (factorial(n - r) * factorial(r));``}` `// Function to find the number of ways to arrange 2*N persons``function` `numberOfWays(n, x, y) {``  ``return` `nCr(2 * n - x - y, n - x) * factorial(n) * factorial(n);``}` `// Driver code``const n = 5, x = 4, y = 2;` `// Function call``console.log(numberOfWays(n, x, y));``// This code is contributed by Sundaram`

Output
`57600`

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