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 inways 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
Input : N = 3, X = 1, Y = 2
Output : 108
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 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 <bits/stdc++.h>using namespace std;// Function to find factorial of a numberint factorial(int n){ if (n <= 1) return 1; return n * factorial(n - 1);}// Function to find nCrint nCr(int n, int r){ return factorial(n) / (factorial(n - r) * factorial(r));}// Function to find the number of ways to arrange 2*N personsint NumberOfWays(int n, int x, int y){ return nCr(2*n-x-y, n-x) * factorial(n) * factorial(n); }// Driver codeint main(){ int n = 5, x = 4, y = 2; // Function call cout << NumberOfWays(n, x, y); return 0;} |
Java
// Java implementation for the above approachimport 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 numberdef factorial(n): if (n <= 1): return 1; return n * factorial(n - 1);# Function to find nCrdef nCr(n, r): return (factorial(n) / (factorial(n - r) * factorial(r)));# Function to find the number of ways # to arrange 2*N personsdef NumberOfWays(n, x, y): return (nCr(2 * n - x - y, n - x) * factorial(n) * factorial(n)); # Driver coden, x, y = 5, 4, 2;# Function callprint(int(NumberOfWays(n, x, y)));# This code is contributed by PrinciRaj1992 |
C#
// C# implementation for the above approachusing 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
<?php // PHP implementation for the above approach// Function to find factorial of a number function factorial($n) { if ($n <= 1) return 1; return $n * factorial($n - 1); } // 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 $n = 5;$x = 4;$y = 2; // Function call echo (NumberOfWays($n, $x, $y)); // This code is contributed by Naman_garg. ?> |
Javascript
<script>// JavaScript implementation for the above approach// Function to returns factorial of n function factorial(n) { if (n <= 1) return 1; return n * factorial(n - 1); } // 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 var n = 5, x = 4, y = 2; // Function call document.write(NumberOfWays(n, x, y));// This code contributed by Rajput-Ji</script> |
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 <bits/stdc++.h>using namespace std;// Iterative function to find factorial of a numberint factorial(int n) { int fact = 1; for (int i = 1; i <= n; i++) { fact = fact * i; } return fact;}// Function to find nCrint nCr(int n, int r) { return factorial(n) / (factorial(n - r) * factorial(r));}// Function to find the number of ways to arrange 2*N personsint NumberOfWays(int n, int x, int y) { return nCr(2 * n - x - y, n - x) * factorial(n) * factorial(n);}// Driver codeint main() { int n = 5, x = 4, y = 2; // Function call cout << NumberOfWays(n, x, y); return 0;} |
Java
// Java code for above mentioned solutionpublic 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 numberdef factorial(n): fact = 1 for i in range(1, n+1): fact = fact * i return fact# Function to find nCrdef nCr(n, r): return factorial(n) // (factorial(n - r) * factorial(r))# Function to find the number of ways to arrange 2*N personsdef NumberOfWays(n, x, y): return nCr(2*n - x - y, n - x) * factorial(n) * factorial(n)# Driver coden, x, y = 5, 4, 2# Function callprint(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 numberfunction factorial(n) { let fact = 1; for (let i = 1; i <= n; i++) { fact = fact * i; } return fact;}// Function to find nCrfunction nCr(n, r) { return factorial(n) / (factorial(n - r) * factorial(r));}// Function to find the number of ways to arrange 2*N personsfunction numberOfWays(n, x, y) { return nCr(2 * n - x - y, n - x) * factorial(n) * factorial(n);}// Driver codeconst n = 5, x = 4, y = 2;// Function callconsole.log(numberOfWays(n, x, y));// This code is contributed by Sundaram |
Output
57600
Time Complexity: O(N)
Auxiliary Space: O(1)
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