n个男孩和女孩在圆桌旁的座位安排

有n个男孩和n个女孩围成一圈坐在圆桌旁。任务是找到让n个男孩和n个女孩交替坐在圆桌旁的几种方式。给定n <10

例子：

Input: n = 5 
Output: 2880

Input: n = 1
Output: 1

方法：

1. 首先找到在圆桌会议上安排男孩子的总数。
   在桌子上布置男孩的方式数量= (n-1)！
2. 在安排了男孩之后，现在安排了女孩。在男生就座之后，他们之间没有空间。因此，有n个职位和n个女孩。因此，女孩坐在男孩之间的安排总数为n！ 。
3. 因此，方法总数=(男孩的布置数量)*(男孩中女孩坐在那里的方法数量)= (n-1)！ *(n！)

下面是上述方法的实现：

C++

// C++ program to find number of ways in which
// n boys and n girls can sit alternatively
// sound a round table.
#include 
using namespace std;
  
#define ll long int
  
int main()
{
  
    // Get n
    ll n = 5;
  
    // find fac1 = (n-1)!
    ll fac1 = 1;
    for (int i = 2; i <= n - 1; i++)
        fac1 = fac1 * i;
  
    // Find fac2 = n!
    ll fac2 = fac1 * n;
  
    // Find total number of ways
    ll totalWays = fac1 * fac2;
  
    // Print the total number of ways
    cout << totalWays << endl;
  
    return 0;
}

Java

// Java program to find number of ways
// in which n boys and n girls can sit 
// alternatively sound a round table.
import java .io.*;
  
class GFG
{
public static void main(String[] args)
{
  
    // Get n
    long n = 5;
  
    // find fac1 = (n-1)!
    long fac1 = 1;
    for (int i = 2; i <= n - 1; i++)
        fac1 = fac1 * i;
  
    // Find fac2 = n!
    long fac2 = fac1 * n;
  
    // Find total number of ways
    long totalWays = fac1 * fac2;
  
    // Print the total number of ways
    System.out.println(totalWays);
}
}
  
// This code is contributed
// by anuj_67..

Python3

# Python3 program to find number 
# of ways in which n boys and n 
# girls can sit alternatively 
# sound a round table.
  
# Get n 
n = 5
  
# find fac1 = (n-1)! 
fac1 = 1
for i in range(2, n):
    fac1 = fac1 * i
  
# Find fac2 = n! 
fac2 = fac1 * n
  
# Find total number of ways
totalWays = fac1 * fac2
  
# Print the total number of ways 
print(totalWays)
  
# This code is contributed 
# by sahilshelangia

C#

// C# program to find number of ways
// in which n boys and n girls can sit 
// alternatively sound a round table.
using System;
  
class GFG
{
public static void Main()
{
  
    // Get n
    long n = 5;
  
    // find fac1 = (n-1)!
    long fac1 = 1;
    for (int i = 2; i <= n - 1; i++)
        fac1 = fac1 * i;
  
    // Find fac2 = n!
    long fac2 = fac1 * n;
  
    // Find total number of ways
    long totalWays = fac1 * fac2;
  
    // Print the total number of ways
    Console.WriteLine(totalWays);
}
}
  
// This code is contributed
// by Akanksha Rai(Abby_akku)

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输出：

2880