在一张桌子的两边安排 2*N 个人，X 和 Y 个人相对的方式数量

原文:https://www . geeksforgeeks . org/排列方式数-在 x 和 y 人相对的桌子两侧排列 2n 人/

给定三个整数 N 、 X 和 Y 。任务是找到将 2*N 人沿桌子两侧排列的方法数量，每侧的椅子数量为 N 个，使得 X 人在一侧， Y 人在另一侧。 注:X 和 Y 均小于或等于 N. 例:

输入: N = 5，X = 4，Y = 2 输出: 57600 说明: 总人数 10 人。一边 x 人，另一边 Y 人，则剩下 10–4–2 = 4 人。我们可以通过4\choose 1 方式选择其中 5–4 = 1 人坐在一边，其余人员将自动坐在另一边。每一面的布置都是在 5！方法。途径数为4\choose 1 .5！5! 输入: N = 3，X = 1，Y = 2 输出: 108

进场: 总人数 2N，让我们称双方为 A 和 B，A 面 X 人，B 面 Y 人，则剩下 2 * N–X–Y 人。我们可以通过2*N - X - Y\choose N-X 的方式为 A 面选择其中的 N-X 个，剩下的人会自动坐在 b 面的另一边，每一面的排列都是在 N 中完成的！方法。沿桌子两边安排 2N** 人的方式数量为2*N - X - Y\choose N-X 。n！n！ 以下是上述方法的实施:

C++

#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?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.
?>

java 描述语言

<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

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