一次将n个事物与m个事物一起排列的排列永远不会融合在一起

给定n和m ，任务是找到n个不同事物的排列数目，一次将它们全部吸收，以使m个特定事物永远不会在一起。
例子：

Input  : 7, 3
Output : 420

Input  : 9, 2
Output : 282240

方法：
公式的推导–
使用n个不同的对象一次取走的所有布置的总数= $n!$
当m个特定的事物总是在一起时，一次包含全部n个不同事物的排列数是$(n-m+1)! × m!$
因此， $n$不同的事物一次全部占据， $m$特殊的事情永远不会融合在一起–

Permutations = n! - (n-m+1)! × m!

以下是上述方法的Python实现–

C++

#include
using namespace std;
 
int factorial(int n)
{
    int fact = 1;
    for (int i = 2; i <= n; i++)
        fact = fact * i ;
    return (fact);
}
 
int result(int n, int m)
{
    return(factorial(n) -  
           factorial(n - m + 1) *
           factorial(m));
}
 
// Driver Code
int main()
{
    cout(result(5, 3));
}
 
// This code is contributed by Mohit Kumar

Java

class GFG
{
static int factorial(int n)
{
    int fact = 1;
    for (int i = 2; i <= n; i++)
        fact = fact * i ;
    return (fact);
}
 
static int result(int n, int m)
{
    return(factorial(n) -
           factorial(n - m + 1) *
           factorial(m));
}
 
// Driver Code
public static void main(String args[])
{
    System.out.println(result(5, 3));
}
}
 
// This code is contributed by Arnab Kundu

Python3

def factorial(n):
    fact = 1;
    for i in range(2, n + 1):
        fact = fact * i
    return (fact)
 
def result(n, m):
    return(factorial(n) - factorial(n - m + 1) * factorial(m))
 
# driver code
print(result(5, 3))

C#

using System;
 
class GFG
{
    static int factorial(int n)
    {
        int fact = 1;
        for (int i = 2; i <= n; i++)
            fact = fact * i ;
        return (fact);
    }
     
    static int result(int n, int m)
    {
        return(factorial(n) -
               factorial(n - m + 1) *
               factorial(m));
    }
     
    // Driver Code
    public static void Main()
    {
        Console.WriteLine(result(5, 3));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Below is the JavaScript implementation of above approach
 
function factorial(n)
{
    let fact = 1;
    for (let i = 2; i <= n; i++)
        fact = fact * i ;
    return (fact);
}
 
function result(n, m)
{
    return(factorial(n) -
        factorial(n - m + 1) *
        factorial(m));
}
 
// Driver Code
 
document.write(result(5, 3));
 
// This code is contributed by Surbhi Tyagi.
 

输出 ：

84