n 个事物一次全部排列，m 个事物永远不会在一起
给定 n 和 m ，任务是找到 n 个不同事物的排列数量，一次取全部，使得 m 个特定事物永远不会在一起。 举例:
Input : 7, 3
Output : 420
Input : 9, 2
Output : 282240
方法:
公式推导–
使用 n 个不同的物体一次取全部的可能排列总数= 
当 m 个特定的事物总是聚集在一起时，n 个不同的事物一次取全部的排列数是
因此，当
个特定的事物从不聚集在一起时–
个不同的事物一次取全部的排列数
Permutations = n! - (n-m+1)! × m!
下面是上述方法的 Python 实现–
C++
#include<bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
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
Python 3
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
java 描述语言
<script>
// 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.
</script>
输出:
84
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