Permutations of n things taken all at a time with m things never come together

Given n and m, the task is to find the number of permutations of n distinct things taking them all at a time such that m particular things never come together.

Examples:

Input  : 7, 3
Output : 420

Input  : 9, 2
Output : 282240

Approach:

Derivation of the formula –
Total number of arrangements possible using n distinct objects taking all at a time = $n!$
Number of arrangements of n distinct things taking all at a time, when m particular things always come together, is $(n-m+1)! × m!$
Hence, the number of permutations of $n$ distinct things taking all at a time, when $m$ particular things never come together –

Permutations = n! - (n-m+1)! × m!

Below is the Python implementation of above approach –

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

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 

Output :

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My Personal Notes

C++

Java

Python3

C#

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