Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Find the number of good permutations

  • Last Updated : 20 May, 2021

Given two integers N and K. The task is to find the number of good permutations of the first N natural numbers. A permutation is called good if there exist at least N – K indices i (1 ≤ i ≤ N) such that Pi = i.

Examples: 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: N = 4, K = 1 
Output:
{1, 2, 3, 4} is the only possible good permutation.



Input: N = 5, K = 2 
Output: 11 

Approach: Let’s iterate on m which is the number of indices such that Pi does not equal i. Obviously, 0 ≤ m ≤ k
In order to count the number of permutations with fixed m, we need to choose the indices that have the property Pi not equals to i – there are nCm ways to do this, then we need to construct a permutation Q for chosen indices such that for every chosen index Qi is not equaled to i. Permutations with this property are called derangements and the number of derangements of fixed size can be calculated using an exhaustive search for m ≤ 4.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of good permutations
int Permutations(int n, int k)
{
    // For m = 0, ans is 1
    int ans = 1;
 
    // If k is greater than 1
    if (k >= 2)
        ans += (n) * (n - 1) / 2;
 
    // If k is greater than 2
    if (k >= 3)
        ans += (n) * (n - 1) * (n - 2) * 2 / 6;
 
    // If k is greater than 3
    if (k >= 4)
        ans += (n) * (n - 1) * (n - 2) * (n - 3) * 9 / 24;
 
    return ans;
}
 
// Driver code
int main()
{
    int n = 5, k = 2;
    cout << Permutations(n, k);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function to return the count of good permutations
static int Permutations(int n, int k)
{
    // For m = 0, ans is 1
    int ans = 1;
 
    // If k is greater than 1
    if (k >= 2)
        ans += (n) * (n - 1) / 2;
 
    // If k is greater than 2
    if (k >= 3)
        ans += (n) * (n - 1) * (n - 2) * 2 / 6;
 
    // If k is greater than 3
    if (k >= 4)
        ans += (n) * (n - 1) * (n - 2) * (n - 3) * 9 / 24;
 
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 5, k = 2;
    System.out.println(Permutations(n, k));
}
}
 
// This code contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to return the count
# of good permutations
def Permutations(n, k):
 
    # For m = 0, ans is 1
    ans = 1
 
    # If k is greater than 1
    if k >= 2:
        ans += (n) * (n - 1) // 2
 
    # If k is greater than 2
    if k >= 3:
        ans += ((n) * (n - 1) *
                (n - 2) * 2 // 6)
 
    # If k is greater than 3
    if k >= 4:
        ans += ((n) * (n - 1) * (n - 2) *
                      (n - 3) * 9 // 24)
 
    return ans
 
# Driver code
if __name__ == "__main__":
 
    n, k = 5, 2
    print(Permutations(n, k))
     
# This code is contributed
# by Rituraj Jain

C#




// C# implementation of the above approach.
using System;
 
class GFG
{
 
// Function to return the count of good permutations
static int Permutations(int n, int k)
{
    // For m = 0, ans is 1
    int ans = 1;
 
    // If k is greater than 1
    if (k >= 2)
        ans += (n) * (n - 1) / 2;
 
    // If k is greater than 2
    if (k >= 3)
        ans += (n) * (n - 1) * (n - 2) * 2 / 6;
 
    // If k is greater than 3
    if (k >= 4)
        ans += (n) * (n - 1) * (n - 2) * (n - 3) * 9 / 24;
 
    return ans;
}
 
// Driver code
public static void Main()
{
    int n = 5, k = 2;
    Console.WriteLine(Permutations(n, k));
}
}
 
/* This code contributed by PrinciRaj1992 */

PHP




<?php
// PHP implementation of the approach
 
// Function to return the count
// of good permutations
function Permutations($n, $k)
{
    // For m = 0, ans is 1
    $ans = 1;
 
    // If k is greater than 1
    if ($k >= 2)
        $ans += ($n) * ($n - 1) / 2;
 
    // If k is greater than 2
    if ($k >= 3)
        $ans += ($n) * ($n - 1) *
                       ($n - 2) * 2 / 6;
 
    // If k is greater than 3
    if ($k >= 4)
        $ans += ($n) * ($n - 1) * ($n - 2) *
                       ($n - 3) * 9 / 24;
 
    return $ans;
}
 
// Driver code
$n = 5; $k = 2;
echo(Permutations($n, $k));
 
// This code contributed by Code_Mech.
?>

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return the count of good permutations
function Permutations(n, k)
{
     
    // For m = 0, ans is 1
    var ans = 1;
 
    // If k is greater than 1
    if (k >= 2)
        ans += (n) * (n - 1) / 2;
 
    // If k is greater than 2
    if (k >= 3)
        ans += (n) * (n - 1) *
           (n - 2) * 2 / 6;
 
    // If k is greater than 3
    if (k >= 4)
        ans += (n) * (n - 1) * (n - 2) *
                     (n - 3) * 9 / 24;
 
    return ans;
}
 
// Driver Code
var n = 5, k = 2;
document.write(Permutations(n, k));
   
// This code is contributed by Khushboogoyal499
 
</script>
Output: 
11

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :