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 P_i = i.

Examples:

Input: N = 4, K = 1
Output: 1
{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 P_i 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 P_i not equals to i – there are ⁿC_m ways to do this, then we need to construct a permutation Q for chosen indices such that for every chosen index Q_i 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:

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