Count the number of special permutations

Given two positive integers n and k, the task is to count the number of special permutations. A special permutation P is defined as a permutation of first n natural numbers in which there exists at least (n – k) indices such that P_i = i.
Prerequisite : Derangements

Examples:

Input: n = 4, k = 2
Output: 7
{1, 2, 3, 4}, {1, 2, 4, 3}, {4, 2, 3, 1}, {2, 1, 3, 4}, {1, 4, 3, 2}, {1, 3, 2, 4} and {3, 2, 1, 4} are the only possible special permutations.

Input: n = 5, k = 3
Output: 31

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Let the function f_x denote the number of special permutations in which there exists exactly x indices such that P_i = i. Hence, the required answer will be:

f_{(n – k)} + f_{(n – k + 1)} + f_{(n – k + 2)} + … + f_{(n – 1)} + f_n

Now, f_x can be calculated by choosing x indices from n where P_i = i and calculating the number of derangements for (n – i) other indices as for them P_i should not be equal to i then multiplying the result by ⁿC_x as there can be different ways to select x indices from n.

Below is the implementation of the above approach:

C++

// C++ program to count the number 
// of required permutations 
#include <bits/stdc++.h> 
using namespace std; 
#define ll long long int 
  
// Function to return the number of ways 
// to choose r objects out of n objects 
int nCr(int n, int r) 
{ 
    int ans = 1; 
    if (r > n - r) 
        r = n - r; 
    for (int i = 0; i < r; i++) { 
        ans *= (n - i); 
        ans /= (i + 1); 
    } 
    return ans; 
} 
  
// Function to return the number 
// of derangements of n 
int countDerangements(int n) 
{ 
    int der[n + 1]; 
  
    der[0] = 1; 
    der[1] = 0; 
    der[2] = 1; 
  
    for (int i = 3; i <= n; i++) 
        der[i] = (i - 1) * (der[i - 1] 
                         + der[i - 2]); 
    return der[n]; 
} 
  
// Function to return the required 
// number of permutations 
ll countPermutations(int n, int k) 
{ 
    ll ans = 0; 
    for (int i = n - k; i <= n; i++) { 
  
        // Ways to choose i indices from n indices 
        int ways = nCr(n, i); 
  
        // Dearangements of (n - i) indices 
        ans += ways * countDerangements(n - i); 
    } 
    return ans; 
} 
  
// Driver Code to test above functions 
int main() 
{ 
    int n = 5, k = 3; 
    cout << countPermutations(n, k); 
    return 0; 
} 

Java

// Java program to count the number  
// of required permutations 
      
public class GFG{ 
      
    // Function to return the number of ways  
    // to choose r objects out of n objects  
    static int nCr(int n, int r)  
    {  
        int ans = 1;  
        if (r > n - r)  
            r = n - r;  
        for (int i = 0; i < r; i++) {  
            ans *= (n - i);  
            ans /= (i + 1);  
        }  
        return ans;  
    }  
      
    // Function to return the number  
    // of derangements of n  
    static int countDerangements(int n)  
    {  
        int der[] = new int[ n + 3];  
      
        der[0] = 1;  
        der[1] = 0;  
        der[2] = 1;  
      
        for (int i = 3; i <= n; i++)  
            der[i] = (i - 1) * (der[i - 1]  
                            + der[i - 2]);  
        return der[n];  
    }  
      
    // Function to return the required  
    // number of permutations  
    static int countPermutations(int n, int k)  
    {  
        int ans = 0;  
        for (int i = n - k; i <= n; i++) {  
      
            // Ways to choose i indices from n indices  
            int ways = nCr(n, i);  
      
            // Dearangements of (n - i) indices  
            //System.out.println(ans) ; 
            ans += (ways * countDerangements(n- i));  
        }  
        return ans;  
    }  
      
    // Driver Code to test above functions  
    public static void main(String []args) 
    {  
        int n = 5, k = 3;  
        System.out.println(countPermutations(n, k)) ; 
    }  
    // This code is contributed by Ryuga 
} 

Python3

# Python 3 program to count the number 
# of required premutation 
  
# function to return the number of ways  
# to chooser objects out of n objects 
def nCr(n, r): 
    ans = 1
    if r > n - r: 
        r = n - r 
    for i in range(r): 
        ans *= (n - i) 
        ans /= (i + 1) 
    return ans 
  
# function to return the number of 
# degrangemnets of n 
def countDerangements(n): 
    der = [0 for i in range(n + 3)] 
      
    der[0] = 1
    der[1] = 0
    der[2] = 1
      
    for i in range(3, n + 1): 
        der[i] = (i - 1) * (der[i - 1] + 
                            der[i - 2]) 
          
    return der[n] 
  
# function to return the required  
# number of premutations 
def countPermutations(n, k): 
    ans = 0
    for i in range(n - k, n + 1): 
          
        # ways to chose i indices  
        # from n indices 
        ways = nCr(n, i) 
          
        # Dearrangements of (n-i) indices 
        ans += ways * countDerangements(n - i) 
    return ans 
  
# Driver Code 
n, k = 5, 3
  
print(countPermutations(n, k)) 
  
# This code is contributed by 
# Mohit kumar 29 (IIIT gwalior) 

C#

      
// C# program to count the number  
// of required permutations 
using System;     
public class GFG{ 
       
    // Function to return the number of ways  
    // to choose r objects out of n objects  
    static int nCr(int n, int r)  
    {  
        int ans = 1;  
        if (r > n - r)  
            r = n - r;  
        for (int i = 0; i < r; i++) {  
            ans *= (n - i);  
            ans /= (i + 1);  
        }  
        return ans;  
    }  
       
    // Function to return the number  
    // of derangements of n  
    static int countDerangements(int n)  
    {  
        int []der = new int[ n + 3];  
       
        der[0] = 1;  
        der[1] = 0;  
        der[2] = 1;  
       
        for (int i = 3; i <= n; i++)  
            der[i] = (i - 1) * (der[i - 1]  
                            + der[i - 2]);  
        return der[n];  
    }  
       
    // Function to return the required  
    // number of permutations  
    static int countPermutations(int n, int k)  
    {  
        int ans = 0;  
        for (int i = n - k; i <= n; i++) {  
       
            // Ways to choose i indices from n indices  
            int ways = nCr(n, i);  
       
            // Dearangements of (n - i) indices  
            //System.out.println(ans) ; 
            ans += (ways * countDerangements(n- i));  
        }  
        return ans;  
    }  
       
    // Driver Code to test above functions  
    public static void Main() 
    {  
        int n = 5, k = 3;  
        Console.WriteLine(countPermutations(n, k)) ; 
    }  
} 
// This code is contributed by 29AjayKumar 

PHP

<?php 
// PHP program to count the number 
// of required permutations 
  
// Function to return the number of ways 
// to choose r objects out of n objects 
function nCr($n, $r) 
{ 
    $ans = 1; 
    if ($r > $n - $r) 
        $r = $n - $r; 
    for ($i = 0; $i < $r; $i++) 
    { 
        $ans *= ($n - $i); 
        $ans /= ($i + 1); 
    } 
    return $ans; 
} 
  
// Function to return the number 
// of derangements of n 
function countDerangements($n) 
{ 
    $der = array($n + 1); 
  
    $der[0] = 1; 
    $der[1] = 0; 
    $der[2] = 1; 
  
    for ($i = 3; $i <= $n; $i++) 
        $der[$i] = ($i - 1) *  
                   ($der[$i - 1] +  
                    $der[$i - 2]); 
    return $der[$n]; 
} 
  
// Function to return the required 
// number of permutations 
function countPermutations($n, $k) 
{ 
    $ans = 0; 
    for ($i = $n - $k; $i <= $n; $i++) 
    { 
  
        // Ways to choose i indices 
        // from n indices 
        $ways = nCr($n, $i); 
  
        // Dearangements of (n - i) indices 
        $ans += $ways * countDerangements($n - $i); 
    } 
    return $ans; 
} 
  
// Driver Code 
$n = 5; 
$k = 3; 
echo (countPermutations($n, $k)); 
  
// This code is contributed  
// by Shivi_Aggarwal  
?> 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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