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Count the number of special permutations

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  • Difficulty Level : Medium
  • Last Updated : 24 Jun, 2022
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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 Pi = i
Prerequisite: Derangements

Examples: 

Input: n = 4, k = 2 
Output:
{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 

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

f(n – k) + f(n – k + 1) + f(n – k + 2) + … + f(n – 1) + fn 
 

Now, fx can be calculated by choosing x indices from n where Pi = i and calculating the number of derangements for (n – i) other indices as for them Pi should not be equal to i then multiplying the result by nCx 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);
 
        // Derangements 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);
     
            // Derangements 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 permutation
 
# 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 permutations
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)
         
        # Derangements 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);
      
            // Derangements 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);
 
        // Derangements 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
?>

Javascript




<script>
 
      // JavaScript 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) {
        var ans = 1;
        if (r > n - r) r = n - r;
        for (var 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) {
        var der = [...Array(n + 1)];
 
        der[0] = 1;
        der[1] = 0;
        der[2] = 1;
 
        for (var 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) {
        var ans = 0;
        for (var i = n - k; i <= n; i++) {
          // Ways to choose i indices from n indices
          var ways = nCr(n, i);
 
          // Derangements of (n - i) indices
          ans += ways * countDerangements(n - i);
        }
        return ans;
      }
 
      // Driver Code to test above functions
      var n = 5,
        k = 3;
      document.write(countPermutations(n, k));
       
</script>

Output: 

31

 

Time Complexity: O(N^2), since there runs a loop inside another loop.
Auxiliary Space: O(N), since N extra space has been taken.


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