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 Pi = 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

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++

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// 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;
}

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Java














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// 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 


} 



















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Python3














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# 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) 



















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C#














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// 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 



















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PHP














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<?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  


?> 



















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Output:


31








    

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