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++
// 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
// 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
# 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#
// 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
<?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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