Find the number of good permutations
Given two integers N and K. The task is to find the number of good permutations of first N natural numbers. A permutation is called good if there exist at least N – K indices i (1 ≤ i ≤ N) such that Pi = 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 Pi not equals to 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 Pi not equals to i – there are nCm ways to do this then we need to construct a permutation Q for chosen indices such that for every chosen index Qi not equals to i. Permutations with this property are called derangements and the number of derangements of fixed size can be calculated using exhaustive search since 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; } |
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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 |
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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 |
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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 */ |
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PHP
= 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.
?>
Output:
11
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