Permutation refers to the process of arranging all the members of a given set to form a sequence. The number of permutations on a set of n elements is given by n! , where “!” represents factorial.

The **Permutation Coefficient** represented by P(n, k) is used to represent the number of ways to obtain an ordered subset having k elements from a set of n elements.

Mathematically it’s given as:

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Examples:

P(10, 2) = 90 P(10, 3) = 720 P(10, 0) = 1 P(10, 1) = 10

The coefficient can also be computed recursively using the below recursive formula:

P(n, k) = P(n-1, k) + k* P(n-1, k-1)

If we observe closely, we can analyze that the problem has overlapping substructure, hence we can apply dynamic programming here. Below is a C++ program implementing the same idea.

// A Dynamic Programming based solution that uses // table P[][] to calculate the Permutation Coefficient #include<bits/stdc++.h> // Returns value of Permutation Coefficient P(n, k) int permutationCoeff(int n, int k) { int P[n+1][k+1]; // Caculate value of Permutation Coefficient in // bottom up manner for (int i =0; i <= n; i++) { for (int j = 0; j <= std::min(i,k); j++) { // Base Cases if (j == 0) P[i][j] = 1; // Calculate value using previosly stored values else P[i][j] = P[i-1][j] + (j*P[i-1][j-1]); // This step is important as P(i,j)=0 for j>i P[i][j+1] = 0; } } return P[n][k]; } // Driver program to test above function int main() { int n = 10, k = 2; printf("Value of P(%d, %d) is %d ", n, k, permutationCoeff(n, k) ); return 0; }

Output:

Value of P(10, 2) is 90

Here as we can see the time complexity is O(n*k) and space complexity is O(n*k) as the program uses an auxiliary matrix to store the result.

**Can we do it in O(n) time ?**

Let us suppose we maintain a single 1D array to compute the factorials up to n. We can use computed factorial value and apply the formula P(n, k) = n! / (n-k)!. Below is a C++ program illustrating the same concept.

// A O(n) solution that uses table fact[] to // calculate the Permutation Coefficient #include<bits/stdc++.h> // Returns value of Permutation Coefficient P(n, k) int permutationCoeff(int n, int k) { int fact[n+1]; // base case fact[0] = 1; // Caculate value factorials up to n for (int i=1; i<=n; i++) fact[i] = i*fact[i-1]; // P(n,k)= n!/(n-k)! return fact[n]/fact[n-k]; } // Driver program to test above function int main() { int n = 10, k = 2; printf ("Value of P(%d, %d) is %d ", n, k, permutationCoeff(n, k) ); return 0; }

Output:

Value of P(10, 2) is 90

**A O(n) time and O(1) Extra Space Solution**

// A O(n) time and O(1) extra space solution to // calculate the Permutation Coefficient #include <iostream> using namespace std; int PermutationCoeff(int n, int k) { int Fn = 1, Fk; // Compute n! and (n-k)! for (int i=1; i<=n; i++) { Fn *= i; if (i == n-k) Fk = Fn; } int coeff = Fn/Fk; return coeff; } // Driver program to test above function int main() { int n=10, k=2; cout << "Value of P(" << n << ", " << k << ") is " << PermutationCoeff(n, k); return 0; }

Output:

Value of P(10, 2) is 90

Thanks to Shiva Kumar for suggesting this solution.

This article is contributed by Ashutosh Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above