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:
Image Source : Wiki
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)
The recursive formula for permutation-coefficient is :
P(n, k) = P(n-1, k) + k* P(n-1, k-1)
But how ??
here is the proof,
We already know,
The binomial coefficient is nCk = n!
k! (n-k)!
and, permutation-coefficient nPr = n!
(n-k)!
So, I can write
nCk = nPk
k!
=> k! * nCk = nPk ———————- eq.1
The recursive formula for the Binomial coefficient nCk can be written as,
nCk(n,k) = nCk(n-1,k-1) + nCk(n-1,k) you can refer to the following article for more details
https://www.geeksforgeeks.org/binomial-coefficient-dp-9/
Basically, it makes use of Pascal’s triangle which states that in order to fill the value at nCk[n][k] you need the summation of nCk[n-1][k-1] and nCk[n-1][k] along with some base cases. i.e,
nCk[n][k] = nCk[n-1][k-1]+nCk[n-1][k].
nCk[n][0] = nCk[n][n] = 1 (Base Case)
Anyways, let’s proceed with our eq.1
=> k! * nCk = nPk
=> k! * ( nCk(n-1,k-1) + nCk(n-1,k) ) = nPk [ as, nCk = nCk(n-1,k-1)+nCk(n-1,k) ]
=> k! * ( (n-1)! + (n-1)! ) = nPk
((n-1)-(k-1))! * (k-1)! (n-1-k)! * k!
=> k! * (n-1)! + k! * (n-1)! = nPk
((n-1)-(k-1))! * (k-1)! (n-1-k)! * k!
=> k * (k-1)! * (n-1)! + k! * (n-1)! = nPk [ as, k! = k*(k-1)! ]
((n-1)-(k-1))! * (k-1)! (n-1-k)! * k!
=> k * (n-1)! + (n-1)! = nPk
((n-1)-(k-1))! (n-1-k)!
(n-1)! can be replaced by nPk(n-1,k-1) as per the permutation-coefficient
((n-1) – (k-1))!
Similarly, (n-1)! can be replaced by nPk(n-1,k).
(n-1-k)!
Therefore,
=> k * nPk(n-1, k-1) + nPk(n-1, k) = nPk
Finally, that is where our recursive formula came from.
P(n, k) = P(n-1, k) + k* P(n-1, k-1)
If we observe closely, we can analyze that the problem has an overlapping substructure, hence we can apply dynamic programming here. Below is a program implementing the same idea.
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// 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]; // Calculate 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 // previously 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 Codeint main(){ int n = 10, k = 2; cout << "Value of P(" << n <<" " << k<< ") is " << permutationCoeff(n, k); return 0;}// This code is contributed by code_hunt. |
C
// 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]; // Calculate 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 // previously 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 Code int main() { int n = 10, k = 2; printf("Value of P(%d, %d) is %d ", n, k, permutationCoeff(n, k)); return 0; } |
Java
// Java code for Dynamic Programming based // solution that uses table P[][] to // calculate the Permutation Coefficient import java.io.*; import java.math.*; class GFG { // Returns value of Permutation // Coefficient P(n, k) static int permutationCoeff(int n, int k) { int P[][] = new int[n + 2][k + 2]; // Calculate value of Permutation // Coefficient in bottom up manner for (int i = 0; i <= n; i++) { for (int j = 0; j <= Math.min(i, k); j++) { // Base Cases if (j == 0) P[i][j] = 1; // Calculate value using previously // 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 Code public static void main(String args[]) { int n = 10, k = 2; System.out.println("Value of P( " + n + ","+ k +")" + " is " + permutationCoeff(n, k) ); } } // This code is contributed by Nikita Tiwari. |
Python3
# A Dynamic Programming based # solution that uses # table P[][] to calculate the # Permutation Coefficient # Returns value of Permutation # Coefficient P(n, k) def permutationCoeff(n, k): P = [[0 for i in range(k + 1)] for j in range(n + 1)] # Calculate value of Permutation # Coefficient in # bottom up manner for i in range(n + 1): for j in range(min(i, k) + 1): # Base cases if (j == 0): P[i][j] = 1 # Calculate value using # previously 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 if (j < k): P[i][j + 1] = 0 return P[n][k] # Driver Code n = 10k = 2print("Value of P(", n, ", ", k, ") is ", permutationCoeff(n, k), sep = "") # This code is contributed by Soumen Ghosh. |
C#
// C# code for Dynamic Programming based // solution that uses table P[][] to // calculate the Permutation Coefficient using System; class GFG { // Returns value of Permutation // Coefficient P(n, k) static int permutationCoeff(int n, int k) { int [,]P = new int[n + 2,k + 2]; // Calculate value of Permutation // Coefficient in bottom up manner for (int i = 0; i <= n; i++) { for (int j = 0; j <= Math.Min(i, k); j++) { // Base Cases if (j == 0) P[i,j] = 1; // Calculate value using previously // 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 Code public static void Main() { int n = 10, k = 2; Console.WriteLine("Value of P( " + n + ","+ k +")" + " is " + permutationCoeff(n, k) ); } } // This code is contributed by anuj_67.. |
PHP
<?php // A Dynamic Programming based // solution that uses table P[][] // to calculate the Permutation // Coefficient // Returns value of Permutation // Coefficient P(n, k) function permutationCoeff( $n, $k) { $P = array(array()); // Calculate value of Permutation // Coefficient in bottom up manner for($i = 0; $i <= $n; $i++) { for($j = 0; $j <= min($i, $k); $j++) { // Base Cases if ($j == 0) $P[$i][$j] = 1; // Calculate value using // previously 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 Code $n = 10; $k = 2; echo "Value of P(",$n," ,",$k,") is ", permutationCoeff($n, $k); // This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript code for Dynamic Programming based // solution that uses table P[][] to // calculate the Permutation Coefficient // Returns value of Permutation // Coefficient P(n, k) function permutationCoeff(n, k) { let P = new Array(n + 2); for(let i = 0; i < n + 2; i++) { P[i] = new Array(k + 2); } // Calculate value of Permutation // Coefficient in bottom up manner for (let i = 0; i <= n; i++) { for (let j = 0; j <= Math.min(i, k); j++) { // Base Cases if (j == 0) P[i][j] = 1; // Calculate value using previously // 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]; } let n = 10, k = 2; document.write("Value of P(" + n + ","+ k +")" + " is " + permutationCoeff(n, k) ); // This code is contributed by decode2207.</script> |
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 program illustrating the same concept.
C++
// A O(n) solution that uses // table fact[] to calculate // the Permutation Coefficient #include<bits/stdc++.h> using namespace std; // Returns value of Permutation // Coefficient P(n, k) int permutationCoeff(int n, int k) { int fact[n + 1]; // Base case fact[0] = 1; // Calculate 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 Code int main() { int n = 10, k = 2; cout << "Value of P(" << n << ", " << k << ") is " << permutationCoeff(n, k); return 0; } // This code is contributed by shubhamsingh10 |
C
// 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; // Calculate 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 Code int main() { int n = 10, k = 2; printf ("Value of P(%d, %d) is %d ", n, k, permutationCoeff(n, k) ); return 0; } |
Java
// A O(n) solution that uses // table fact[] to calculate // the Permutation Coefficient import java .io.*; public class GFG { // Returns value of Permutation // Coefficient P(n, k) static int permutationCoeff(int n, int k) { int []fact = new int[n+1]; // base case fact[0] = 1; // Calculate 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 Code static public void main (String[] args) { int n = 10, k = 2; System.out.println("Value of" + " P( " + n + ", " + k + ") is " + permutationCoeff(n, k) ); } } // This code is contributed by anuj_67. |
Python3
# A O(n) solution that uses # table fact[] to calculate # the Permutation Coefficient # Returns value of Permutation # Coefficient P(n, k) def permutationCoeff(n, k): fact = [0 for i in range(n + 1)] # base case fact[0] = 1 # Calculate value # factorials up to n for i in range(1, n + 1): fact[i] = i * fact[i - 1] # P(n, k) = n!/(n-k)! return int(fact[n] / fact[n - k]) # Driver Code n = 10k = 2print("Value of P(", n, ", ", k, ") is ", permutationCoeff(n, k), sep = "") # This code is contributed # by Soumen Ghosh |
C#
// A O(n) solution that uses // table fact[] to calculate // the Permutation Coefficient using System; public class GFG { // Returns value of Permutation // Coefficient P(n, k) static int permutationCoeff(int n, int k) { int []fact = new int[n+1]; // base case fact[0] = 1; // Calculate 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 Code static public void Main () { int n = 10, k = 2; Console.WriteLine("Value of" + " P( " + n + ", " + k + ") is " + permutationCoeff(n, k) ); } } // This code is contributed by anuj_67. |
PHP
<?php // A O(n) Solution that // uses table fact[] to // calculate the Permutation // Coefficient // Returns value of Permutation // Coefficient P(n, k) function permutationCoeff($n, $k) { $fact = array(); // base case $fact[0] = 1; // Calculate value // factorials up to n for ($i = 1; $i <= $n; $i++) $fact[$i] = $i * $fact[$i - 1]; // P(n,k)= n!/(n-k)! return $fact[$n] / $fact[$n - $k]; } // Driver Code $n = 10; $k = 2; echo"Value of P(",$n," ", $k,") is ", permutationCoeff($n, $k) ; // This code is contributed by anuj_67. ?> |
Javascript
<script> // A O(n) solution that uses // table fact[] to calculate // the Permutation Coefficient // Returns value of Permutation // Coefficient P(n, k) function permutationCoeff(n, k) { let fact = new Array(n+1); // base case fact[0] = 1; // Calculate value // factorials up to n for (let i = 1; i <= n; i++) fact[i] = i * fact[i - 1]; // P(n,k) = n! / (n - k)! return parseInt(fact[n] / fact[n - k], 10); } let n = 10, k = 2; document.write("Value of" + " P(" + n + ", " + k + ") is " + permutationCoeff(n, k) ); </script> |
Output :
Value of P(10, 2) is 90
A O(n) time and O(1) Extra Space Solution
C++
// 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 P = 1; // Compute n*(n-1)*(n-2)....(n-k+1) for (int i = 0; i < k; i++) P *= (n-i) ; return P;}// Driver Codeint main(){ int n = 10, k = 2; cout << "Value of P(" << n << ", " << k << ") is " << PermutationCoeff(n, k); return 0;} |
Java
// A O(n) time and O(1) extra // space solution to calculate// the Permutation Coefficientimport java.io.*;class GFG { static int PermutationCoeff(int n, int k) { int Fn = 1, Fk = 1; // 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 Code public static void main(String args[]) { int n = 10, k = 2; System.out.println("Value of P( " + n + "," + k +") is " + PermutationCoeff(n, k) ); }} // This code is contributed by Nikita Tiwari. |
C#
// A O(n) time and O(1) extra // space solution to calculate// the Permutation Coefficientusing System;class GFG { static int PermutationCoeff(int n, int k) { int Fn = 1, Fk = 1; // 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 Code public static void Main() { int n = 10, k = 2; Console.WriteLine("Value of P( " + n + "," + k +") is " + PermutationCoeff(n, k) ); }} // This code is contributed by anuj_67. |
PHP
<?php// A O(n) time and O(1) extra // space PHP solution to calculate // the Permutation Coefficientfunction PermutationCoeff( $n, $k){ $Fn = 1; $Fk; // Compute n! and (n-k)! for ( $i = 1; $i <= $n; $i++) { $Fn *= $i; if ($i == $n - $k) $Fk = $Fn; } $coeff = $Fn / $Fk; return $coeff;}// Driver Code$n = 10; $k = 2;echo "Value of P(" , $n , ", " , $k , ") is " , PermutationCoeff($n, $k);// This code is contributed by anuj_67. ?> |
Javascript
<script>// A O(n) time and O(1) extra// space solution to calculate// the Permutation Coefficient function PermutationCoeff(n, k){ let P = 1; // Compute n*(n-1)*(n-2)....(n-k+1) for(let i = 0; i < k; i++) P *= (n - i); return P;}// Driver codelet n = 10, k = 2;document.write("Value of P(" + n + ", " + k + ") is " + PermutationCoeff(n, k)); // This code is contributed by divyesh072019</script> |
Python3
# A O(n) solution that uses# table fact[] to calculate# the Permutation Coefficient# Returns value of Permutation# Coefficient P(n, k)def permutationCoeff(n, k): f=1 for i in range(k): #P(n,k)=n*(n-1)*(n-2)*....(n-k-1) f*=(n-i) return f #This code is contributed by Suyash Saxena# Driver Coden = 10k = 2print("Value of P(", n, ", ", k, ") is ", permutationCoeff(n, k)) |
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
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