Permutation Coefficient

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:
permu

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)   

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

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];
  
    // 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 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];
      
        // Caculate 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 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 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 = 10
k = 2
print("Value fo 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];
      
        // Caculate 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 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 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());
  
    // Caculate 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
            // 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 Code
    $n = 10; $k = 2;
    echo "Value of P(",$n," ,",$k,") is ",
               permutationCoeff($n, $k);
  
// This code is contributed by anuj_67.
?>


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>
  
// 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 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;
      
        // 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 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 = 10
k = 2
print("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;
      
        // 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 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.             
?>


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 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 Code
int 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 Coefficient
import 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.

Python3

# A O(n) time and O(1) extra 
# space solution to calculate
# the Permutation Coefficient
  
def PermutationCoeff(n, k):
    Fn = 1
  
    # Compute n! and (n-k)!
    for i in range(1, n + 1):
        Fn *= i
        if (i == n - k):
            Fk = Fn
  
    coeff = Fn // Fk
    return coeff
  
# Driver Code
n = 10
k = 2
print("Value of P(", n, ", ", k, ") is ",
       PermutationCoeff(n, k), sep = "")
  
# This code is contributed
# by Soumen Ghosh.

C#

// A O(n) time and O(1) extra 
// space solution to calculate
// the Permutation Coefficient
using 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 Coefficient
  
function 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. 
?>


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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Improved By : vt_m