Program to calculate the value of nPr

Given two integers n and r, find the value of nPr, the number of ways to arrange r elements selected from n distinct elements. The value of nPr is given by: nPr = n! / (n − r)!, where ! denotes the factorial of a number.

Examples:

Input: n = 5, r = 2
Output: 20
Explanation: ⁵P₂ = 5! / (5 - 2)! = 20

Input: n = 6, r = 3
Output: 120
Explanation: ⁶P₃ = 6! / (6 - 3)! = 120

[Naive Approach] Using Factorial Formula - O(n) Time and O(1) Space

The idea is to compute n! and (n-r)! separately using an iterative factorial function and then apply the permutation formula: nPr = n! / (n-r)!

• If r > n, return 0.
• Compute n! using an iterative factorial function.
• Compute (n-r)! using the same function.
• Return n! / (n-r)!.

#include <iostream>
using namespace std;

// Function to calculate factorial
long long fact(int n)
{
    long long result = 1;
    for (int i = 2; i <= n; i++)
    {
        result *= i;
    }
    return result;
}

// Function to calculate nPr
long long nPr(int n, int r)
{
    if (r > n)
        return 0;

    return fact(n) / fact(n - r);
}

int main()
{
    int n = 5;
    int r = 2;

    cout << nPr(n, r) << endl;

    return 0;
}

#include <stdio.h>

// Function to calculate factorial
long long fact(int n)
{
    long long result = 1;
    for (int i = 2; i <= n; i++)
    {
        result *= i;
    }
    return result;
}

// Function to calculate nPr
long long nPr(int n, int r)
{
    if (r > n)
        return 0;

    return fact(n) / fact(n - r);
}

int main()
{
    int n = 5;
    int r = 2;

    printf("%lld\n", nPr(n, r));

    return 0;
}

class Solution {

    // Function to calculate factorial
    static long fact(int n) {
        long result = 1;
        for (int i = 2; i <= n; i++) {
            result *= i;
        }
        return result;
    }

    // Function to calculate nPr
    static long nPr(int n, int r) {
        if (r > n)
            return 0;

        return fact(n) / fact(n - r);
    }

    public static void main(String[] args) {
        int n = 5;
        int r = 2;

        System.out.println(nPr(n, r));
    }
}

# Function to calculate factorial
def fact(n: int) -> int:
    result = 1
    for i in range(2, n + 1):
        result *= i
    return result

# Function to calculate nPr
def nPr(n: int, r: int) -> int:
    return fact(n) // fact(n - r)


if __name__ == "__main__":
    n = 5
    r = 2

    print(nPr(n, r))

using System;

class GFG
{
    // Function to calculate factorial
    static long fact(int n)
    {
        long result = 1;
        for (int i = 2; i <= n; i++)
        {
            result *= i;
        }
        return result;
    }

    // Function to calculate nPr
    static long nPr(int n, int r)
    {
        if (r > n)
            return 0;

        return fact(n) / fact(n - r);
    }

    static void Main()
    {
        int n = 5;
        int r = 2;

        Console.WriteLine(nPr(n, r));
    }
}

// Function to calculate factorial
function fact(n) {
    let result = 1;
    for (let i = 2; i <= n; i++) {
        result *= i;
    }
    return result;
}

// Function to calculate nPr
function nPr(n, r) {
    if (r > n)
        return 0;

    return fact(n) / fact(n - r);
}


// Driver code
    let n = 5;
    let r = 2;

    console.log(nPr(n, r));

20

[Expected Approach] Multiply the Required Terms Directly - O(r) Time and O(1) Space

Instead of computing n! and (n-r)! separately, use the relation:
nPr = n × (n - 1) × (n - 2) × ... × (n - r + 1).

This computes only the required r terms, avoiding unnecessary factorial calculations and making the solution more efficient.

#include <iostream>
using namespace std;

long long nPr(int n, int r)
{
    if (r > n)
        return 0;

    long long ans = 1;

    // Compute n × (n-1) × ... × (n-r+1)
    for (int i = 0; i < r; i++)
        ans *= (n - i);

    return ans;
}

int main()
{
    int n = 5;
    int r = 2;

    cout << nPr(n, r) << endl;

    return 0;
}

#include <stdio.h>

long long nPr(int n, int r)
{
    if (r > n)
        return 0;

    long long ans = 1;

    // Compute n × (n-1) × ... × (n-r+1)
    for (int i = 0; i < r; i++)
        ans *= (n - i);

    return ans;
}

int main()
{
    int n = 5;
    int r = 2;

    printf("%lld\n", nPr(n, r));

    return 0;
}

class GFG {

    static long nPr(int n, int r) {
        if (r > n)
            return 0;

        long ans = 1;

        // Compute n × (n-1) × ... × (n-r+1)
        for (int i = 0; i < r; i++)
            ans *= (n - i);

        return ans;
    }

    public static void main(String[] args) {
        int n = 5;
        int r = 2;

        System.out.println(nPr(n, r));
    }
}

def nPr(n: int, r: int) -> int:
    if r > n:
        return 0

    ans = 1

    # Compute n × (n-1) × ... × (n-r+1)
    for i in range(r):
        ans *= (n - i)

    return ans


if __name__ == "__main__":
    n = 5
    r = 2

    print(nPr(n, r))

using System;

class GFG
{
    static long nPr(int n, int r)
    {
        if (r > n)
            return 0;

        long ans = 1;

        // Compute n × (n-1) × ... × (n-r+1)
        for (int i = 0; i < r; i++)
            ans *= (n - i);

        return ans;
    }

    static void Main()
    {
        int n = 5;
        int r = 2;

        Console.WriteLine(nPr(n, r));
    }
}

function nPr(n, r) {
    if (r > n)
        return 0;

    let ans = 1;

    // Compute n × (n-1) × ... × (n-r+1)
    for (let i = 0; i < r; i++)
        ans *= (n - i);

    return ans;
}

// Driver code
    let n = 5;
    let r = 2;

    console.log(nPr(n, r));

20