程序来计算nPr的值

给定两个数字n和r，任务是找到nPr的值。
nPr代表n个置换r，其计算公式为n！/(nk)!。 。排列是指将给定集合的所有成员排列成一个序列的过程。一组n个元素上的排列数由n！在哪里 ”！”代表阶乘。

nPr = n! / (n - r)! 

程序：

C++

// CPP program to calculate nPr
#include
using namespace std;
 
int fact(int n)
{
    if (n <= 1)
        return 1;
    return n * fact(n - 1);
}
 
int nPr(int n, int r)
{
    return fact(n) / fact(n - r);
}
 
// Driver code
int main()
{
    int n = 5;
    int r = 2;
 
    cout << n << "P" << r << " = " << nPr(n, r);
}
 
// This code is contributed by
// Surendra_Gangwar

Java

// Java program to calculate nPr
 
import java.util.*;
 
public class GFG {
 
    static int fact(int n)
    {
        if (n <= 1)
            return 1;
        return n * fact(n - 1);
    }
 
    static int nPr(int n, int r)
    {
        return fact(n) / fact(n - r);
    }
 
    public static void main(String args[])
    {
        int n = 5;
        int r = 2;
 
        System.out.println(n + "P" + r + " = "
                           + nPr(n, r));
    }
}

Python3

# Python3 program to calculate nPr
import math
def fact(n):
    if (n <= 1):
        return 1
         
    return n * fact(n - 1)
 
def nPr(n, r):
     
    return math.floor(fact(n) /
                fact(n - r))
     
# Driver code
n = 5
r = 2
 
print(n, "P", r, "=", nPr(n, r))
 
# This code contributed by Rajput-Ji

C#

// C# program to calculate nPr
using System;
 
class GFG
{
 
    static int fact(int n)
    {
        if (n <= 1)
            return 1;
        return n * fact(n - 1);
    }
 
    static int nPr(int n, int r)
    {
        return fact(n) / fact(n - r);
    }
 
    public static void Main()
    {
        int n = 5;
        int r = 2;
 
        Console.WriteLine(n + "P" + r + " = "
                        + nPr(n, r));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

// Javascript program to calculate nPr
function fact(n)
{
    if (n <= 1)
        return 1;
         
    return n * fact(n - 1);
}
 
function nPr(n, r)
{
    return Math.floor(fact(n) /
                fact(n - r));
}
     
// Driver code
let n = 5;
let r = 2;
 
document.write(n, "P", r, " = ", nPr(n, r));
 
// This code is contributed by gfgking

输出：

5P2 = 20

优化多个nPr查询
如果对nPr有多个查询，我们可以预先计算阶乘值，并在每次调用时使用相同的值。这将避免一次又一次地计算相同的阶乘值。