从总共M个男性和W个女性中选择X个男性和Y个女性的总方法

给定四个整数X ， Y ， M和W。任务是找到从M个男性和W个女性中选择X个男性和Y个女性的方法。

例子：

Input: X = 1, Y = 2, M = 1, W = 3
Output: 3
Way 1: Choose the only man and 1^st and 2^nd women.
Way 2: Choose the only man and 2^nd and 3^rd women.
Way 3: Choose the only man and 1^st and 3^rd women.

Input: X = 4, Y = 3, M = 6, W = 5
Output: 150

方法：从总共M个男人中选择X个男人的总数为^M C _X ，从W个女人中选择Y个女人的总数为^W C _Y。因此，组合方式的总数将为^M C _X * ^W C _Y。

下面是上述方法的实现：

C++

// C++ implementataion of the approach
#include 
using namespace std;
  
// Function to return the
// value of ncr effectively
int ncr(int n, int r)
{
  
    // Initialize the answer
    int ans = 1;
  
    for (int i = 1; i <= r; i += 1) {
  
        // Divide simultaneously by
        // i to avoid overflow
        ans *= (n - r + i);
        ans /= i;
    }
    return ans;
}
  
// Function to return the count of required ways
int totalWays(int X, int Y, int M, int W)
{
    return (ncr(M, X) * ncr(W, Y));
}
  
int main()
{
    int X = 4, Y = 3, M = 6, W = 5;
  
    cout << totalWays(X, Y, M, W);
  
    return 0;
}

Java

// JAVA implementataion of the approach 
import java.io.*;
  
class GFG 
{
          
    // Function to return the 
    // value of ncr effectively 
    static int ncr(int n, int r) 
    { 
      
        // Initialize the answer 
        int ans = 1; 
      
        for (int i = 1; i <= r; i += 1) 
        { 
      
            // Divide simultaneously by 
            // i to avoid overflow 
            ans *= (n - r + i); 
            ans /= i; 
        } 
        return ans; 
    } 
      
    // Function to return the count of required ways 
    static int totalWays(int X, int Y, int M, int W) 
    { 
        return (ncr(M, X) * ncr(W, Y)); 
    } 
      
    // Driver code
    public static void main (String[] args) 
    {
        int X = 4, Y = 3, M = 6, W = 5; 
      
        System.out.println(totalWays(X, Y, M, W)); 
    }
}
  
// This code is contributed by ajit_23

Python3

# Python3 implementataion of the approach
  
# Function to return the
# value of ncr effectively
def ncr(n, r):
    # Initialize the answer
    ans = 1
  
    for i in range(1,r+1):
  
        # Divide simultaneously by
        # i to avoid overflow
        ans *= (n - r + i)
        ans //= i
    return ans
  
# Function to return the count of required ways
def totalWays(X, Y, M, W):
  
    return (ncr(M, X) * ncr(W, Y))
  
X = 4
Y = 3
M = 6
W = 5
  
print(totalWays(X, Y, M, W))
  
# This code is contributed by mohit kumar 29

C#

// C# implementataion of the approach 
using System;
  
class GFG
{
      
    // Function to return the 
    // value of ncr effectively 
    static int ncr(int n, int r) 
    { 
      
        // Initialize the answer 
        int ans = 1; 
      
        for (int i = 1; i <= r; i += 1) 
        { 
      
            // Divide simultaneously by 
            // i to avoid overflow 
            ans *= (n - r + i); 
            ans /= i; 
        } 
        return ans; 
    } 
      
    // Function to return the count of required ways 
    static int totalWays(int X, int Y, int M, int W) 
    { 
        return (ncr(M, X) * ncr(W, Y)); 
    } 
      
    // Driver code
    static public void Main ()
    {
        int X = 4, Y = 3, M = 6, W = 5; 
      
        Console.WriteLine(totalWays(X, Y, M, W)); 
    }
}
  
// This code is contributed by AnkitRai01

输出：

150