Total ways of choosing X men and Y women from a total of M men and W women

Given four integers X, Y, M and W. The task is to find the number of ways to choose X men and Y women from total M men and W women.

Examples:

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

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

Approach: The total number of ways of choosing X men from a total of M men is MCX and the total number of ways of choosing Y women from W women is WCY. Hence, the total number of combined ways will be MCX * WCY.

Below is the implementation of the above approach:

C++

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// C++ implementataion of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

150


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