Ways of selecting men and women from a group to make a team

Given four integers n, w, m and k where,

m is the total number of men.
w is the total number of women.
n is the total number of people that need to be selected to form the team.
k is the minimum number of men that have to be selected.

The task is to find the number of ways in which the team can be formed.

Examples:

Input: m = 2, w = 2, n = 3, k = 1
Output: 4
There are 2 men, 2 women. We need to make a team of size 3 with at least one man and one woman. We can make the team in following ways.
m1 m2 w1
m1 w1 w2
m2 w1 w2
m1 m2 w2

Input: m = 7, w = 6, n = 5, k = 3
Output: 756

Input: m = 5, w = 6, n = 6, k = 3
Output: 281

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since, we have to take at least k men.

Totals ways = Ways when ‘k’ men are selected + Ways when ‘k+1’ men are selected + … + when ‘n’ men are selected

.
Taking the first example from above where out of 7 men and 6 women, total 5 people need to be selected with at least 3 men,
Number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2)
= 525 + 7 x 6 x 5 x 6 + 7 x 6
= (525 + 210 + 21)
= 756

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns factorial 
// of the number 
int fact(int n) 
{ 
    int fact = 1; 
    for (int i = 2; i <= n; i++) 
        fact *= i; 
    return fact; 
} 
  
// Function to calculate ncr 
int ncr(int n, int r) 
{ 
    int ncr = fact(n) / (fact(r) * fact(n - r)); 
    return ncr; 
} 
  
// Function to calculate 
// the total possible ways 
int ways(int m, int w, int n, int k) 
{ 
  
    int ans = 0; 
    while (m >= k) { 
        ans += ncr(m, k) * ncr(w, n - k); 
        k += 1; 
    } 
  
    return ans; 
} 
  
// Driver code 
int main() 
{ 
  
    int m, w, n, k; 
    m = 7; 
    w = 6; 
    n = 5; 
    k = 3; 
    cout << ways(m, w, n, k); 
} 

Java

// Java implementation of the approach 
  
import java.io.*; 
  
class GFG { 
  
// Returns factorial 
// of the number 
static int fact(int n) 
{ 
    int fact = 1; 
    for (int i = 2; i <= n; i++) 
        fact *= i; 
    return fact; 
} 
  
// Function to calculate ncr 
static int ncr(int n, int r) 
{ 
    int ncr = fact(n) / (fact(r) * fact(n - r)); 
    return ncr; 
} 
  
// Function to calculate 
// the total possible ways 
static int ways(int m, int w, int n, int k) 
{ 
  
    int ans = 0; 
    while (m >= k) { 
        ans += ncr(m, k) * ncr(w, n - k); 
        k += 1; 
    } 
  
    return ans; 
} 
  
// Driver code 
    public static void main (String[] args) { 
          
    int m, w, n, k; 
    m = 7; 
    w = 6; 
    n = 5; 
    k = 3; 
    System.out.println( ways(m, w, n, k)); 
    } 
} 
// This Code is contributed 
// by shs 

Python3

# Python 3 implementation of the approach  
  
# Returns factorial of the number  
def fact(n):  
    fact = 1
    for i in range(2, n + 1):  
        fact *= i  
    return fact 
  
# Function to calculate ncr  
def ncr(n, r): 
    ncr = fact(n) // (fact(r) * fact(n - r))  
    return ncr 
  
# Function to calculate  
# the total possible ways  
def ways(m, w, n, k): 
    ans = 0
    while (m >= k):  
        ans += ncr(m, k) * ncr(w, n - k)  
        k += 1
  
    return ans; 
  
# Driver code  
m = 7
w = 6
n = 5
k = 3
print(ways(m, w, n, k)) 
  
# This code is contributed by sahishelangia 

C#

// C# implementation of the approach 
  
class GFG { 
  
// Returns factorial 
// of the number 
static int fact(int n) 
{ 
    int fact = 1; 
    for (int i = 2; i <= n; i++) 
        fact *= i; 
    return fact; 
} 
  
// Function to calculate ncr 
static int ncr(int n, int r) 
{ 
    int ncr = fact(n) / (fact(r) * fact(n - r)); 
    return ncr; 
} 
  
// Function to calculate 
// the total possible ways 
static int ways(int m, int w, int n, int k) 
{ 
  
    int ans = 0; 
    while (m >= k) { 
        ans += ncr(m, k) * ncr(w, n - k); 
        k += 1; 
    } 
  
    return ans; 
} 
  
// Driver code 
    static void Main () { 
          
    int m, w, n, k; 
    m = 7; 
    w = 6; 
    n = 5; 
    k = 3; 
    System.Console.WriteLine( ways(m, w, n, k)); 
    } 
} 
// This Code is contributed by mits 

PHP

<?php 
// PHP implementation of the approach 
  
// Returns factorial of the number 
function fact($n) 
{ 
    $fact = 1; 
    for ($i = 2; $i <= $n; $i++) 
        $fact *= $i; 
    return $fact; 
} 
  
// Function to calculate ncr 
function ncr($n, $r) 
{ 
    $ncr = (int)(fact($n) / (fact($r) *  
                 fact($n - $r))); 
    return $ncr; 
} 
  
// Function to calculate the total  
// possible ways 
function ways($m, $w, $n, $k) 
{ 
    $ans = 0; 
    while ($m >= $k)  
    { 
        $ans += ncr($m, $k) * 
                ncr($w, $n - $k); 
        $k += 1; 
    } 
  
    return $ans; 
} 
  
// Driver code 
$m = 7; 
$w = 6; 
$n = 5; 
$k = 3; 
echo ways($m, $w, $n, $k); 
  
// This Code is contributed 
// by Mukul Singh 

Output:

Further Optimization : The above code can be optimized using faster algorithms for binomial coefficient computation.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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