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 

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

Algorithm:

• Create a method named “fact” that takes a parameter n and returns the factorial of n.
  • Create a variable named “fact” and initialize it to 1.
  • Start a for loop and traverse through the range 2 to n (inclusive) and multiply each number with the “fact” variable.
  •  Return the value of “fact”.
• Create another method named “ncr” that takes two parameters n and r, and returns the value of nCr.
  • Inside the “ncr” function, use the “fact” function to calculate the factorial of n, r, and n-r.
  • Calculate nCr using the formula n! / (r! * (n-r)!).
  • Return the value of nCr. 
• Create another method “ways” that takes four parameters m, w, n, and k, and returns the total possible ways.
  • Create a variable named “ans” and initialize it to 0.
  • Start a while loop to iterate until m is greater than or equal to k.
    • Inside the while loop, add the value of nCr(m, k) multiplied by nCr(w, n-k) to the “ans” variable.
    •  Increment k by 1 in each iteration.                                                                                                                                    
•  Return the value of “ans”.

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

Javascript

<script>
// javascript implementation of the approach

// Returns factorial
// of the number
function fact(n)
{
    var fact = 1;
    for (i = 2; i <= n; i++)
        fact *= i;
    return fact;
}

// Function to calculate ncr
function ncr(n , r)
{
    var ncr = fact(n) / (fact(r) * fact(n - r));
    return parseInt(ncr);
}

// Function to calculate
// the total possible ways
function ways(m , w , n , k)
{

    var ans = 0;
    while (m >= k)
    {
        ans += ncr(m, k) * ncr(w, n - k);
        k += 1;
    }

    return parseInt(ans);
}

// Driver code
var m, w, n, k;
m = 7;
w = 6;
n = 5;
k = 3;
document.write( ways(m, w, n, k));

// This code is contributed by 29AjayKumar.
</script>

Output:

756

Time Complexity: O(n)
Auxiliary Space: O(1)

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