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

Below is the implementation of the above approach:

C++

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// 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);
}

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Java

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// 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

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Python3

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# Python 3 implementation of the approach 
  
# Returns factorial of the number 
def fact(n): 
    fact = 1
    for i in range(2, n + 1): 
        fact *=
    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

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C#

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// 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

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PHP

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<?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

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Output:

756

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

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