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 *= 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 



















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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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