从小组中选择男性和女性组成团队的方法

给定四个整数n，w，m和k，其中，

• m是总人数。
• w是妇女总数。
• n是组成团队所需的人员总数。
• k是必须选择的最小人数。

任务是找到组建团队的方式。
例子：

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 
 

方法：从那以后，我们必须至少要招募k个人。

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

。
从上面的第一个例子中，在7名男性和6名女性中，总共需要选择5个人，其中至少3位男性，
路数=(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
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

= $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

// 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.

输出：

756

进一步优化：可以使用更快的算法对上述代码进行优化，以进行二项式系数计算。