从 P Boys 和 Q Girls 中选择包含至少 4 个男孩和 1 个女孩的 N 人的方法计数 |设置 2

给定整数N、P 和 Q ，任务是从P 个男孩和Q 个女孩中找出由N 个人组成的一组至少有 4 个男孩和1 个女孩的方法的数量。

例子：

Input:  P = 5, Q = 2, N = 5
Output: 10
Explanation:  Suppose given pool is {m1, m2, m3, m4, m5} and {w1, w2}. Then possible combinations are:
m1 m2 m3 m4 w1
m2 m3 m4 m5 w1
m1 m3 m4 m5 w1
m1 m2 m4 m5 w1
m1 m2 m3 m5 w1
m1 m2 m3 m4 w2
m2 m3 m4 m5 w2
m1 m3 m4 m5 w2
m1 m2 m4 m5 w2
m1 m2 m3 m5 w2

Hence the count is 10.

Input:  P = 5, Q = 2, N = 6
Output: 7

朴素方法：这个问题是基于组合学的，朴素方法的细节已经在这个问题的Set-1中讨论过。

对于 P、Q 和 N 的一些一般值，我们可以使用以下公式计算所有可能的方式：

$_{4}^{P}\textrm{C} \ast _{N-4}^{Q}\textrm{C} + _{5}^{P}\textrm{C} \ast _{N-5}^{Q}\textrm{C} + . . . + _{N-2}^{P}\textrm{C} \ast _{2}^{Q}\textrm{C} + _{N-1}^{P}\textrm{C} \ast _{1}^{Q}\textrm{C}$

where 

$_{r}^{n}\textrm{C} = \frac{n!}{r!*(n-r)!}$

在这种方法中，我们每一步都在计算每种可能方式的价值。
时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：为了有效地解决这个问题，我们可以使用帕斯卡三角属性来计算$_{r}^{n}\textrm{C}$ ， IE

1
1 1
1 2 1
1 3 3 1
.
.
.

这不过是

$_{0}^{0}\textrm{C}$
$_{0}^{1}\textrm{C}$  $_{1}^{1}\textrm{C}$
$_{0}^{2}\textrm{C}$  $_{1}^{2}\textrm{C}$  $_{2}^{2}\textrm{C}$
$_{0}^{3}\textrm{C}$  $_{1}^{3}\textrm{C}$  $_{2}^{3}\textrm{C}$  $_{3}^{3}\textrm{C}$
.
.
.

请按照以下步骤操作：

• 使用帕斯卡三角形预先计算组合的值。
• 开始从i = 4到i = P迭代一个循环，并为每次迭代执行以下操作。
• 检查(Ni) ≥ 1和(Ni) ≤ Q是否。
• 如果满足条件，则计算i men和(Ni) women的可能方式，否则，跳过该步骤。
• 将计数与总路数相加。
• 返回总计数作为您的答案。

下面是该方法的实现：

C++

#include 
using namespace std;
 
long long int pascal[31][31];
 
// Function to calculate the pascal triangle
void pascalTriangle()
{
    pascal[0][0] = 1;
    pascal[1][0] = 1;
    pascal[1][1] = 1;
 
    // Loop to calculate values of
    // pascal triangle
    for (int i = 2; i < 31; i++) {
        pascal[i][0] = 1;
        for (int j = 1; j < i; j++)
            pascal[i][j]
                = pascal[i - 1][j]
                  + pascal[i - 1][j - 1];
        pascal[i][i] = 1;
    }
}
 
// Function to calculate the number of ways
long long int countWays(int n, int p, int q)
{
 
    // Variable to store the answer
    long long int sum = 0;
 
    // Loop to calculate the number of ways
    for (long long int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q)
            sum += pascal[p][i]
                   * pascal[q][n - i];
    }
    return sum;
}
 
// Driver code
int main()
{
    pascalTriangle();
 
    int P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    cout << countWays(N, P, Q) << endl;
    return 0;
}

Java

import java.util.*;
public class GFG
{
     
static long [][]pascal = new long[31][31];
 
// Function to calculate the pascal triangle
static void pascalTriangle()
{
    pascal[0][0] = 1;
    pascal[1][0] = 1;
    pascal[1][1] = 1;
 
    // Loop to calculate values of
    // pascal triangle
    for (int i = 2; i < 31; i++) {
        pascal[i][0] = (int)1;
        for (int j = 1; j < i; j++)
            pascal[i][j]
                = pascal[i - 1][j]
                  + pascal[i - 1][j - 1];
        pascal[i][i] = 1;
    }
}
 
// Function to calculate the number of ways
static long countWays(int n, int p, int q)
{
 
    // Variable to store the answer
    long sum = 0;
 
    // Loop to calculate the number of ways
    for (int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q) {
            sum += (int)pascal[p][i]
                   * (int)pascal[q][n - i];
        }
    }
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    pascalTriangle();
 
    int P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    System.out.print(countWays(N, P, Q));
 
}
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 program for the above approach
import numpy as np
 
pascal = np.zeros((31,31));
 
# Function to calculate the pascal triangle
def pascalTriangle() :
     
    pascal[0][0] = 1;
    pascal[1][0] = 1;
    pascal[1][1] = 1;
 
    # Loop to calculate values of
    # pascal triangle
    for i in range(2, 31) :
         
        pascal[i][0] = 1;
        for j in range(1, i) :
            pascal[i][j] = pascal[i - 1][j] + pascal[i - 1][j - 1];
         
        pascal[i][i] = 1;
   
 
# Function to calculate the number of ways
def countWays(n, p, q) :
 
    # Variable to store the answer
    sum = 0;
 
    # Loop to calculate the number of ways
    for i in range(4, p + 1) :
         
        if (n - i >= 1 and n - i <= q) :
             
            sum += pascal[p][i] * pascal[q][n - i];
 
    return int(sum);
 
# Driver code
if __name__ ==  "__main__" :
 
    pascalTriangle();
 
    P = 5; Q = 2; N = 5;
 
    # Calculate possible ways for given
    # N, P, and Q
    print(countWays(N, P, Q));
 
    # This code is contributed by AnkThon

C#

using System;
class GFG
{
     
static long [,]pascal = new long[31, 31];
 
// Function to calculate the pascal triangle
static void pascalTriangle()
{
    pascal[0, 0] = 1;
    pascal[1, 0] = 1;
    pascal[1, 1] = 1;
 
    // Loop to calculate values of
    // pascal triangle
    for (int i = 2; i < 31; i++) {
        pascal[i, 0] = (int)1;
        for (int j = 1; j < i; j++)
            pascal[i, j]
                = pascal[i - 1, j]
                  + pascal[i - 1, j - 1];
        pascal[i, i] = 1;
    }
}
 
// Function to calculate the number of ways
static long countWays(int n, int p, int q)
{
 
    // Variable to store the answer
    long sum = 0;
 
    // Loop to calculate the number of ways
    for (int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q) {
            sum += (int)pascal[p, i]
                   * (int)pascal[q, n - i];
        }
    }
    return sum;
}
 
// Driver code
public static void Main()
{
    pascalTriangle();
 
    int P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    Console.Write(countWays(N, P, Q));
 
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript

 
    let pascal = new Array(31).fill(0).map(() => new Array(31).fill(0));
 
    // Function to calculate the pascal triangle
    const pascalTriangle = () => {
        pascal[0][0] = 1;
        pascal[1][0] = 1;
        pascal[1][1] = 1;
 
        // Loop to calculate values of
        // pascal triangle
        for (let i = 2; i < 31; i++) {
            pascal[i][0] = 1;
            for (let j = 1; j < i; j++)
                pascal[i][j]
                    = pascal[i - 1][j]
                    + pascal[i - 1][j - 1];
            pascal[i][i] = 1;
        }
    }
 
    // Function to calculate the number of ways
    const countWays = (n, p, q) => {
 
        // Variable to store the answer
        let sum = 0;
 
        // Loop to calculate the number of ways
        for (let i = 4; i <= p; i++) {
            if (n - i >= 1 && n - i <= q)
                sum += pascal[p][i]
                    * pascal[q][n - i];
        }
        return sum;
    }
 
    // Driver code
    pascalTriangle();
 
    let P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    document.write(countWays(N, P, Q));
 
// This code is contributed by rakeshsahni
 

输出

10

时间复杂度： O(N)
辅助空间： O(N ² )