从N个不同的事物(没有大小A或B的集合)中找到形成集合的方法的数量

给定三个数字N，A，B 。任务是计算选择事物的方式数量，以使A或B都不存在大小集。答案可能非常大。因此，输出答案取模10 ⁹ +7 。

注意：空集不被视为一种方法。

例子：

Input: N = 4, A = 1, B = 3
Output: 7
Explanation:
The number of ways to form sets of size 2 are 6 (⁴C₂).
The number of ways to form sets of size 4 are 1 (⁴C₄).

Input: N = 10, A = 4, B = 9
Output: 803

方法：想法是首先找到一些方法，包括大小集(包括A，B和空集)。然后删除大小为A，B的套数和空套数。

下面是上述方法的实现：

CPP

// C++ program to find number of sets without size A and B
#include 
using namespace std;
#define mod (int)(1e9 + 7)
  
// Function to find a^m1
int power(int a, int m1)
{
    if (m1 == 0)
        return 1;
    else if (m1 == 1)
        return a;
    else if (m1 == 2)
        return (1LL * a * a) % mod;
    // If m1 is odd, then return a * a^m1/2 * a^m1/2
    else if (m1 & 1)
        return (1LL * a * power(power(a, m1 / 2), 2)) % mod;
    else
        return power(power(a, m1 / 2), 2) % mod;
}
  
// Function to find factorial of a number
int factorial(int x)
{
    int ans = 1;
    for (int i = 1; i <= x; i++)
        ans = (1LL * ans * i) % mod;
  
    return ans;
}
  
// Function to find inverse of x
int inverse(int x)
{
    return power(x, mod - 2);
}
  
// Function to find nCr
int binomial(int n, int r)
{
    if (r > n)
        return 0;
  
    int ans = factorial(n);
  
    ans = (1LL * ans * inverse(factorial(r))) % mod;
  
    ans = (1LL * ans * inverse(factorial(n - r))) % mod;
  
    return ans;
}
  
// Function to find number of sets without size a and b
int number_of_sets(int n, int a, int b)
{
    // First calculate all sets
    int ans = power(2, n);
  
    // Remove sets of size a
    ans = ans - binomial(n, a);
  
    if (ans < 0)
        ans += mod;
  
    // Remove sets of size b
    ans = ans - binomial(n, b);
  
    // Remove empty set
    ans--;
  
    if (ans < 0)
        ans += mod;
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int N = 4, A = 1, B = 3;
  
    // Function call
    cout << number_of_sets(N, A, B);
  
    return 0;
}

Java

// Java program to find number of sets without size A and B
import java.util.*;
  
class GFG{
static final int mod =(int)(1e9 + 7);
   
// Function to find a^m1
static int power(int a, int m1)
{
    if (m1 == 0)
        return 1;
    else if (m1 == 1)
        return a;
    else if (m1 == 2)
        return (int) ((1L * a * a) % mod);
    // If m1 is odd, then return a * a^m1/2 * a^m1/2
    else if (m1 % 2 == 1)
        return (int) ((1L * a * power(power(a, m1 / 2), 2)) % mod);
    else
        return power(power(a, m1 / 2), 2) % mod;
}
   
// Function to find factorial of a number
static int factorial(int x)
{
    int ans = 1;
    for (int i = 1; i <= x; i++)
        ans = (int) ((1L * ans * i) % mod);
   
    return ans;
}
   
// Function to find inverse of x
static int inverse(int x)
{
    return power(x, mod - 2);
}
   
// Function to find nCr
static int binomial(int n, int r)
{
    if (r > n)
        return 0;
   
    int ans = factorial(n);
   
    ans = (int) ((1L * ans * inverse(factorial(r))) % mod);
   
    ans = (int) ((1L * ans * inverse(factorial(n - r))) % mod);
   
    return ans;
}
   
// Function to find number of sets without size a and b
static int number_of_sets(int n, int a, int b)
{
    // First calculate all sets
    int ans = power(2, n);
   
    // Remove sets of size a
    ans = ans - binomial(n, a);
   
    if (ans < 0)
        ans += mod;
   
    // Remove sets of size b
    ans = ans - binomial(n, b);
   
    // Remove empty set
    ans--;
   
    if (ans < 0)
        ans += mod;
   
    // Return the required answer
    return ans;
}
   
// Driver code
public static void main(String[] args)
{
    int N = 4, A = 1, B = 3;
   
    // Function call
    System.out.print(number_of_sets(N, A, B));
   
}
}
  
// This code contributed by sapnasingh4991

Python3

# Python3 program to find number of 
# sets without size A and B
mod = 10**9 + 7
  
# Function to find a^m1
def power(a, m1):
    if (m1 == 0):
        return 1
    elif (m1 == 1):
        return a
    elif (m1 == 2):
        return (a * a) % mod
           
    # If m1 is odd, then return a * a^m1/2 * a^m1/2
    elif (m1 & 1):
        return (a * power(power(a, m1 // 2), 2)) % mod
    else:
        return power(power(a, m1 // 2), 2) % mod
  
# Function to find factorial of a number
def factorial(x):
    ans = 1
    for i in range(1, x + 1):
        ans = (ans * i) % mod
  
    return ans
  
# Function to find inverse of x
def inverse(x):
    return power(x, mod - 2)
  
# Function to find nCr
def binomial(n, r):
    if (r > n):
        return 0
  
    ans = factorial(n)
  
    ans = (ans * inverse(factorial(r))) % mod
  
    ans = (ans * inverse(factorial(n - r))) % mod
  
    return ans
  
# Function to find number of sets without size a and b
def number_of_sets(n, a, b):
      
    # First calculate all sets
    ans = power(2, n)
  
    # Remove sets of size a
    ans = ans - binomial(n, a)
  
    if (ans < 0):
        ans += mod
  
    # Remove sets of size b
    ans = ans - binomial(n, b)
  
    # Remove empty set
    ans -= 1
  
    if (ans < 0):
        ans += mod
  
    # Return the required answer
    return ans
  
# Driver code
if __name__ == '__main__':
    N = 4
    A = 1
    B = 3
  
    # Function call
    print(number_of_sets(N, A, B))
  
# This code is contributed by mohit kumar 29

C#

// C# program to find number of sets without size A and B
using System;
  
class GFG{
static readonly int mod =(int)(1e9 + 7);
    
// Function to find a^m1
static int power(int a, int m1)
{
    if (m1 == 0)
        return 1;
    else if (m1 == 1)
        return a;
    else if (m1 == 2)
        return (int) ((1L * a * a) % mod);
    // If m1 is odd, then return a * a^m1/2 * a^m1/2
    else if (m1 % 2 == 1)
        return (int) ((1L * a * power(power(a, m1 / 2), 2)) % mod);
    else
        return power(power(a, m1 / 2), 2) % mod;
}
    
// Function to find factorial of a number
static int factorial(int x)
{
    int ans = 1;
    for (int i = 1; i <= x; i++)
        ans = (int) ((1L * ans * i) % mod);
    
    return ans;
}
    
// Function to find inverse of x
static int inverse(int x)
{
    return power(x, mod - 2);
}
    
// Function to find nCr
static int binomial(int n, int r)
{
    if (r > n)
        return 0;
    
    int ans = factorial(n);
    
    ans = (int) ((1L * ans * inverse(factorial(r))) % mod);
    
    ans = (int) ((1L * ans * inverse(factorial(n - r))) % mod);
    
    return ans;
}
    
// Function to find number of sets without size a and b
static int number_of_sets(int n, int a, int b)
{
    // First calculate all sets
    int ans = power(2, n);
    
    // Remove sets of size a
    ans = ans - binomial(n, a);
    
    if (ans < 0)
        ans += mod;
    
    // Remove sets of size b
    ans = ans - binomial(n, b);
    
    // Remove empty set
    ans--;
    
    if (ans < 0)
        ans += mod;
    
    // Return the required answer
    return ans;
}
    
// Driver code
public static void Main(String[] args)
{
    int N = 4, A = 1, B = 3;
    
    // Function call
    Console.Write(number_of_sets(N, A, B));
}
}
   
// This code is contributed by PrinciRaj1992

输出：

7