选择球的方式，使得至少选择一个球

给定整数N ，任务是找到从给定的N个球中选择一些球的方法，以便至少选择一个球。由于该值可能很大，因此请打印模1000000007的值。
例子：

Input: N = 2 
Output: 3 
The three ways are “*.”, “.*” and “**” where ‘*’ denotes 
the chosen ball and ‘.’ denotes the ball which didn’t get chosen.
Input: N = 30000 
Output: 165890098 
 

方法：有N个球，每个球都可以选择或不选择。不同配置的总数为2 * 2 * 2 *…* N。我们可以写成2 ^N。但是必须从答案中减去没有选择球的状态。因此，结果将是(2 ^N – 1)％1000000007 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
const int MOD = 1000000007;
 
// Function to return the count of
// ways to choose the balls
int countWays(int n)
{
 
    // Calculate (2^n) % MOD
    int ans = 1;
    for (int i = 0; i < n; i++) {
        ans *= 2;
        ans %= MOD;
    }
 
    // Subtract the only where
    // no ball was chosen
    return ((ans - 1 + MOD) % MOD);
}
 
// Driver code
int main()
{
    int n = 3;
 
    cout << countWays(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
static int MOD = 1000000007;
 
// Function to return the count of
// ways to choose the balls
static int countWays(int n)
{
 
    // Calculate (2^n) % MOD
    int ans = 1;
    for (int i = 0; i < n; i++)
    {
        ans *= 2;
        ans %= MOD;
    }
 
    // Subtract the only where
    // no ball was chosen
    return ((ans - 1 + MOD) % MOD);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
 
    System.out.println(countWays(n));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
MOD = 1000000007
 
# Function to return the count of
# ways to choose the balls
def countWays(n):
     
    # Return ((2 ^ n)-1) % MOD
    return (((2**n) - 1) % MOD)
 
# Driver code
n = 3
print(countWays(n))

C#

// C# implementation of the approach
using System;
     
class GFG
{
static int MOD = 1000000007;
 
// Function to return the count of
// ways to choose the balls
static int countWays(int n)
{
 
    // Calculate (2^n) % MOD
    int ans = 1;
    for (int i = 0; i < n; i++)
    {
        ans *= 2;
        ans %= MOD;
    }
 
    // Subtract the only where
    // no ball was chosen
    return ((ans - 1 + MOD) % MOD);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
 
    Console.WriteLine(countWays(n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// javascript implementation of the approach
 
     MOD = 1000000007;
 
    // Function to return the count of
    // ways to choose the balls
    function countWays(n) {
 
        // Calculate (2^n) % MOD
        var ans = 1;
        for (i = 0; i < n; i++) {
            ans *= 2;
            ans %= MOD;
        }
 
        // Subtract the only where
        // no ball was chosen
        return ((ans - 1 + MOD) % MOD);
    }
 
    // Driver code
     
        var n = 3;
 
        document.write(countWays(n));
 
// This code contributed by gauravrajput1

输出：

7