Ways to choose balls such that at least one ball is chosen

Given an integer N, the task is to find the ways to choose some balls out of the given N balls such that at least one ball is chosen. Since the value can be large so print the value modulo 1000000007.
Example:

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

Approach: There are N balls and each ball can either be chosen or not chosen. Total number of different configurations is 2 * 2 * 2 * … * N. We can write this as 2^N. But the state where no ball is chosen has to be subtracted from the answer. So, the result will be (2^N – 1) % 1000000007.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

<script>
// 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
</script>