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# 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:
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 2N. But the state where no ball is chosen has to be subtracted from the answer. So, the result will be (2N – 1) % 1000000007.
Below is the implementation of the above approach:

## 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

 ``

Output:

`7`

Time Complexity : O(n)
Auxiliary Space : O(1)