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

  • Last Updated : 03 Mar, 2022

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 <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>
Output: 
7

 

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


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