Number of ways to select exactly K even numbers from given Array

Last Updated : 14 Sep, 2020

Given an array arr[] of n integers and an integer K, the task is to find the number of ways to select exactly K even numbers from the given array.

Examples:

Input: arr[] = {1, 2, 3, 4} k = 1
Output: 2
Explanation:
The number of ways in which we can select one even number is 2.
Input: arr[] = {61, 65, 99, 26, 57, 68, 23, 2, 32, 30} k = 2
Output:10
Explanation:
The number of ways in which we can select 2 even number is 10.

Approach: The idea is to apply the rule of combinatorics. For choosing r objects from the given n objects, the total number of ways of choosing is given by ⁿC_r. Below are the steps:

Count the total number of even elements from the given array(say cnt).
Check if the value of K is greater than cnt then the number of ways will be equal to 0.
Otherwise, the answer will be ⁿC_k.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
long long f[12];
  
// Function for calculating factorial
void fact()
{
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
  
    for (int i = 2; i <= 10; i++)
        f[i] = i * 1LL * f[i - 1];
}
  
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
void solve(int arr[], int n, int k)
{
    fact();
  
    // Count even numbers
    int even = 0;
    for (int i = 0; i < n; i++) {
  
        // Check if the current
        // number is even
        if (arr[i] % 2 == 0)
            even++;
    }
  
    // Check if the even numbers to be
    // choosen is greater than n. Then,
    // there is no way to pick it.
    if (k > even)
        cout << 0 << endl;
  
    else {
        // The number of ways will be nCk
        cout << f[even] / (f[k] * f[even - k]);
    }
}
  
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof arr / sizeof arr[0];
  
    // Given count of even elements
    int k = 1;
  
    // Function Call
    solve(arr, n, k);
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
static int []f = new int[12];
  
// Function for calculating factorial
static void fact()
{
      
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
  
    for(int i = 2; i <= 10; i++)
        f[i] = i * 1 * f[i - 1];
}
  
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
static void solve(int arr[], int n, int k)
{
    fact();
  
    // Count even numbers
    int even = 0;
    for(int i = 0; i < n; i++)
    {
          
        // Check if the current
        // number is even
        if (arr[i] % 2 == 0)
            even++;
    }
  
    // Check if the even numbers to be
    // choosen is greater than n. Then,
    // there is no way to pick it.
    if (k > even)
        System.out.print(0 + "\n");
  
    else 
    {
        // The number of ways will be nCk
        System.out.print(f[even] / 
                        (f[k] * f[even - k]));
    }
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
    int n = arr.length;
  
    // Given count of even elements
    int k = 1;
  
    // Function call
    solve(arr, n, k);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
f = [0] * 12
  
# Function for calculating factorial
def fact():
      
    # Factorial of n defined as:
    # n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1
  
    for i in range(2, 11):
        f[i] = i * 1 * f[i - 1]
  
# Function to find the number of ways to
# select exactly K even numbers
# from the given array
def solve(arr, n, k):
      
    fact()
  
    # Count even numbers
    even = 0
    for i in range(n):
  
        # Check if the current
        # number is even
        if (arr[i] % 2 == 0):
            even += 1
      
    # Check if the even numbers to be
    # choosen is greater than n. Then,
    # there is no way to pick it.
    if (k > even):
        print(0)
    else:
          
        # The number of ways will be nCk
        print(f[even] // (f[k] * f[even - k]))
      
# Driver Code
  
# Given array arr[]
arr = [ 1, 2, 3, 4 ]
  
n = len(arr)
  
# Given count of even elements
k = 1
  
# Function call
solve(arr, n, k)
  
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
class GFG{
      
static int []f = new int[12];
  
// Function for calculating factorial
static void fact()
{
      
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
  
    for(int i = 2; i <= 10; i++)
        f[i] = i * 1 * f[i - 1];
}
  
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
static void solve(int []arr, int n, int k)
{
    fact();
  
    // Count even numbers
    int even = 0;
    for(int i = 0; i < n; i++)
    {
          
        // Check if the current
        // number is even
        if (arr[i] % 2 == 0)
            even++;
    }
  
    // Check if the even numbers to be
    // choosen is greater than n. Then,
    // there is no way to pick it.
    if (k > even)
        Console.Write(0 + "\n");
  
    else 
    {
        // The number of ways will be nCk
        Console.Write(f[even] / 
                        (f[k] * f[even - k]));
    }
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given array []arr
    int []arr = { 1, 2, 3, 4 };
    int n = arr.Length;
  
    // Given count of even elements
    int k = 1;
  
    // Function call
    solve(arr, n, k);
}
}
  
// This code is contributed by sapnasingh4991

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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