# Count of K length subsequence whose product is even

Given an array arr[] and an integer K, the task is to find number of non empty subsequence of length K from the given array arr of size N such that the product of subsequence is a even number.
Example:

Input: arr[] = [2, 3, 1, 7], K = 3
Output:
Explanation:
There are 3 subsequences of length 3 whose product is even number {2, 3, 1}, {2, 3, 7}, {2, 1, 7}.

Input: arr[] = [2, 4], K = 1
Output:
Explanation:
There are 2 subsequence of length 1 whose product is even number {2} {4}.

Approach:
To solve the problem mentioned above we have to find the total number of subsequence of length K and subtract the count of K length subsequence whose product is odd.

1. For making a product of the subsequence odd we must choose K numbers as odd.
2. So the number of subsequences of length K whose product is odd is possibly finding k odd numbers, i.e., “o choose k” or where o is the count of odd numbers in the subsequence.
3. where n and o is the count of total numbers and odd numbers respectively.

Below is the implementation of above program:

## C++

 // C++ implementation to Count of K  // length subsequence whose  // Product is even     #include  using namespace std;     int fact(int n);     // Function to calculate nCr  int nCr(int n, int r)  {      if (r > n)          return 0;      return fact(n)             / (fact(r)                * fact(n - r));  }     // Returns factorial of n  int fact(int n)  {      int res = 1;      for (int i = 2; i <= n; i++)          res = res * i;      return res;  }     // Function for finding number  // of K length subsequences  // whose product is even number  int countSubsequences(      int arr[], int n, int k)  {      int countOdd = 0;         // counting odd numbers in the array      for (int i = 0; i < n; i++) {          if (arr[i] & 1)              countOdd++;      }      int ans = nCr(n, k)                - nCr(countOdd, k);         return ans;  }     // Driver code  int main()  {         int arr[] = { 2, 4 };      int K = 1;         int N = sizeof(arr) / sizeof(arr);         cout << countSubsequences(arr, N, K);         return 0;  }

## Java

 // Java implementation to count of K  // length subsequence whose product   // is even  import java.util.*;     class GFG{         // Function to calculate nCr  static int nCr(int n, int r)  {      if (r > n)          return 0;      return fact(n) / (fact(r) *                        fact(n - r));  }     // Returns factorial of n  static int fact(int n)  {      int res = 1;      for(int i = 2; i <= n; i++)          res = res * i;                 return res;  }     // Function for finding number  // of K length subsequences  // whose product is even number  static int countSubsequences(int arr[],                               int n, int k)  {      int countOdd = 0;         // Counting odd numbers in the array      for(int i = 0; i < n; i++)      {          if (arr[i] % 2 == 1)              countOdd++;      }      int ans = nCr(n, k) - nCr(countOdd, k);         return ans;  }     // Driver code  public static void main(String args[])  {      int arr[] = { 2, 4 };      int K = 1;         int N = arr.length;         System.out.println(countSubsequences(arr, N, K));  }  }     // This code is contributed by ANKITKUMAR34

## Python3

 # Python3 implementation to Count of K  # length subsequence whose  # Product is even     # Function to calculate nCr  def nCr(n, r):             if (r > n):          return 0     return fact(n) // (fact(r) *                         fact(n - r))     # Returns factorial of n  def fact(n):             res = 1     for i in range(2, n + 1):          res = res * i                 return res     # Function for finding number  # of K length subsequences  # whose product is even number  def countSubsequences(arr, n, k):             countOdd = 0        # Counting odd numbers in the array      for i in range(n):          if (arr[i] & 1):              countOdd += 1;         ans = nCr(n, k) - nCr(countOdd, k);         return ans         # Driver code  arr = [ 2, 4 ]  K = 1    N = len(arr)     print(countSubsequences(arr, N, K))     # This code is contributed by ANKITKUAR34

## C#

 // C# implementation to count of K  // length subsequence whose product   // is even  using System;     class GFG{         // Function to calculate nCr  static int nCr(int n, int r)  {      if (r > n)          return 0;                 return fact(n) / (fact(r) *                        fact(n - r));  }     // Returns factorial of n  static int fact(int n)  {      int res = 1;      for(int i = 2; i <= n; i++)          res = res * i;                 return res;  }     // Function for finding number  // of K length subsequences  // whose product is even number  static int countSubsequences(int []arr,                               int n, int k)  {      int countOdd = 0;         // Counting odd numbers in the array      for(int i = 0; i < n; i++)      {          if (arr[i] % 2 == 1)              countOdd++;      }      int ans = nCr(n, k) - nCr(countOdd, k);         return ans;  }     // Driver code  public static void Main(String []args)  {      int []arr = { 2, 4 };      int K = 1;         int N = arr.Length;         Console.WriteLine(countSubsequences(arr, N, K));  }  }     // This code is contributed by Princi Singh

Output:

2


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