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 = 3Output:3Explanation:

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 = 1Output:2Explanation:

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.

- For making a product of the subsequence odd we must choose K numbers as odd.
- 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.

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 <bits/stdc++.h>` `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[0]);` ` ` `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` |

## Javascript

`<script>` `// javascript implementation to Count of K` `// length subsequence whose` `// Product is even` `// Function to calculate nCr` `function` `nCr(n, r)` `{` ` ` `if` `(r > n)` ` ` `return` `0;` ` ` `return` `fact(n)` ` ` `/ (fact(r)` ` ` `* fact(n - r));` `}` `// Returns factorial of n` `function` `fact(n)` `{` ` ` `var` `res = 1;` ` ` `for` `(` `var` `i = 2; i <= n; i++)` ` ` `res = res * i;` ` ` `return` `res;` `}` `// Function for finding number` `// of K length subsequences` `// whose product is even number` `function` `countSubsequences( arr, n, k)` `{` ` ` `var` `countOdd = 0;` ` ` `// counting odd numbers in the array` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `if` `(arr[i] & 1)` ` ` `countOdd++;` ` ` `}` ` ` `var` `ans = nCr(n, k)` ` ` `- nCr(countOdd, k);` ` ` `return` `ans;` `}` `// Driver code` `var` `arr = [ 2, 4 ];` `var` `K = 1;` `var` `N = arr.length;` `document.write( countSubsequences(arr, N, K));` `</script>` |

**Output:**

2

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **DSA Live Classes**