# Count of subsequences which consists exactly K prime numbers

Given an integer K and an array arr[], the task is to find the number of subsequences from the given array such that each subsequence consists exactly K prime numbers.

Example:

Input: K = 2, arr = [2, 3, 4, 6]
Output: 4
Explanation:
There are 4 subsequences which consists exaclty 2 prime numbers {2, 3} {2, 3, 4} {2, 3, 6} {2, 3, 4, 6}

Input: K = 3, arr = [1, 2, 3, 4, 5, 6, 7]
Output: 32
Explanation:
There are 32 subsequences which consists exaclty 3 prime numbers.

Approach: To solve the problem mentioned above we have to find the count of prime numbers in the given array.

1. Let the count of prime numbers is m.
2. We can choose any K integers among m prime numbers.
3. So the possible combination of choosing prime numbers in the subsequence is and in the subsequence we can add any number of non – prime numbers because there is no restriction on the non prime numbers where count of non prime numbers will be (N – m).
4. We can choose any subset of the (N – m) for nonprime numbers in our subsequence.
5. Possibility of choosing all subset of size (N – m) is pow(2, (m – N)).
6. For generating subsequences, we will multiply prime number possibility with nonprime number possibility.

Sub sequence count = (Count of prime numbers) C (K) * pow(2, count of non-prime numbers)

Below is the implementation of the above approach:

 `// C++ implementation to find ` `// the count of subsequences ` `// which consist exactly K primes ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Returns factorial of n ` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = 1; ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function to return total ` `// number of combinations ` `int` `nCr(``int` `n, ``int` `r) ` `{ ` `    ``return` `fact(n) ` `           ``/ (fact(r) ` `              ``* fact(n - r)); ` `} ` ` `  `// Function check whether a number ` `// is prime or not ` `bool` `isPrime(``int` `n) ` `{ ` `    ``// Corner case ` `    ``if` `(n <= 1) ` `        ``return` `false``; ` ` `  `    ``// Check from 2 to n-1 ` `    ``for` `(``int` `i = 2; i < n; i++) ` `        ``if` `(n % i == 0) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function for finding number of subsequences ` `// which consists exactly K primes ` `int` `countSubsequences(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``int` `countPrime = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(isPrime(arr[i])) ` `            ``countPrime++; ` `    ``} ` `    ``// if number of primes are less thn k ` `    ``if` `(countPrime < k) ` `        ``return` `0; ` ` `  `    ``return` `nCr(countPrime, k) ` `           ``* ``pow``(2, (n - countPrime)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7 }; ` `    ``int` `K = 3; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << countSubsequences(arr, n, K); ` `    ``return` `0; ` `} `

Output:

```32
```

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