# K length words that can be formed from given characters without repetition

Given an integer k and a string str consisting of lowercase English alphabets, the task is to count how many k character words (with or without meaning) can be formed from the characters of str when repetition is not allowed.

Examples:

Input: str = “cat”, k = 3
Output: 6
Required words are “cat”, “cta”, “act”, “atc”, “tca” and “tac”.

Input: str = “geeksforgeeks”, k = 3
Output: 840

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Count the number of distinct characters in str and store it in cnt, now the task is to arrange k characters out of cnt characters i.e. nPr = n! / (n – r)!.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the required count ` `int` `findPermutation(string str, ``int` `k) ` `{ ` `    ``bool` `has[26] = { ``false` `}; ` ` `  `    ``// To store the count of distinct characters in str ` `    ``int` `cnt = 0; ` ` `  `    ``// Traverse str character by character ` `    ``for` `(``int` `i = 0; i < str.length(); i++) { ` ` `  `        ``// If current character is appearing ` `        ``// for the first time in str ` `        ``if` `(!has[str[i] - ``'a'``]) { ` ` `  `            ``// Increment the distinct character count ` `            ``cnt++; ` ` `  `            ``// Update the appearance of the current character ` `            ``has[str[i] - ``'a'``] = ``true``; ` `        ``} ` `    ``} ` ` `  `    ``long` `long` `int` `ans = 1; ` ` `  `    ``// Since P(n, r) = n! / (n - r)! ` `    ``for` `(``int` `i = 2; i <= cnt; i++) ` `        ``ans *= i; ` ` `  `    ``for` `(``int` `i = cnt - k; i > 1; i--) ` `        ``ans /= i; ` ` `  `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"geeksforgeeks"``; ` `    ``int` `k = 4; ` `    ``cout << findPermutation(str, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `solution ` `{ ` `// Function to return the required count ` `static` `int` `findPermutation(String str, ``int` `k) ` `{ ` `    ``boolean``[] has = ``new` `boolean``[``26``]; ` `    ``Arrays.fill(has,``false``); ` ` `  `    ``// To store the count of distinct characters in str ` `    ``int` `cnt = ``0``; ` ` `  `    ``// Traverse str character by character ` `    ``for` `(``int` `i = ``0``; i < str.length(); i++) { ` ` `  `        ``// If current character is appearing ` `        ``// for the first time in str ` `        ``if` `(!has[str.charAt(i) - ``'a'``])  ` `        ``{ ` ` `  `            ``// Increment the distinct character count ` `            ``cnt++; ` ` `  `            ``// Update the appearance of the current character ` `            ``has[str.charAt(i) - ``'a'``] = ``true``; ` `        ``} ` `    ``} ` ` `  `    ``int` `ans = ``1``; ` ` `  `    ``// Since P(n, r) = n! / (n - r)! ` `    ``for` `(``int` `i = ``2``; i <= cnt; i++) ` `        ``ans *= i; ` ` `  `    ``for` `(``int` `i = cnt - k; i > ``1``; i--) ` `        ``ans /= i; ` ` `  `    ``// Return the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``String str = ``"geeksforgeeks"``; ` `    ``int` `k = ``4``; ` `    ``System.out.println(findPermutation(str, k)); ` ` `  `} ` `} ` `// This code is contributed by ` `// Sanjit_prasad `

## Python3

 `# Python3 implementation of the approach ` `import` `math as mt  ` ` `  `# Function to return the required count ` `def` `findPermutation(string, k): ` ` `  `    ``has ``=` `[``False` `for` `i ``in` `range``(``26``)] ` ` `  `    ``# To store the count of distinct  ` `    ``# characters in str ` `    ``cnt ``=` `0` ` `  `    ``# Traverse str character by character ` `    ``for` `i ``in` `range``(``len``(string)): ` `         `  `        ``# If current character is appearing ` `        ``# for the first time in str ` `        ``if` `(has[``ord``(string[i]) ``-` `ord``(``'a'``)] ``=``=` `False``): ` ` `  `            ``# Increment the distinct ` `            ``# character count ` `            ``cnt ``+``=` `1` ` `  `            ``# Update the appearance of the  ` `            ``# current character ` `            ``has[``ord``(string[i]) ``-` `ord``(``'a'``)] ``=` `True` `         `  `    ``ans ``=` `1` ` `  `    ``# Since P(n, r) = n! / (n - r)! ` `    ``for` `i ``in` `range``(``2``, cnt ``+` `1``): ` `        ``ans ``*``=` `i ` ` `  `    ``for` `i ``in` `range``(cnt ``-` `k, ``1``, ``-``1``): ` `        ``ans ``/``/``=` `i ` ` `  `    ``# Return the answer ` `    ``return` `ans ` ` `  `# Driver code ` `string ``=` `"geeksforgeeks"` `k ``=` `4` `print``(findPermutation(string, k)) ` ` `  `# This code is contributed  ` `# by Mohit kumar 29 `

## C#

 `// C# implementation of the approach  ` ` `  `using` `System; ` ` `  `class` `solution  ` `{  ` `// Function to return the required count  ` `static` `int` `findPermutation(``string` `str, ``int` `k)  ` `{  ` `    ``bool` `[]has = ``new` `bool``[26];  ` `     `  `    ``for` `(``int` `i = 0; i < 26 ; i++)  ` `        ``has[i] = ``false``; ` ` `  `    ``// To store the count of distinct characters in str  ` `    ``int` `cnt = 0;  ` ` `  `    ``// Traverse str character by character  ` `    ``for` `(``int` `i = 0; i < str.Length; i++) {  ` ` `  `        ``// If current character is appearing  ` `        ``// for the first time in str  ` `        ``if` `(!has[str[i] - ``'a'``])  ` `        ``{  ` ` `  `            ``// Increment the distinct character count  ` `            ``cnt++;  ` ` `  `            ``// Update the appearance of the current character  ` `            ``has[str[i] - ``'a'``] = ``true``;  ` `        ``}  ` `    ``}  ` ` `  `    ``int` `ans = 1;  ` ` `  `    ``// Since P(n, r) = n! / (n - r)!  ` `    ``for` `(``int` `i = 2; i <= cnt; i++)  ` `        ``ans *= i;  ` ` `  `    ``for` `(``int` `i = cnt - k; i > 1; i--)  ` `        ``ans /= i;  ` ` `  `    ``// Return the answer  ` `    ``return` `ans;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``string` `str = ``"geeksforgeeks"``;  ` `    ``int` `k = 4;  ` `    ``Console.WriteLine(findPermutation(str, k));  ` ` `  `}  ` `// This code is contributed by Ryuga ` `}  `

## PHP

 ` 1; ``\$i``--) ` `        ``\$ans` `/= ``\$i``; ` ` `  `    ``// Return the answer ` `    ``return` `\$ans``; ` `} ` ` `  `// Driver code ` `\$str` `= ``"geeksforgeeks"``; ` `\$k` `= 4; ` `echo` `findPermutation(``\$str``, ``\$k``); ` ` `  ` `  `// This code is contributed by ihritik ` `?> `

Output:

```840
```

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