# Number of ways to arrange a word such that all vowels occur together

Given a word containing vowels and consonants. The task is to find that in how many different ways word can be arranged so that the vowels always come together. Given that the length of the word <10

Examples:

```Input: str = "geek"
Output: 6
Ways such that both 'e' comes together are 6
i.e. geek, gkee, kgee, eekg, eegk, keeg

Input: str = "corporation"
Output: 50400
```

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

Approach: Since word contains vowels and consonant together. All vowels are needed to remain together then we will take all vowels as a single letter.

As, in the word ‘geeksforgeeks’, we can treat the vowels “eeoee” as one letter.
Thus, we have gksfrgks (eeoee).
This has 9 (8 + 1) letters of which g, k, s each occurs 2 times and the rest are different.

The number of ways arranging these letters = 9!/(2!)x(2!)x(2!) = 45360 ways

Now, 5 vowels in which ‘e’ occurs 4 times and ‘o’ occurs 1 time, can be arranged in 5! /4! = 5 ways.

Required number of ways = (45360 x 5) = 226800

Below is the implementation of the above approach:

 `// CPP program to calculate the no. of ways ` `// to arrange the word having vowels together ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Factorial of a number ` `ll fact(``int` `n) ` `{ ` `    ``ll f = 1; ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``f = f * i; ` `    ``return` `f; ` `} ` ` `  `// calculating ways for arranging consonants ` `ll waysOfConsonants(``int` `size1, ``int` `freq[]) ` `{ ` `    ``ll ans = fact(size1); ` `    ``for` `(``int` `i = 0; i < 26; i++) { ` ` `  `        ``// Ignore vowels ` `        ``if` `(i == 0 || i == 4 || i == 8 || i == 14 || i == 20) ` `            ``continue``; ` `        ``else` `            ``ans = ans / fact(freq[i]); ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// calculating ways for arranging vowels ` `ll waysOfVowels(``int` `size2, ``int` `freq[]) ` `{ ` `    ``return` `fact(size2) / (fact(freq[0]) * fact(freq[4]) * fact(freq[8]) ` `                    ``* fact(freq[14]) * fact(freq[20])); ` `} ` ` `  `// Function to count total no. of ways ` `ll countWays(string str) ` `{ ` ` `  `    ``int` `freq[26] = { 0 }; ` `    ``for` `(``int` `i = 0; i < str.length(); i++) ` `        ``freq[str[i] - ``'a'``]++; ` ` `  `    ``// Count vowels and consonant ` `    ``int` `vowel = 0, consonant = 0; ` `    ``for` `(``int` `i = 0; i < str.length(); i++) { ` ` `  `        ``if` `(str[i] != ``'a'` `&& str[i] != ``'e'` `&& str[i] != ``'i'` `            ``&& str[i] != ``'o'` `&& str[i] != ``'u'``) ` `            ``consonant++; ` `        ``else` `            ``vowel++; ` `    ``} ` ` `  `    ``// total no. of ways ` `    ``return` `waysOfConsonants(consonant+1, freq) *  ` `           ``waysOfVowels(vowel, freq); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"geeksforgeeks"``; ` ` `  `    ``cout << countWays(str) << endl; ` ` `  `    ``return` `0; ` `} `

Output:

```226800
```

Further Optimizations : We can pre-compute required factorial values to avoid re-computations.

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