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Permutations of string such that no two vowels are adjacent
• Difficulty Level : Expert
• Last Updated : 10 Jan, 2019

Given a string consisting of vowels and consonants. The task is to find the number of ways in which the characters of the string can be arranged such that no two vowels are adjacent to each other.

Note: Given that No. of vowels <= No. of consonants.

Examples:

Input: str = "permutation"
Output : 907200

Input: str = "geeksforgeeks"
Output: 3175200


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

Approach:
Consider the above example string “permutation”:

• First place all of the consonants at the alternate places like below:
-- p -- r -- m -- t -- t -- n --


Number of ways to place consonants = 6! / 2!. as appears twice and should be considered once.

• Then place the vowels at the remaining positions. We have 7 remaining positions and 5 vowels to fill these 7 places.
Therefore, number of ways to fill vowels = .
Total no. of ways =
= 907200


Suppose in a string number of vowels is vowelCount and the number of consonants is consonantCount.

Therefore,

Total ways = (consonantCount! / duplicateConsonant!) * C(consonantCount+1 , vowelCount) * (vowelCount! / duplicateVowel!)

Below is the implementation of the above approach:

## C++

 // CPP program to count permutations of string// such that no two vowels are adjacent  #include using namespace std;  // Factorial of a numberint factorial(int n){      int fact = 1;    for (int i = 2; i <= n; i++)        fact = fact * i;      return fact;}  // Function to find c(n, r)int ncr(int n, int r){    return factorial(n) / (factorial(r) * factorial(n - r));}  // Function to count permutations of string// such that no two vowels are adjacentint countWays(string str){    int freq[26] = { 0 };    int nvowels = 0, nconsonants = 0;      int vplaces, cways, vways;      // Finding the frequencies of    // the characters    for (int i = 0; i < str.length(); i++)        ++freq[str[i] - 'a'];      // finding the no. of vowels and    // consonants in given word    for (int i = 0; i < 26; i++) {          if (i == 0 || i == 4 || i == 8            || i == 14 || i == 20)            nvowels += freq[i];        else            nconsonants += freq[i];    }    // finding places for the vowels    vplaces = nconsonants + 1;      // ways to fill consonants 6! / 2!    cways = factorial(nconsonants);    for (int i = 0; i < 26; i++) {        if (i != 0 && i != 4 && i != 8 && i != 14            && i != 20 && freq[i] > 1) {              cways = cways / factorial(freq[i]);        }    }      // ways to put vowels 7C5 x 5!    vways = ncr(vplaces, nvowels) * factorial(nvowels);    for (int i = 0; i < 26; i++) {        if (i == 0 || i == 4 || i == 8 || i == 14            || i == 20 && freq[i] > 1) {            vways = vways / factorial(freq[i]);        }    }      return cways * vways;}  // Driver codeint main(){    string str = "permutation";      cout << countWays(str) << endl;      return 0;}

## Java

 // Java program to count permutations of string// such that no two vowels are adjacent  class GFG{                      // Factorial of a number        static int factorial(int n)        {                      int fact = 1;            for (int i = 2; i <= n; i++)                fact = fact * i;                      return fact;        }                  // Function to find c(n, r)        static int ncr(int n, int r)        {            return factorial(n) / (factorial(r) * factorial(n - r));        }                  // Function to count permutations of string        // such that no two vowels are adjacent        static int countWays(String str)        {            int freq[]=new int[26];                          for(int i=0;i<26;i++)            {                freq[i]=0;            }                          int nvowels = 0, nconsonants = 0;                      int vplaces, cways, vways;                      // Finding the frequencies of            // the characters            for (int i = 0; i < str.length(); i++)                ++freq[str.charAt(i) - 'a'];                      // finding the no. of vowels and            // consonants in given word            for (int i = 0; i < 26; i++) {                          if (i == 0 || i == 4 || i == 8                    || i == 14 || i == 20)                    nvowels += freq[i];                else                    nconsonants += freq[i];            }            // finding places for the vowels            vplaces = nconsonants + 1;                      // ways to fill consonants 6! / 2!            cways = factorial(nconsonants);            for (int i = 0; i < 26; i++) {                if (i != 0 && i != 4 && i != 8 && i != 14                    && i != 20 && freq[i] > 1) {                              cways = cways / factorial(freq[i]);                }            }                      // ways to put vowels 7C5 x 5!            vways = ncr(vplaces, nvowels) * factorial(nvowels);            for (int i = 0; i < 26; i++) {                if (i == 0 || i == 4 || i == 8 || i == 14                    || i == 20 && freq[i] > 1) {                    vways = vways / factorial(freq[i]);                }            }                      return cways * vways;        }                  // Driver code        public static void main(String []args)        {            String str = "permutation";                      System.out.println(countWays(str));                            }}  // This code is contributed// by ihritik

## Python3

 # Python3 program to count permutations of # string such that no two vowels are adjacent   # Factorial of a number def factorial(n) :       fact = 1;     for i in range(2, n + 1) :        fact = fact * i      return fact  # Function to find c(n, r) def ncr(n, r) :          return factorial(n) // (factorial(r) *                             factorial(n - r))  # Function to count permutations of string # such that no two vowels are adjacent def countWays(string) :       freq = [0] * 26    nvowels, nconsonants = 0, 0      # Finding the frequencies of     # the characters     for i in range(len(string)) :        freq[ord(string[i]) - ord('a')] += 1      # finding the no. of vowels and     # consonants in given word     for i in range(26) :          if (i == 0 or i == 4 or i == 8            or i == 14 or i == 20) :            nvowels += freq[i]         else :            nconsonants += freq[i]          # finding places for the vowels     vplaces = nconsonants + 1      # ways to fill consonants 6! / 2!     cways = factorial(nconsonants)    for i in range(26) :        if (i != 0 and i != 4 and i != 8 and            i != 14 and i != 20 and freq[i] > 1) :               cways = cways // factorial(freq[i])      # ways to put vowels 7C5 x 5!     vways = ncr(vplaces, nvowels) * factorial(nvowels)    for i in range(26) :        if (i == 0 or i == 4 or i == 8 or i == 14            or i == 20 and freq[i] > 1) :            vways = vways // factorial(freq[i])       return cways * vways;   # Driver code if __name__ == "__main__" :      string = "permutation"      print(countWays(string))  # This code is contributed by Ryuga

## C#

 // C# program to count permutations of string// such that no two vowels are adjacent  using System;class GFG{                      // Factorial of a number        static int factorial(int n)        {                      int fact = 1;            for (int i = 2; i <= n; i++)                fact = fact * i;                      return fact;        }                  // Function to find c(n, r)        static int ncr(int n, int r)        {            return factorial(n) / (factorial(r) * factorial(n - r));        }                  // Function to count permutations of string        // such that no two vowels are adjacent        static int countWays(String str)        {            int []freq=new int[26];                          for(int i=0;i<26;i++)            {                freq[i]=0;            }                          int nvowels = 0, nconsonants = 0;                      int vplaces, cways, vways;                      // Finding the frequencies of            // the characters            for (int i = 0; i < str.Length; i++)                ++freq[str[i] - 'a'];                      // finding the no. of vowels and            // consonants in given word            for (int i = 0; i < 26; i++) {                          if (i == 0 || i == 4 || i == 8                    || i == 14 || i == 20)                    nvowels += freq[i];                else                    nconsonants += freq[i];            }            // finding places for the vowels            vplaces = nconsonants + 1;                      // ways to fill consonants 6! / 2!            cways = factorial(nconsonants);            for (int i = 0; i < 26; i++) {                if (i != 0 && i != 4 && i != 8 && i != 14                    && i != 20 && freq[i] > 1) {                              cways = cways / factorial(freq[i]);                }            }                      // ways to put vowels 7C5 x 5!            vways = ncr(vplaces, nvowels) * factorial(nvowels);            for (int i = 0; i < 26; i++) {                if (i == 0 || i == 4 || i == 8 || i == 14                    || i == 20 && freq[i] > 1) {                    vways = vways / factorial(freq[i]);                }            }                      return cways * vways;        }                  // Driver code        public static void Main()        {            String str = "permutation";                      Console.WriteLine(countWays(str));                            }}  // This code is contributed// by ihritik

## PHP

  1) {               $cways = $cways / factorial($freq[$i]);        }    }       // ways to put vowels 7C5 x 5!    $vways = ncr($vplaces, $nvowels) * factorial($nvowels);    for ($i = 0; $i < 26; $i++) { if ($i == 0 || $i == 4 || $i == 8 || $i == 14 || $i == 20 && $freq[$i] > 1) {            $vways = $vways / factorial($freq[$i]);        }    }    return $cways * $vways;}   // Driver code  $str = "permutation"; echo countWays($str)."\n";return 0;// this code is contributed by Ita_c.?>

Output:

907200


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