# Arrangement of the characters of a word such that all vowels are at odd places

Given a string ‘S’ containing vowels and consonants of lowercase English alphabets. The task is to find the number of ways in which the characters of the word can be arranged such that the vowels occupy only the odd positions.

Examples:

Input: geeks
Output: 36 Input: publish
Output: 1440 ## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

First find the total no. of odd places and even places in the given word.

Total number of even places = floor(word length/2)
Total number of odd places = word length – total even places

Let’s consider the string “contribute” then there are 10 letters in the given word and there are 5 odd places, 5 even places, 4 vowels and 6 consonants.

Let us mark these positions as under:
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

Now, 4 vowels can be placed at any of the five places, marked 1, 3, 5, 7, 9.
The number of ways of arranging the vowels = 5_P_4 = 5! = 120

Also, the 6 consonants can be arranged at the remaining 6 positions.
Number of ways of these arrangements = 6_P_6 = 6! = 720.

Total number of ways = (120 x 720) = 86400

Below is the implementation of the above approach:

## C++

 // C++ program to find the number of ways  // in which the characters of the word  // can be arranged such that the vowels  // occupy only the odd positions  #include  using namespace std;     // Function to return the  // factorial of a number  int fact(int n)  {      int f = 1;      for (int i = 2; i <= n; i++) {          f = f * i;      }         return f;  }     // calculating nPr  int npr(int n, int r)  {      return fact(n) / fact(n - r);  }     // Function to find the number of ways  // in which the characters of the word  // can be arranged such that the vowels  // occupy only the odd positions  int countPermutations(string str)  {      // Get total even positions      int even = floor(str.length() / 2);         // Get total odd positions      int odd = str.length() - even;         int ways = 0;         // Store frequency of each character of      // the string      int freq = { 0 };      for (int i = 0; i < str.length(); i++) {          ++freq[str[i] - 'a'];      }         // Count total number of vowels      int nvowels          = freq + freq            + freq + freq            + freq;         // Count total number of consonants      int nconsonants          = str.length() - nvowels;         // Calculate the total number of ways      ways = npr(odd, nvowels) * npr(nconsonants, nconsonants);         return ways;  }     // Driver code  int main()  {      string str = "geeks";         cout << countPermutations(str);         return 0;  }

## Java

 // Java program to find the number of ways  // in which the characters of the word  // can be arranged such that the vowels  // occupy only the odd positions  class GFG{  // Function to return the  // factorial of a number  static int fact(int n)  {      int f = 1;      for (int i = 2; i <= n; i++) {          f = f * i;      }         return f;  }     // calculating nPr  static int npr(int n, int r)  {      return fact(n) / fact(n - r);  }     // Function to find the number of ways  // in which the characters of the word  // can be arranged such that the vowels  // occupy only the odd positions  static int countPermutations(String str)  {      // Get total even positions      int even = (int)Math.floor((double)(str.length() / 2));         // Get total odd positions      int odd = str.length() - even;         int ways = 0;         // Store frequency of each character of      // the string      int[] freq=new int;      for (int i = 0; i < str.length(); i++) {          freq[(int)(str.charAt(i)-'a')]++;      }         // Count total number of vowels      int nvowels= freq + freq+ freq                   + freq+ freq;         // Count total number of consonants      int nconsonants= str.length() - nvowels;         // Calculate the total number of ways      ways = npr(odd, nvowels) * npr(nconsonants, nconsonants);         return ways;  }     // Driver code  public static void main(String[] args)  {      String str = "geeks";         System.out.println(countPermutations(str));  }  }  // This code is contributed by mits

## Python3

 # Python3 program to find the number   # of ways in which the characters   # of the word can be arranged such   # that the vowels occupy only the   # odd positions  import math     # Function to return the factorial  # of a number  def fact(n):      f = 1;      for i in range(2, n + 1):          f = f * i;         return f;     # calculating nPr  def npr(n, r):      return fact(n) / fact(n - r);     # Function to find the number of   # ways in which the characters of   # the word can be arranged such   # that the vowels occupy only the   # odd positions  def countPermutations(str):         # Get total even positions      even = math.floor(len(str) / 2);         # Get total odd positions      odd = len(str) - even;         ways = 0;         # Store frequency of each       # character of the string      freq =  * 26;      for i in range(len(str)):          freq[ord(str[i]) - ord('a')] += 1;         # Count total number of vowels      nvowels = (freq + freq + freq +                 freq + freq);         # Count total number of consonants      nconsonants = len(str) - nvowels;         # Calculate the total number of ways      ways = (npr(odd, nvowels) *              npr(nconsonants, nconsonants));         return int(ways);     # Driver code  str = "geeks";     print(countPermutations(str));         # This code is contributed by mits

## C#

 // C# program to find the number of ways  // in which the characters of the word  // can be arranged such that the vowels  // occupy only the odd positions  using System;  class GFG{  // Function to return the  // factorial of a number  static int fact(int n)  {      int f = 1;      for (int i = 2; i <= n; i++) {          f = f * i;      }         return f;  }     // calculating nPr  static int npr(int n, int r)  {      return fact(n) / fact(n - r);  }     // Function to find the number of ways  // in which the characters of the word  // can be arranged such that the vowels  // occupy only the odd positions  static int countPermutations(String str)  {      // Get total even positions      int even = (int)Math.Floor((double)(str.Length / 2));         // Get total odd positions      int odd = str.Length - even;         int ways = 0;         // Store frequency of each character of      // the string      int[] freq=new int;      for (int i = 0; i < str.Length; i++) {          freq[(int)(str[i]-'a')]++;      }         // Count total number of vowels      int nvowels= freq + freq+ freq                   + freq+ freq;         // Count total number of consonants      int nconsonants= str.Length - nvowels;         // Calculate the total number of ways      ways = npr(odd, nvowels) * npr(nconsonants, nconsonants);         return ways;  }     // Driver code  static void Main()  {      String str = "geeks";         Console.WriteLine(countPermutations(str));  }  }  // This code is contributed by mits

## PHP

 

Output:

36


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