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

• Last Updated : 07 May, 2021

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 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 numberint fact(int n){    int f = 1;    for (int i = 2; i <= n; i++) {        f = f * i;    }     return f;} // calculating nPrint 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 positionsint 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 codeint 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 positionsclass GFG{// Function to return the// factorial of a numberstatic int fact(int n){    int f = 1;    for (int i = 2; i <= n; i++) {        f = f * i;    }     return f;} // calculating nPrstatic 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 positionsstatic 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 codepublic 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 positionsimport math # Function to return the factorial# of a numberdef fact(n):    f = 1;    for i in range(2, n + 1):        f = f * i;     return f; # calculating nPrdef 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 positionsdef 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 codestr = "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 positionsusing System;class GFG{// Function to return the// factorial of a numberstatic int fact(int n){    int f = 1;    for (int i = 2; i <= n; i++) {        f = f * i;    }     return f;} // calculating nPrstatic 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 positionsstatic 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 codestatic void Main(){    String str = "geeks";     Console.WriteLine(countPermutations(str));}}// This code is contributed by mits

## PHP

 

## Javascript

 
Output:
36

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