可以使用给定字符串的准确P个辅音和Q个元音构成的单词数

给定一个字符串str和两个整数P和Q。任务是找到可以通过从给定的字符串精确选择P个辅音和Q个元音来形成的单词总数。
例子：

Input: str = “geek”, P = 1, Q = 1 
Output: 8 
“ge”, “ge”, “eg”, “ek”, “eg”, “ek”, 
“ke” and “ke” are the possible words.
Input: str = “crackathon”, P = 4, Q = 3 
Output: 176400 
 

方法：由于必须从给定字符串的辅音和元音的原始计数中选择P个辅音和Q个元音。因此，可以使用二项式系数来计算选择这些字符的组合，并且可以使用其计数的阶乘将选择的字符排列在自身中。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
#define lli long long int
 
// Function to return the value of nCk
lli binomialCoeff(lli n, lli k)
{
    if (k == 0 || k == n)
        return 1;
 
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}
 
// Function to return the factorial of n
lli fact(lli n)
{
    if (n >= 1)
        return n * fact(n - 1);
    else
        return 1;
}
 
// Function that returns true if ch is a vowel
bool isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' || ch == 'i'
        || ch == 'o' || ch == 'u') {
        return true;
    }
 
    return false;
}
 
// Function to return the number of words possible
lli countWords(string s, int p, int q)
{
 
    // To store the count of vowels and
    // consonanats in the given string
    lli countc = 0, countv = 0;
    for (int i = 0; i < s.length(); i++) {
 
        // If current character is a vowel
        if (isVowel(s[i]))
            countv++;
        else
            countc++;
    }
 
    // Find the total possible words
    lli a = binomialCoeff(countc, p);
    lli b = binomialCoeff(countv, q);
    lli c = fact(p + q);
    lli ans = (a * b) * c;
    return ans;
}
 
// Driver code
int main()
{
    string s = "crackathon";
    int p = 4, q = 3;
 
    cout << countWords(s, p, q);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
     
    // Function to return the value of nCk
    static long binomialCoeff(long n, long k)
    {
        if (k == 0 || k == n)
            return 1;
     
        return binomialCoeff(n - 1, k - 1) +
               binomialCoeff(n - 1, k);
    }
     
    // Function to return the factorial of n
    static long fact(long n)
    {
        if (n >= 1)
            return n * fact(n - 1);
        else
            return 1;
    }
     
    // Function that returns true if ch is a vowel
    static boolean isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' ||
                         ch == 'o' || ch == 'u')
        {
            return true;
        }
     
        return false;
    }
     
    // Function to return the number of words possible
    static long countWords(String s, int p, int q)
    {
     
        // To store the count of vowels and
        // consonanats in the given string
        long countc = 0, countv = 0;
        for (int i = 0; i < s.length(); i++)
        {
     
            // If current character is a vowel
            if (isVowel(s.charAt(i)))
                countv++;
            else
                countc++;
        }
     
        // Find the total possible words
        long a = binomialCoeff(countc, p);
        long b = binomialCoeff(countv, q);
        long c = fact(p + q);
        long ans = (a * b) * c;
        return ans;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String s = "crackathon";
        int p = 4, q = 3;
     
        System.out.println(countWords(s, p, q));
    }
}
 
// This Code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to return the value of nCk
def binomialCoeff(n, k):
    if (k == 0 or k == n):
        return 1
 
    return binomialCoeff(n - 1, k - 1) + \
           binomialCoeff(n - 1, k)
 
# Function to return the factorial of n
def fact(n):
    if (n >= 1):
        return n * fact(n - 1)
    else:
        return 1
 
# Function that returns true if ch is a vowel
def isVowel(ch):
 
    if (ch == 'a' or ch == 'e' or
        ch == 'i' or ch == 'o' or ch == 'u'):
        return True
 
    return False
 
# Function to return the number of words possible
def countWords(s, p, q):
 
    # To store the count of vowels and
    # consonanats in the given string
    countc = 0
    countv = 0
    for i in range(len(s)):
 
        # If current character is a vowel
        if (isVowel(s[i])):
            countv += 1
        else:
            countc += 1
 
    # Find the total possible words
    a = binomialCoeff(countc, p)
    b = binomialCoeff(countv, q)
    c = fact(p + q)
    ans = (a * b) * c
    return ans
 
# Driver code
s = "crackathon"
p = 4
q = 3
 
print(countWords(s, p, q))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
     
    // Function to return the value of nCk
    static long binomialCoeff(long n, long k)
    {
        if (k == 0 || k == n)
            return 1;
     
        return binomialCoeff(n - 1, k - 1) +
               binomialCoeff(n - 1, k);
    }
     
    // Function to return the factorial of n
    static long fact(long n)
    {
        if (n >= 1)
            return n * fact(n - 1);
        else
            return 1;
    }
     
    // Function that returns true if ch is a vowel
    static bool isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' ||
                         ch == 'o' || ch == 'u')
        {
            return true;
        }
        return false;
    }
     
    // Function to return the number of words possible
    static long countWords(String s, int p, int q)
    {
     
        // To store the count of vowels and
        // consonanats in the given string
        long countc = 0, countv = 0;
        for (int i = 0; i < s.Length; i++)
        {
     
            // If current character is a vowel
            if (isVowel(s[i]))
                countv++;
            else
                countc++;
        }
     
        // Find the total possible words
        long a = binomialCoeff(countc, p);
        long b = binomialCoeff(countv, q);
        long c = fact(p + q);
        long ans = (a * b) * c;
        return ans;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        String s = "crackathon";
        int p = 4, q = 3;
     
        Console.WriteLine(countWords(s, p, q));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript implementation of the approach
 
// Function to return the value of nCk
function binomialCoeff(n, k)
{
    if (k == 0 || k == n)
        return 1;
 
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}
 
// Function to return the factorial of n
function fact(n)
{
    if (n >= 1)
        return n * fact(n - 1);
    else
        return 1;
}
 
// Function that returns true if ch is a vowel
 
function isVowel(ch)
{
    if (ch == 'a' || ch == 'e' || ch == 'i'
        || ch == 'o' || ch == 'u') {
        return true;
    }
 
    return false;
}
 
// Function to return the number of words possible
 
function countWords(s, p, q)
{
 
    // To store the count of vowels and
    // consonanats in the given string
    var countc = 0, countv = 0;
    for (var i = 0; i < s.length; i++) {
 
        // If current character is a vowel
        if (isVowel(s[i]))
            countv++;
        else
            countc++;
    }
 
    // Find the total possible words
    var a = binomialCoeff(countc, p);
    var b = binomialCoeff(countv, q);
    var c = fact(p + q);
    var ans = (a * b) * c;
    return ans;
}
 
// Driver code
var s = "crackathon";
var p = 4, q = 3;
document.write(countWords(s, p, q));
 

输出：

176400