Open In App

Find the number of words of X vowels and Y consonants that can be formed from M vowels and N consonants

Improve
Improve
Like Article
Like
Save
Share
Report

Given four integers X, Y, M and N. The task is to find the number of ways to form a word by choosing X number of vowels and Y number of consonants from total numbers of M vowels and N consonants.
Examples: 
 

Input : X = 2, Y = 2, M = 3, N = 3 
Output : 216 
The total number of ways of choosing 2 vowels from a total number of 3 vowels is 3\choose2   i.e 3 
The total number of ways of choosing 2 consonants from a total number of 3 consonants is 3\choose2   i.e 3. 
The total number of ways of selecting 2 consonants from 3 and 2 vowels from 3 is 3\choose2   3\choose2   = 9 
The total number of ways of arranging 4 letters among themselves = 4! = 24 
Hence, the required number of ways = 24 * 9 = 216
Input : X = 1, Y = 2, M = 2, N = 3 
Output : 36 
 


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

  • The total number of ways of choosing X vowels from a total number of M vowels is M\choose X
  • The total number of ways of choosing Y consonants from a total number of N consonants is N\choose Y
  • The total number of ways of selecting Y consonants from N and X vowels from M is N\choose Y   M\choose X
  • The total number of ways of arranging (X+Y) letters among themselves = (X+Y)!
  • Hence, the required number of ways = (X+Y)! * N\choose Y   M\choose X


Below is the implementation of the above approach: 
 

C++

// CPP program to find the number of words
// of X vowels  and Y consonants can be
// formed from M vowels and N consonants
#include <bits/stdc++.h>
using namespace std;
 
// Function to returns factorial of n
int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Function to find nCr
int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
 
// Function to find the number of words
// of X vowels  and Y consonants can be
// formed from M vowels and N consonants
int NumberOfWays(int X, int Y, int M, int N)
{
    return fact(X + Y) * nCr(M, X) * nCr(N, Y);
}
 
// Driver code
int main()
{
    int X = 2, Y = 2, M = 3, N = 3;
 
    // Function call
    cout << NumberOfWays(X, Y, M, N);
 
    return 0;
}

                    

Java

// Java program to find the number of words
// of X vowels and Y consonants can be
// formed from M vowels and N consonants
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
     
    // Function to returns factorial of n
    static int fact(int n)
    {
        int res = 1;
        for (int i = 2; i <= n; i++)
            res = res * i;
        return res;
    }
     
    // Function to find nCr
    static int nCr(int n, int r)
    {
        return fact(n) / (fact(r) *
                          fact(n - r));
    }
     
    // Function to find the number of words
    // of X vowels and Y consonants can be
    // formed from M vowels and N consonants
    static int NumberOfWays(int X, int Y,
                            int M, int N)
    {
        return fact(X + Y) * nCr(M, X) *
                             nCr(N, Y);
    }
     
     
    // Driver code
    public static void main (String[] args)
                  throws java.lang.Exception
    {
        int X = 2, Y = 2, M = 3, N = 3;
     
        // Function call
        System.out.println(NumberOfWays(X, Y, M, N));        
    }
}
 
// This code is contributed by Nidhiva

                    

Python3

# Python 3 program to find the number of words
# of X vowels and Y consonants can be
# formed from M vowels and N consonants
 
# Function to returns factorial of n
def fact(n):
    res = 1
    for i in range(2, n + 1, 1):
        res = res * i
    return res
 
# Function to find nCr
def nCr(n, r):
    return fact(n) // (fact(r) * fact(n - r))
 
# Function to find the number of words
# of X vowels and Y consonants can be
# formed from M vowels and N consonants
def NumberOfWays(X, Y, M, N):
    return fact(X + Y) * nCr(M, X) * nCr(N, Y)
 
# Driver code
if __name__ == '__main__':
    X = 2
    Y = 2
    M = 3
    N = 3
 
    # Function call
    print(NumberOfWays(X, Y, M, N))
 
# This code is contributed by
# Surendra_Gangwar

                    

C#

// C# program to find the number of words
// of X vowels and Y consonants can be
// formed from M vowels and N consonants
using System;
 
class GFG
{
     
    // Function to returns factorial of n
    static int fact(int n)
    {
        int res = 1;
        for (int i = 2; i <= n; i++)
            res = res * i;
        return res;
    }
     
    // Function to find nCr
    static int nCr(int n, int r)
    {
        return fact(n) / (fact(r) *
                          fact(n - r));
    }
     
    // Function to find the number of words
    // of X vowels and Y consonants can be
    // formed from M vowels and N consonants
    static int NumberOfWays(int X, int Y,
                            int M, int N)
    {
        return fact(X + Y) * nCr(M, X) *
                             nCr(N, Y);
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int X = 2, Y = 2, M = 3, N = 3;
     
        // Function call
        Console.WriteLine(NumberOfWays(X, Y, M, N));        
    }
}
 
// This code is contributed by 29AjayKumar

                    

Javascript

<script>
      // JavaScript program to find the number of words
      // of X vowels and Y consonants can be
      // formed from M vowels and N consonants
 
      // Function to returns factorial of n
      function fact(n) {
        var res = 1;
        for (var i = 2; i <= n; i++)
            res = res * i;
        return res;
      }
 
      // Function to find nCr
      function nCr(n, r) {
        return fact(n) / (fact(r) * fact(n - r));
      }
 
      // Function to find the number of words
      // of X vowels and Y consonants can be
      // formed from M vowels and N consonants
      function NumberOfWays(X, Y, M, N) {
        return fact(X + Y) * nCr(M, X) * nCr(N, Y);
      }
 
      // Driver code
      var X = 2,
        Y = 2,
        M = 3,
        N = 3;
         
      // Function call
      document.write(NumberOfWays(X, Y, M, N));
       
      // This code is contributed by rdtank.
    </script>

                    

Output: 
216

 


Time Complexity: O(n), time to calculate factorial of n
Auxiliary Space: O(1), as no extra space is required



Last Updated : 21 Jun, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads