The number of words which can be formed with 2 different consonants

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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 

i.e 3 
The total number of ways of choosing 2 consonants from a total number of 3 consonants is 

i.e 3. 
The total number of ways of selecting 2 consonants from 3 and 2 vowels from 3 is 

* 

= 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 : 
 

Below is the implementation of the above approach: 
 

C++

#include <bits/stdc++.h>

using namespace std;

int fact(int n)

{

    int res = 1;

    for (int i = 2; i <= n; i++)

        res = res * i;

    return res;

}

int nCr(int n, int r)

{

    return fact(n) / (fact(r) * fact(n - r));

}

int NumberOfWays(int X, int Y, int M, int N)

{

    return fact(X + Y) * nCr(M, X) * nCr(N, Y);

}

int main()

{

    int X = 2, Y = 2, M = 3, N = 3;

    cout << NumberOfWays(X, Y, M, N);

    return 0;

}

Java

import java.util.*;

import java.lang.*;

import java.io.*;

class GFG

{

    static int fact(int n)

    {

        int res = 1;

        for (int i = 2; i <= n; i++)

            res = res * i;

        return res;

    }

    static int nCr(int n, int r)

    {

        return fact(n) / (fact(r) *

                          fact(n - r));

    }

    static int NumberOfWays(int X, int Y,

                            int M, int N)

    {

        return fact(X + Y) * nCr(M, X) *

                             nCr(N, Y);

    }

    public static void main (String[] args)

                  throws java.lang.Exception

    {

        int X = 2, Y = 2, M = 3, N = 3;

        System.out.println(NumberOfWays(X, Y, M, N));        

    }

}

Python3

def fact(n):

    res = 1

    for i in range(2, n + 1, 1):

        res = res * i

    return res

def nCr(n, r):

    return fact(n) // (fact(r) * fact(n - r))

def NumberOfWays(X, Y, M, N):

    return fact(X + Y) * nCr(M, X) * nCr(N, Y)

if __name__ == '__main__':

    X = 2

    Y = 2

    M = 3

    N = 3

    print(NumberOfWays(X, Y, M, N))

C#

using System;

class GFG

{

    static int fact(int n)

    {

        int res = 1;

        for (int i = 2; i <= n; i++)

            res = res * i;

        return res;

    }

    static int nCr(int n, int r)

    {

        return fact(n) / (fact(r) *

                          fact(n - r));

    }

    static int NumberOfWays(int X, int Y,

                            int M, int N)

    {

        return fact(X + Y) * nCr(M, X) *

                             nCr(N, Y);

    }

    public static void Main (String[] args)

    {

        int X = 2, Y = 2, M = 3, N = 3;

        Console.WriteLine(NumberOfWays(X, Y, M, N));        

    }

}

Javascript

<script>

      function fact(n) {

        var res = 1;

        for (var i = 2; i <= n; i++)

            res = res * i;

        return res;

      }

      function nCr(n, r) {

        return fact(n) / (fact(r) * fact(n - r));

      }

      function NumberOfWays(X, Y, M, N) {

        return fact(X + Y) * nCr(M, X) * nCr(N, Y);

      }

      var X = 2,

        Y = 2,

        M = 3,

        N = 3;

      document.write(NumberOfWays(X, Y, M, N));

    </script>

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