Find the count of M character words which have at least one character repeated

Given two integers N and M, the task is count the total words of M character length formed by the given N distinct characters such that the words have at least one character repeated more than once.

Examples:

Input: N = 3, M = 2
Output: 3
Suppose the characters are {‘a’, ‘b’, ‘c’}
All 2 length words that can be formed with these characters
are “aa”, “ab”, “ac”, “ba”, “bb”, “bc”, “ca”, “cb” and “cc”.
Out of these words only “aa”, “bb” and “cc” have
at least one character repeated more than once.

Input: N = 10, M = 5
Output: 69760

Approach:
Total number of M character words possible from N characters, total = NM.
Total number of M character words possible from N characters where no character repeats itself, noRepeat = NPM.
So, total words where at least a single character appear more than once is total – noRepeat i.e. NMNPM.

Below is the implementation of the above approach:

C++

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// C++ implementation for the above approach
#include <math.h> 
#include <iostream>
using namespace std;
  
// Function to return the
// factorial of a number
int fact(int n)
{
    if (n <= 1)
        return 1;
    return n * fact(n - 1);
}
  
// Function to return the value of nPr
int nPr(int n, int r)
{
    return fact(n) / fact(n - r);
}
  
// Function to return the total number of
// M length words which have at least a
// single character repeated more than once
int countWords(int N, int M)
{
    return pow(N, M) - nPr(N, M);
}
  
// Driver code
int main()
{
    int N = 10, M = 5;
    cout << (countWords(N, M));
    return 0;
}
  
// This code is contributed by jit_t

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Java

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// Java implementation of the approach
class GFG {
  
    // Function to return the
    // factorial of a number
    static int fact(int n)
    {
        if (n <= 1)
            return 1;
        return n * fact(n - 1);
    }
  
    // Function to return the value of nPr
    static int nPr(int n, int r)
    {
        return fact(n) / fact(n - r);
    }
  
    // Function to return the total number of
    // M length words which have at least a
    // single character repeated more than once
    static int countWords(int N, int M)
    {
        return (int)Math.pow(N, M) - nPr(N, M);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int N = 10, M = 5;
        System.out.print(countWords(N, M));
    }
}

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Python3

# Python3 implementation for the above approach

# Function to return the
# factorial of a number
def fact(n):

if (n <= 1): return 1; return n * fact(n - 1); # Function to return the value of nPr def nPr(n, r): return fact(n) // fact(n - r); # Function to return the total number of # M length words which have at least a # single character repeated more than once def countWords(N, M): return pow(N, M) - nPr(N, M); # Driver code N = 10; M = 5; print(countWords(N, M)); # This code is contributed by Code_Mech [tabby title="C#"]

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// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to return the
    // factorial of a number
    static int fact(int n)
    {
        if (n <= 1)
            return 1;
        return n * fact(n - 1);
    }
  
    // Function to return the value of nPr
    static int nPr(int n, int r)
    {
        return fact(n) / fact(n - r);
    }
  
    // Function to return the total number of
    // M length words which have at least a
    // single character repeated more than once
    static int countWords(int N, int M)
    {
        return (int)Math.Pow(N, M) - nPr(N, M);
    }
  
    // Driver code
    static public void Main ()
    {
        int N = 10, M = 5;
        Console.Write(countWords(N, M));
    }
}
  
// This code is contributed by ajit.

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Output:

69760


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Improved By : jit_t, Code_Mech