查找至少重复一个字符的M个字符单词的数量

给定两个整数N和M ，任务是对由给定的N个不同字符形成的M个字符长度的单词总数进行计数，以使单词中至少一个字符重复一次以上。

例子：

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

方法：
N个字符可能存在的M个字符词的总数，总数= N ^M。
N个字符(不重复任何字符可能存在的M个字符单词的总数， noRepeat = ^N P _M。
因此，至少一个字符出现多次的单词总数为– noRepeat，即N ^M – ^N P _M。

下面是上述方法的实现：

C++

// C++ implementation for the above approach
#include  
#include 
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

Java

// 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));
    }
}

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

C#

// 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.

输出：

69760