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 = 2Output: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 = 5Output:69760

**Approach:**

Total number of M character words possible from N characters, **total = N ^{M}**.

Total number of M character words possible from N characters where no character repeats itself,

**noRepeat =**.

^{N}P_{M}So, total words where at least a single character appear more than once is

**total – noRepeat**i.e.

**N**.

^{M}–^{N}P_{M}Below is the implementation of the above approach:

## C++

`// 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` |

## 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.` |

## Javascript

`// javascript implementation of the approach` ` ` ` ` `// Function to return the` ` ` `// factorial of a number` ` ` ` ` `function` `fact(n)` ` ` `{` ` ` `if` `(n <= 1)` ` ` `return` `1;` ` ` `return` `n * fact(n - 1);` ` ` `}` ` ` ` ` `// Function to return the value of nPr` ` ` ` ` `function` `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` ` ` ` ` `function` `countWords( N, M)` ` ` `{` ` ` `return` `Math.pow(N, M) - nPr(N, M);` ` ` `}` ` ` ` ` `// Driver code` ` ` `var` `N = 10 ;` ` ` `var` `M = 5;` ` ` `document.write(countWords(N, M));` ` ` `// This code is contributed by bunnyram19.` |

**Output:**

69760

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **Geeks Classes Live**