GeeksforGeeks App
Open App
Browser
Continue

# 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:
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++

 `// 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.`

## 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`

Time Complexity: O(n)
Auxiliary Space: O(N), for recursive stack space.

My Personal Notes arrow_drop_up