My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       
#include <iostream>
       
 
       
// number of ways to choose n elements from k available
       
// code taken from problem 116

       
unsigned long long choose(unsigned long long n, unsigned long long k)
       
{
       
  // n! / (n-k)!k!
       
  unsigned long long result = 1;
       
  // reduce overflow by dividing as soon as possible to keep numbers small
       
  for (unsigned long long invK = 1; invK <= k; invK++)
       
  {
       
    result *= n;
       
    result /= invK;
       
    n--;
       
  }
       
  return result;
       
}
       
 
       
// count number words with n characters

       
unsigned long long count(unsigned int n, unsigned int alphabet)
       
{
       
  // invalid parameters: must not use each letter of the alphabet more than once
       
  if (n > alphabet)
       
    return 0;
       
 
       
  // count how many word with n characters use the characters 1..n
       
  unsigned long long result = 0;
       
  // there are n places where the "break" between s1 and s2 can occur
       
  // count all possible characters chosen for s1 and s2
       
  for (unsigned int i = 1; i < n; i++)
       
    result += choose(n, i) - 1; // minus 1 because there is always one combination with no break between s1 and s2
       
 
       
  // general case: use characters 1..Alphabet instead of 1..n
       
  return result * choose(alphabet, n);
       
}
       
 
       
int main()
       
{
       
  // bonus feature: user-defined alphabet size and maximum word length
       
  unsigned int alphabet = 26;
       
  unsigned int size     =  3;
       
  std::cin >> alphabet >> size;
       
 
       
  // all "words" with 2..size characters
       
  unsigned long long best = 0;
       
  for (unsigned int i = 2; i <= size; i++)
       
  {
       
    unsigned long long current = count(i, alphabet);
       
    // more than before ?
       
    if (best < current)
       
      best = current;
       
  }
       
 
       
  std::cout << best << std::endl;
       
  return 0;
       
}

This solution contains 7 empty lines, 13 comments and 1 preprocessor command.