eclipse - Coding Challenge in Java: Given Letters and Returning What Rank They are in -

hey practice found coding challenge have been working on few days. have first part, can't seem figure out how continue am. here challenge:

    consider "word" sequence of capital letters a-z (not limited  "dictionary words"). word @ least 2 different letters,  there other words composed of same letters in different order (for  instance, stationarily/antiroyalist, happen both dictionary words;  our purposes "aaiilnorstty" "word" composed of same letters these two).      can assign number every word, based on falls in  alphabetically sorted list of words made of same set of letters. 1  way generate entire list of words , find  desired one, slow if word long.     write program takes word command line argument , prints  standard output number. not use method above of generating entire  list. program should able accept word 20 letters or less in  length (possibly letters repeated), , should use no more 1 gb  of memory , take no more 500 milliseconds run. answer check  fit in 64-bit integer.  sample words, rank: abab = 2  aaab = 1  baaa = 4  question = 24572  bookkeeper = 10743  nonintuitiveness = 8222334634      program judged on how fast runs , how code  written. running program reading source code,  can make process easier appreciated. 

so far, code have can return correct answer if of letters different. here code:

import java.util.arrays; import java.util.scanner; public class athenadility {      public static void main (string[] args) {         //finds word entered         scanner scan = new scanner (system.in);         string word = scan.next();         scan.close();          //added value         int value = 1;          //alphabetical representation         char[] charm = word.tochararray();         char[] alphacharm = word.tochararray();          arrays.sort(alphacharm);          //comparer         (int m = 0; m < word.length(); m++) {             (int c = 0; c < word.length()-1; c++) {                 system.out.println(charm[m] + " " + alphacharm[c]);                  //skips if alphacharm space                 if (alphacharm[c] == '-') {                 }                  //if same letter breaks , begins next                 else if (charm[m] == alphacharm[c]) {                     system.out.println("deleting: " + alphacharm[c]);                     alphacharm[c] = '-'; //delete letter used , cannot used compare @ later points                     break;                 }                  //if letter in alphacharm comes before charm                 else if (charm[m] > alphacharm[c]){                     system.out.println("found!");                      //factorial calculation                     int factorial = 1;                      //takes length of word minus current location , 1 after factorial                     (int f = word.length() - m - 1; f > 0; f--) {                         system.out.print(f + " ");                         factorial *= f;                     }                     //end loop                     //adding others                     system.out.println("\n" + "factorial: " + factorial);                     value += factorial;                 }                 else {                  }             }         }          //result         system.out.println("end: " + value);     }  }    

to try , explain can, creates 2 strings: 1 letters in alphabetical order , 1 original word. program compares each letter @ time , letter comes before 1 in original word alphabetically causes factorial calculation number of combinations exist before first word.

the part need factoring in if strings entered have more 1 of same letter. have literally spent days trying figure out. thank in advance!

p.s. code has lot of system.out.println testing sake