java - Method that walks through combinations and counting those who passes condition -

i should find every 5-number combination of values 0 1 2 3 4 5 0 not first , 1 digit not repeated more 1 time.the method shall return number of valid combinations.

valid combinations:

1 0 0 2 1 1 2 3 4 5 1 1 2 2 3 

invalid combinations:

0 1 2 3 4 (0 cant first) 2 3 3 3 5 (not allowed 3 of same digits) 0 1 2 3 4 5 (6 digit numbers not allowed) 

i did similar task permutations, i'm not quite sure how approach one. right answer 5100 (i solved mathematically).

you need write isvalid function , try following pseudo-code

nb_solution = 0 = 0 ; < 100000; i++     if isvalid(i)         nb_solution ++ 

edit : did, , find same value found.

public static void computeexample() {     int count = 0;     for(int = 0; < 100000; i++) {         if(isvalid(i)) {             count++;         }     }     system.out.println(count); }  public static int[] splitciffers(int number) {     int tmp = number;     int unite = tmp % 10;     tmp = (tmp - unite) / 10;     int dizaine = tmp % 10;     tmp = (tmp - dizaine) / 10;     int centaine = tmp % 10;     tmp = (tmp - centaine) / 10;     int millier = tmp % 10;     tmp = (tmp - millier) / 10;     int dix_millier = tmp % 10;      int[] retour = new int [5];     retour[0] = dix_millier;     retour[1] = millier;     retour[2] = centaine;     retour[3] = dizaine;     retour[4] = unite;      return retour; }  public static boolean isvalid(int number) {     boolean isvalid = true;     int[] digitbydigit = splitciffers(number);     int[] digitallowed = new int[6];     for(int = 0; < 6; i++) {         digitallowed[i] = 0;     }      if(digitbydigit[0] == 0) {         isvalid = false;     }      for(int = 0; < 5; i++) {         if(digitbydigit[i] < 0 || digitbydigit[i] > 5) {             isvalid = false;         } else {             digitallowed[digitbydigit[i]]++;         }     }     for(int = 0; < 6; i++) {         if(digitallowed[i] > 2) {             isvalid = false;         }     }     return isvalid; }