I believe there would be a total of 1,000 combinations possible, if you're counting 000-999. If you're only counting whole numbers 100 and up (numbers in the hundreds) I think there are 900.
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
99999
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
985,958,895,859,589,598. It should be 6. Hope I helped. Possible you want 4 digit numbers using the digits 9 8 and 5 and you want to know how many of those there are?
6 possible 3 digit combonations
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
290
99999
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
It is possible to create infinitely many numbers, of infinitely many different lengths, using the digits of the given number. Using each of the digits, and only once, there are 5! = 120 different permutations.
Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.
985,958,895,859,589,598. It should be 6. Hope I helped. Possible you want 4 digit numbers using the digits 9 8 and 5 and you want to know how many of those there are?
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
6 possible 3 digit combonations
There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.
There are 60480 numbers.
There are 4^6 = 4096 such numbers.