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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.

Q: How many numbers are possible using 3 digits?

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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?

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.

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Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77

290

99999

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.

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

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.