Q: What are the possible 3 digits codes using the numbers 0-9?

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The possible 4 digit codes using the numbers 0-9 are every number between 0 and 9999. For numbers that have less than 4 digits, just precede the number with 0's. 10,000 possibilities

1000

There are 10,000 possible. I'll list a few then you continue. 0000, 0001, 0002, 0003 ... 1001, 1004, ...1967 etc. to 9999

There are 10000 such codes. Each of the numbers 0-9 can be in the first position. With each such first digit, each of the numbers 0-9 can be in the second position. With each such pair of the first two digits, each of the numbers 0-9 can be in the third position. etc.

10,000

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

290

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.

99999

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.

2 * 2 * 2 * 2 It's not possible to make 16 without using the same numbers twice or without using whole numbers.

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.

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.

0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.

6 possible 3 digit combonations

Assuming you have to use all the digits and use each one exactly once: 135679.

if repeating is allowed... 36 (6x6, for the last two digits) If not, 6 (3x2, last two 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.

There are 60480 numbers.

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

Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.

depends on your answer

128

There are 3,265,920 numbers that can be formed using all the digits. The answer to the question will depend on which one of these.

There are 4^6 = 4096 such numbers.

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