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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

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Wiki User

15y ago
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Wiki User

10y ago

There are 10,000 possible. I'll list a few then you continue.

0000, 0001, 0002, 0003 ... 1001, 1004, ...1967 etc. to 9999

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Anonymous

Lvl 1
3y ago

90

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Q: What are all the possible 4 digits codes using the numbers 0 9?
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Related questions

How many 3-digit codes using the digits 0-9 are possible if repetition are allowed?

1000


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

Counting from 000 to 999 would give a possible 1000 combinations.


How many 3-digit codes using digits 0 through 9 are possible if repetitions are allowed?

10,000


How many 2-digit numbers can be formed using the digits 1 3 7 if repetition of digits is allowed?

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


What are all the possible 5 digit codes using the numbers 0-9?

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.


How many 3 digit codes using the digits 0 9 are possible if repetitions are allowed?

900 ,999 ,909


How many 3-digit numbers using the digits 0-7 are possible if repetition are allowed?

290


How many possible 5 digit combinations are there using 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.


How many 13 digit numbers are possible by using the digits 12345 which are divisible by 4 if repetition of digits is not allowed?

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


How many different 3-digit numbers less than 500 can be made using the digits 345 and 6 if the digits can be only used only once?

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.


How do you make 16 with 4 digits?

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


How many 5 digits numbers can be formed using digits?

99999