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0-9 = 10 numbers

every full set of 10 numbers presents a multiple of 10 variations.

consider if it was a 1-digit code, using 0-9.

the code could be:
0
1
2
3
4
5
6
7
8
or
9

that's 10 options.

now if it were a 2-digit code, your 10 options are permutated 10 more times (10^2).
in practise that looks like
00
01
02
03...
08
09
10
11
12
13...
18
19
20
21
and so on.
you're going from 00 to 99, so, all told, it's 100 permutations.
written simply, that's: 10*10=100 or: 10^2=100

now, if it's a 3-digit code, well, that's a whole notherpermutation of 10 for each of those previous 100 combinations.

it looks like:
000
001
002
003...
(all the way up to)...
996
997
998
999

that's 000 to 999, which is 1000 unique permutations.
written simply, that's: 10*10*10=1000 or better yet: 10^3=1000

the logic continues for a 4-digit code, for which there would be 10,000 permutations. (10^4=10,000)

if you're working with only a set of 9 numbers (for example 1-9 instead of 0-9), then you would simply apply exponents to 9 instead of 10. so a 3-digit code with numbers 1-9 has 729 possibile permutations (9^3 = 9*9*9 = 729)

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Q: 3 digit codes with 0 - 9?
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Related questions

How many four digit codes can be formed using the digits 0123456789 if the last digit cannot be 0 or 1 or 2 and repetition of digits is allowed?

If a leading zero is allowed... The first digit can be 0-9, ten possibilities. The second digit can be 0-9, ten possibilities. The third digit can be 0-9, ten possibilities. The fourth digit can be 3-9, 7 possibilities. So there are 10x10x10x7 = 7000 such codes.


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

1000


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

10,000


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

103 = 1000 in all.


What are the most frequent 3 digit numbers between 0 to 9?

There are no 3 digit numbers between 0 and 9 because 0 and 9 are 1 digit numbers.


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

900 ,999 ,909


How many 3 digit codes are possible with out repeating a number using 0-9 only?

10 x 9 x 8 = 720 different "permutations"


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 possible 4 digit combinations are there using 0-9. how many different combination codes are possible?

There are 210 4 digit combinations and 5040 different 4 digit codes.


What digit is divisible by 3?

0, 3, 6 or 9.


How many different 3-digit codes can be created if no digit can be repeated?

Any 3 from 10 in any order = 10 x 9 x 8 = 720It depends on what other restrictions you place on the numbers. For example, in North America (country code +1 = USA, Canada, etc.), area codes cannot have 0 or 1 as the first digit, and cannot have 9 as the second digit.


How many possible 4 digit 0 thru 9 codes?

If the first digit can be zero, there are 5,040. If the first digit can't be zero, there are only 4,536.