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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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15y ago

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Anonymous

4y ago
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Q: 3 digit codes with 0 - 9?
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