== I suggest starting with a pen and a piece of paper. == Any number which is above 9 isn't a digit (in denary)
None of the numbers from 1 to 45 are 7 digits long
3.918208205 X 10^11 I think but I'm stupid so probably wrong
it is 26
There are twelve possible solutions using the rule you stated.
There are actually 8,998 of them . . . all of the counting numbers from 1,000 to 9,999 . The list is too large to present here, but if you can count, then you'll have no trouble generating it on your own.
It can be calculated as factorial 44! = 4x3x2x1= 60
3.918208205 X 10^11 I think but I'm stupid so probably wrong
There are 5,040 combinations.
There are 210 4 digit combinations and 5040 different 4 digit codes.
it is 26
As the number has to start with 15, we have only 3 remaining digits to work with. There are 3 possible options for the first digit. Then out of each of these, 2 possible options for the second digit, and one option for the last. This means that in total there are 3x2x1 (6) possible combinations. These are: 15234 15243 15324 15342 15423 15432
You don't mean "3 possible digit combinations"; you mean "3-digit possible combinations"and you also forgot to specify that the first digit can't be zero.(We wouldn't have known that, but two of your buddies asked the same questionabout 7 hours before you did.)The question is describing all of the counting numbers from 100 to 999.That's all of the counting numbers up to 999, except for the first 99.So there are 900 of them.
There are twelve possible solutions using the rule you stated.
There are 9999 possible combinations starting from 0000 to 9999
if its not alphanumeric, 999999 variations
There are actually 8,998 of them . . . all of the counting numbers from 1,000 to 9,999 . The list is too large to present here, but if you can count, then you'll have no trouble generating it on your own.
It can be calculated as factorial 44! = 4x3x2x1= 60
36*36*36*36*36*36