` Quick Answer `
10,000
` Long Answer `
The numeric values 0-9 provide 10 total possibilities.
To expand this concept and understand how to arrive at the answer to this question in different situations read on...
If we assume a one digit combination, the possibilities as you already know are:
"0 1 2 3 4 5 6 7 8 9" for a total of 10.
If we assume a two digit combination the possibilities are 100, and here's why. Each of the possibilities in the above example with one digit, combine now with another ten possibilities for each. This is easier to understand visually:
0 1 3
______/\_______ ______/\_______ ______/\_______
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
While I only typed the first 3, this pattern is replicated for all the possibilities, in our case, all the numbers 0-9.
This pattern can be expressed verbally by, "total possibilities" to the power of "combination length" or in the formula:
X^n.
In this case the formula is 10^2. 10 because 0-9 is 10 possibilities and 2 because the length of our combination is 2.
Now let's assume that we have a 3 digit combination using numbers.
Formula:
10^3 = 1000 possibilities.
There are 210 4 digit combinations and 5040 different 4 digit codes.
36*36*36*36*36*36
im assuming that any charcter can be a number or a letter: (24letters*10 possible numbers)^(4 digits)= 3317760000 possible combinations.
i would like a list all possible 4 digit combination using 0-9
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
3.918208205 X 10^11 I think but I'm stupid so probably wrong
There are 5,040 combinations.
it is 26
There are 210 4 digit combinations and 5040 different 4 digit codes.
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
== 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
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 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.
if its not alphanumeric, 999999 variations
It can be calculated as factorial 44! = 4x3x2x1= 60
36*36*36*36*36*36