If it could be 3333, then your answer would be four to the fourth power, which is 256.
Can prove that as: The first number has 4 possibilities, and the second has 4 possibilities, so if each of the possibilities in the first number has four possibilities for the second number, the number of possibilities of the two is 16, and that can continue if the third number as four possibilities, so you can have 64 combination of those three numbers, and of those 64 numbers each of them could have 4 different numbers after them, so you have 256 combinations.
If it could not repeat digits it would be 4x3x2x1 which is 24.
can prove that as:
1.3456
2.3465
3.3546
4.3564
5.3645
6.3654
7.4356
8.4365
9.4536
10.4563
11.3635
12.4653
13.5346
14.5364
15.5436
16.5463
17.5643
18.5634
19.6345
20.6354
21.6435
22.6453
23.6534
24.6543
3.918208205 X 10^11 I think but I'm stupid so probably wrong
it is 26
== 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
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.
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
== 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 9999 possible combinations starting from 0000 to 9999
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