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
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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.