How many combinations can be made out from 235789 from 30000 to 90000?

Answer 1

Set = {2, 3, 5, 7, 8, 9}
n = 6 (there are 6 different digits to choose from)

k = 5 (the length of the number if 5 digits long)

Sometimes making "cases" makes this easier to calculate:

The first digit (k=1) can begin with a 3, 5, 7, or an 8

* it can't be a 9, is it would be greater than 90,000 if it started with a 9

The second digit (k=2) can be any of the 6
The third (k=3) can be any of the 6
The fourth (k=4) can be any of the 6
The fifth (k=5) can be any of the 6

Assuming you are allowed to repeat a digit (repetitive) the answer is:
(4)(6)(6)(6)(6) = 4(6)4= 5184 possibilities

However, if you are not allowed to repeat a digit (non-repetitive), you need to subtract one digit each time you use a digit and you have to consider the first three digits used:

If you start with a 3 (we start with 5 possibilities, because we removed a 3):
(1)(5)(5-1)(5-2)(5-3)=3(5)(4)(3)(2) = 3(5!) = 360 possibilities

If you start with a 5 (exactly the same as above):
(Note: I'm showing you what could be in each digit placeholder, not the math)

[5] [n=5] [n=4] [n=3] [n=2] => 360 possibilities

If you start with an 7, again the same: 360 possibilities

If you start with an 8, again the same: 360 possibilities

So, for non-repetitive, the answer is 360 + 360 + 360 + 360 = 1440 possibilities

To repeat the answers:

Repetitive:5184possibilities
Non-repetitive: 1440 possibilities

(and using logic, it makes since that non-repetitive would be smaller than repetitive)