Calculate possible combinations where the total sum equals 30

Update here is my potential idea on how to solve this..... feel free to shoot holes in the idea

find out what the lowest number is that can be made (regardless of the order of numbers e.g 224688 then find the highest number 222222468

then step up each number from 224688 e.g 224688,224689,224690

then using boolean statements i.e does the number contain a 9 if yes then get rid... if not it... does it contain an 8 e.t.c

it would also have to count to make sure at least ONE 2,4,6,8 is contained in the number too as long as the criteria is met is puts these into a column.

next problem is : 224688 technically is the same combination as 886422 if we're ignoring order of numbers.... so would then have to eliminate the duplicates somehow.....

this was my only way of thinking how to solve this issue.... potentially

I think I would start with the case where fifteen 2's make 30. From this permutation, I can substitute one 4 for every two 2's (if that makes sense to you). From the same fifteen 2's I can substitute one 6 for every three 2's, and then substitute one 8 for every four 2's. So the iterations look like:
15 2's
13 2's 1 4's
11 2's 2 4's
...
12 2's 1 6's
9 2's 2 6's
...
11 2's 1 8's
7 2's 2 8's
...
and on and on

then i realised that the numbers 2,4,6 and 8 have to be used at least once...

I obviously wasn't aware of that requirement, though I'm not sure how different the basic algorithm would be.

Start with the same fifteen 2's
Take out four for the required 8
Take out three for the required 6
Take out two for the required 4
Take out one for the required 2

Now you have one of each value totaling 20
There are five 2's left over totaling 10.
Now you are back to the same problem you started with, only what combinations add up to 10 instead of 30.