Probability of balls in the box. 
How to do i.) and v.) ?
and please my answer ii.), iii.) iv.) is correct or not,
if not explain me where i wrong.

Thank you.
 A: Hint on (i)
There are $30$ balls so there are $30!$ arrangements... 
Eh.., no then every particular arrangement is counted more than once. That must be repaired (how?). How many times is a particular arrangement counted? The arrangement does not change if red balls are interchanged with red balls, blue balls with blue balls, et cetera.
Hint on (v)
The probability that the first ball chosen is not red is $\frac{18}{30}$. If this has been done then you are choosing the second ball. This time the probability of a ball that is not red is $\frac{17}{29}$ (why?). Now it is time for the third ball. Where does this lead to? 
A: i.) There is 30 balls but 12 red, 8 blue, 6 white and 4 yellow. Therefore, you can arrange them in $$\frac{30!}{12!8!6!4!}$$ ways (because if you permute two yellow or two white for example, it doesn't change the arrangement).
v.) You don't get red balls if you take no red balls and 3 of the other 18 balls, therefore the probability is $$\frac{\binom{12}{0}\binom{18}{3}}{\binom{30}{3}}.$$
A: I would say (ii) is a bit ambiguous, because $12!$ is just the number of ways to arrange 12 different balls in a row, it doesn't account for their positions in the box: it's unclear if you need it or not. If you do, then first you need to select 12 positions out of 30 for 12 red balls and only then account for the fact that each arrangement is different, so you need
$$
\binom{30}{12} \cdot 12!
$$
