Should I divide this permutation problem into cases or are there any quicker methods? I have got an idea for the second question but I think my approach is too long and I would like to ask whether there are any other quicker methods?
Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
(a) How many different sequences of eight cards are possible?
(b) How many of the sequences in part (a) will contain three picture cards, three odd-numbered cards and two even-numbered cards?
My solutions:
(a) $52^8$
(b) Divide into cases of:
when none of the picture cards are together, when two of the picture cards are together, when all of the picture cards are together;
similarly for the odd and even numbered cards.
I am not sure whether my working is correct but pretty sure there should be a faster way for part (b). Just for reference, the solution is $3.907\times 10^{12}$.
Many thanks for all the helps!

There are $\binom{8}{3}$  ways to choose the places in the sequence of $8$ cards that the picture cards will occupy. For every such choice, the places can be filled in $12^3$ ways. So this part of the job can be done in $\binom{8}{3}\cdot 12^3$ ways. 
For every way of dealing with the picture cards, there are $\binom{5}{3}$ ways to decide where the odd cards will go. These places can be filled in $20^3$ ways. 
For every way of getting this far, there are $\binom{2}{2}$ ways to decide where the even cards will go (of course this is $1$). And there are $20^2$ ways to fill these spaces. 
