S1 Perms / Comb question(s)

Hey guy's, got my S1 exam on fridays and was wondering if anyone can help me out on these questions.

1: 8 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 8 cards are possible?
b: how many of the sequences in part (a) will contain 3 picture cards, 3 odd numbered cards and 2 even numbered cards?

After finding the total possible number of sequences as 52^8, I have no idea how to do part B.

2. The letters of the word POSSESSES are written on 9 cards, one on each card. The cards are shuffled and our of them are selected and arranged in a straight line.

a: How many possible selections are there of 4 letters?
b: how many arrangements are there of 4 letters?

I got part a as 12, but I don't know if I did it correctly. I did 9! / 5!2! which gave me 12.6 but that's a decimal so I think I've made a mistake

No idea how to do part b.

Some help would be much appreciated

Thanks.

1)b) This may be wrong but is it $\ 3.097\times 10^{12}$?

yes that's perfectly correct.

I got the answers in the back of my book, so could you please explain to me how you got to it? Thanks a lot.

Would you mind explaining the reasoning between each part? I only partly understand what you've done.

Using the permutations of distinct objects: n objects p, q , r etc repeated elements $\dfrac{n!}{p!q!r! . . .}$
So there are $\dfrac{8!}{2!3!3!}$ ways that the $20^3 \times 20^2 \times 12^3$ can be arranged - i.e. come out in.
So times them together and hey presto we get ourselves the right answer.

Thanks a lot.

No worries
For your other q I can't see a better way of doing it than by considering the cases that arise to the number of s's

The answer is 115 apparently, but I have no idea how they got there.