How many ways can 12 different books be distributed among four children so that(a) each child gets 3 books? (b) the two oldest gets four books each and the two youngest get two books each?

Question

anonymous · Answer

Both parts involve combination 
(a)There are four children and each
can have 3 books each so, I think I need
to find out the number of ways that the first
-last child can have 3 books. So I ended up using 
the "rule of product" to get
C(12,3) x C(9,3) x C(6,3) x C(3,3)   

(b) I think for this part, I need to use 
the  "rule of Product" since I am trying to 
find out how many ways I can distribute 4 of the 12 books
amongst the 2 oldest children (stage 1), and 
how many ways I can distribute 2 of the 12 books amongst 
the 2 youngest children(stage 2).

So using the product rule I get:
C(12,4) x C(12,2)  

^ Question: Is the problem C(12,4) x C(12,2) instead
of C(12,4) x C(8,2) because it doesn't matter

anonymous · Answer

i did this on paper just now, A would be A. 12c3 and for part B i also did it and it came out to be B. 4c2 + 2c2 .  Its super lengthy and truly you wouldnt understand it if I tried to show you, so maybe someone may properly show you in a more "Simple" description.

anonymous · Answer

you good?

anonymous · Answer

Is (a) just C(12,3) because we just want to find how much each child can get 3 books, there are no restrictions or additional parts like in part(b). 
For part (b) the sum rule is used because you can either give the two oldest 4 books or the youngest 2 books, but not the other way around. And I guess its C(4,2) and C(2,2) instead of what I did because its restricted to 4 books for the 2 oldest children and 2 books for the 2 youngest children.