How many ways can $2$ a's, $2$ b's and $8$ c's be arranged so that there is a c on both sides of each a and b? How many ways can 2 a's, 2 b's and 8 c's be arranged so that there is a c on both sides of each a and b?
I'm really unsure of how to even begin tackling this. I would say treat each 'cac' and 'cbc' as their own object and permute, but that's not taking into account that the "before" c of a 'cac' combo could be the "after" c of a 'cbc' combo, aka "cbcac". How would I procede?
 A: To ensure that there is a $c$ on both sides of each $a$ and each $b$, first place the eight $c$'s in a row.  This leaves seven gaps between successive $c$'s.
$$c \square c \square c \square c \square c \square c \square c \square c$$
Select two of these seven gaps in which to place the $a$'s, then choose two of the five remaining gaps in which to place the $b$'s.  This can be done in 
$$\binom{7}{2}\binom{5}{2}$$ 
ways.   
A: Use the stars and bars method here.
If you remove the a's and b's from the arrangement you just have 8 consecutive c's. By your restrictions, there cannot be an a or b to the left or to the right of all the c's, and the a's and b's cannot be next to each other.
Therefore we consider the 7 places between consecutive c's. We must place the 2 a's and the 2 b's in these 7 places, with each place taking at most one a or b. So choose 2 places for the a's: there are ${7 \choose 2}$ possibilities. Then choose 2 places for the b's from the remaining places: there are ${5 \choose 2}$ possibilities. Therefore the total number of arrangements is
$${7 \choose 2}{5 \choose 2}=21\cdot 10=210$$
