Define a 'word' to be any distinct rearrangement of the letters.
(a) How many words can be created from AAAABBBCCDE (4 A's, 3 B's, 2 C's, 1 D, 1 E)?
(b) How many words contains no B's next to each other?
(c) How many words contain the D before the E?
The only way I can think of is to start counting the number of possible choices from a 1 letter word to a 11 letter word, which seems horrifying.
For part a the answer simply goes as $\frac {11!}{4!3!2!}$
For the second part, first try arranging all the A's, C's, D and E in $\frac {8!}{4!2!}$ and then select any three gaps ( for the B's to be placed) between any two letters of the formed arrangement( note that you can also select the word starting and ending with B itself). Hence the answer to second part becomes $\frac {8!}{4!2!}* \binom {9}{3}$.
For the third part the probability that D comes before E is simply $\frac {1}{2}$ of the total arrangements. Hence answer to third becomes $\frac {11!}{4!3!2!*2}$