Strategies for solving permutations of a word So I'm trying to prepare for exams, and am having some trouble with permutations, and was wondering what's a good strategy to solve this task is:

Given the set of letters $\text{AAABBBBCCDEEFG}$ find:
$(*)$ The number of unique permutations of these letters.
$(**)$ The number of unique permutations where the vowels ${\color{red}{AAA}}$ appears before ${\color{green}{EE}}$. (E.g. ${\color{red}{AA}}\text{C}{\color{red}{A}}\text{C}{\color{green}{E}}\text{BBG}{\color{green}{E}}\text{FDBB}$, but not $\text{B}{\color{red}{AA}}\text{C}{\color{green}{E}}\text{C}{\color{red}{A}}\text{BBG}{\color{green}{E}}\text{FDB}$)
$(*\!*\!*)$ The number of unique permutations where ${\color{red}{D}}{\color{green}{F}}{\color{blue}{G}}$ appears in that order. (E.g. $\text{AACACEBB}{\color{red}{D}\text{E}{\color{green}{F}}\text{BB}{\color{blue}{G}}}$ but not $\text{AACACEBB}{\color{blue}{G}}\text{E}{\color{green}{F}}{\color{red}{D}}\text{BB}$)
$(*\!*\!**)$ The number of permutations where no vowels appear in a pair. (E.g. $\text{B}{\color{magenta}{A}}\text{C}{\color{magenta}{E}}\text{C}{\color{magenta}{A}}\text{BBG}{\color{magenta}{E}}\text{FDB}{\color{magenta}{A}}$, but not $\text{B}{\color{orange}{AA}}\text{C}{\color{magenta}{E}}\text{C}{\color{magenta}{A}}\text{BBG}{\color{magenta}{E}}\text{FDB}$)

So $(*)$ is $\frac{14!}{3!4!2!1!2!1!1!}$, but I'm not sure how I should proceed with the others...
Thanks in advance!
 A: For ** and ***, you can think of what fraction of the answer to *  are acceptable.  For the second, think of how many ways you can permute the A's and E's in place.  Only one of these permutations is acceptable.  For the fourth one, it is not clear to me if you allow AE together or not.  If not, order the consonants, order the vowels separately, then choose five of the slots before, between, or after the consonants to put the vowels into.  The last is a stars and bars argument.  Multiply the numbers.
A: For the first question just find how many letters there are of each type. You shall get a fraction where the top is the total quantity of letters factorial and the bottom is the product of the factorial of each independent letter:
$\dfrac{14!}{(3!)(4!)(2!)(1!)(2!)(1!)(1!)}$ in this case

for the second part fix the letters $AAAEE$ and look at them as separators. Then figure out there are $\dfrac{9!}{(4!)(2!)(1!)(1!)(1!)}$ ways to fix the remaining letters. And there are $\binom{9+5}{5}$ ways to place the letters of the new word between the seperators by stars and bars.

For the third part do the same as part 2. That is fixing $DFG$ and looking at those letters as separators.

For the fourth part there are $\dfrac{10}{4!2!1!1!1!}$ ways to put the consonants into a word.Notice the consonants leave $11$ slots to put the vowels in. There are $12\cdot 11 \cdot 10 \cdot9\cdot 8$ ways to put the vowels into the slots, except we must divide by $3!\cdot2!$ since all the $A's$ and the $e's$ are identical.
