Count the anagrams from ARARAQUARA, in such a way that there are not 2 As adjacent? 
How many anagrams can be made from the word ARARAQUARA, in such a way that  there are not 2 As adjacent?

My proposed solution: There are 10 letters, 5 As, 3 Rs,1 Q and 1 U. The general approach will be positioning the 5 As with spaces separating them (or before or after them) and counting the ways to fill these spaces with one or two letters.
I divided the problem in 2 cases:
(a) there is only 1 letter different from A between A's. There are two cases here, starting with non-A, or finishing with non-A. Total number of possibilities for the 2 cases, considering the 3 repeated Rs:
$$2\times \frac{5!}{3!}=40.$$
(b) there is one pair of non-A letters between 2 A's (only one pair is possible). Cases for the pairs (considering order): RR, RQ, QR, RU,UR, QU, UQ.
So the anagram possibilities for each pair is given by:
RR - $4!=24$ possibilities;
RQ - $4!/2!=12$ possibilities;
QR - $4!/2!=12$ possibilities;
UR - $4!/2!=12$ possibilities;
RU - $4!/2!=12$ possibilities;
UQ - $4!/3!=4$ possibilities; and,
QU - $4!/3!=4$ possibilities.
So the total number of anagrams considering restrictions is: 40+24+412+24=120.
My problem: the answer given is 128. Am I missing cases? or the answer is wrong?
 A: You can arrange the other letters $RRRQU$ in $\frac{5!}{3!}=20$ ways, and then you can insert $AAAAAX$ (where $X$ means no $A$ is inserted) into the six spaces before between and after the these non-$A$s in 6 ways.
So the total is $120$ as you suspect
A: Your answer is correct.  We will confirm your answer with a different approach.
We first arrange the five letters R, R, R, Q, U, then insert the five As in the six spaces this creates so that no two As are adjacent.
Arrangements of R, R, R, Q, U:  We have five positions to fill.  There are
$$\binom{5}{3}2!$$
arrangements of the letters R, R, R, Q, U since we must select three of the five positions for the three Rs, then arrange the distinct letters Q and U in the remaining two positions.
An arrangement of the letters R, R, R, Q, U creates six spaces in which we can place the As, four between successive letters and two at the ends of the row.
$$\square L \square L \square L \square L \square L \square$$
To ensure that no two of the As are adjacent, we choose five of these spaces in which to place an A, which can be done in
$$\binom{6}{5}$$
ways.
Hence, the number of anagrams of ARARAQUARA in which no two As are adjacent is
$$\binom{5}{3}2!\binom{6}{5} = 120$$
as you found.
