You were headed in the right direction, but you went way too far. The $C(35,7)$ is much too much. It counts, for example, "-M---I-T..." and "--M-I--T..." as different, even though both are basically "MIT..."
Here's a way to think of it. Starting from TTAA, allow yourself to insert the other letters, C, E, H, I, M, M, and S, one at a time, anywhere before, between, or after letters already in position. (It might help to picture the second M as, temporarily, an N, remembering to divide by $2$ when you're done.) The C have $5$ places it can go, the E then has $6$, the H has $7$, and so forth, for a total of
$$(5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11)/2$$
Can you now do the other examples you asked about (both A's before both M's or both M's before the E)? The first of those should actually be very easy!
Let's do it smaller with $PPE$. If the letters $P$ are given an index then there are $3!=6$ possibilities:
- $P_1P_2E$
- $P_2P_1E$
- $P_1EP_2$
- $P_2EP_1$
- $EP_1P_2$
- $EP_2P_1$
If the indices are taken away then $P_1P_2E$ and $P_2P_2E$ both become $PPE$. It appears that possibility $PPE$ has been counted $2!=2$ times. To repair this we must divide by $2!$ and get $3$ as answer. This agrees with the fact that there are $3$ possibilities:
Likewise in $PEPPER$ every permutation is originally counted $3!2!$ times, so we must divide $6!$ by $3!2!$ to repair.
Best Answer
We can bypass Inclusion/Exclusion by using a minor modification of Stars and Bars.
First arrange the $8$ non-A's. The number of arrangements is the multinomial coefficient $$\binom{8}{2,2,2,1,1}.$$ Alternately, put a sticker on one of the two L, one of the two S, and one of the two E, to distinguish them. The $8$ distinct letters can be arranged in $8!$ ways. Remove the stickers, one at a time. Each time we remove a sticker, $2!$ arrangements collapse into $1$. So the number of arrangements of the $8$ letters is $$\frac{8!}{2!2!2!}.$$
The $8$ letters determine $9$ "holes," one at the left end, one at the right end, and seven inter-letter holes. We need to choose $3$ of these holes to put the A's into. This can be done in $\binom{9}{3}$ ways.
It follows that the number of words with no two A's adjacent is $$\binom{8}{2,2,2,1,1}\binom{9}{3} \qquad\text{or equivalently}\qquad\frac{8!}{2!2!2!}\binom{9}{3} .$$
Compute. We get $(5040)(84)$, which is $423360$.
Comment: In the same way, we can get a simple expression for the answer if there are many more letters, including many more A's. Inclusion/Exclusion also generalizes, but gets gradually more unpleasant.