How many permutations are there of M, M, A, A, A, T, T, E, I, K, so that no two consecutive letters are the same? How many permutations are there of 
$$ M, M, A, A, A, T, T, E, I, K $$
so that there are no two consecutive letters are the same?
I would use the Inclusion-exclusion principle where
$$ A_{i} = \{ \text{on} \ i-\text{th} \text{ and }(i+1)-\text{th} \text{ position, there are two same consecutive letters} \}. $$
So my answer would be
$$ \frac{10!}{2!3!2!} - 9 \cdot \left(\frac{9!}{3!2!} + \frac{9!}{3!2!} +\frac{9!}{2!2!}\right)+ 8 \cdot \left(\frac{8!}{2!2!}\right) +8 \cdot 7 \cdot \left(\frac{8!}{3!} + \frac{8!}{2!} +\frac{8!}{2!}\right) - 7 \cdot 6 \cdot 5 \cdot 7!$$

As you observed, there are $10$ letters, of which $3$ are $A$s, $2$ are $M$s, $2$ are $T$s, $1$ is an $E$, $1$ is an $I$, and $1$ is a $K$.  
If there were no restrictions, we would choose three of the ten positions for the $A$s, two of the remaining seven positions for the $M$s, two of the remaining five positions for the $T$s, and arrange the $E$, $I$, $K$ in the remaining three positions in 
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! = \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{2!3!} \cdot 3! = \frac{10!}{3!2!2!}$$
in agreement with your answer.
From these, we must subtract those arrangements in which or more pairs of identical letters are adjacent.
Arrangements with a pair of adjacent identical letters:  We have to consider cases, depending on whether the identical letters are $A$s, $M$s, or $T$s.
A pair of $A$s are adjacent:  We have nine objects to arrange: $AA, A, M, M, T, T, E, I, K$.  Choose two of the nine positions for the $M$s, two of the remaining seven positions for the $T$s, and then arrange the five distinct objects $AA$, $A$, $E$, $I$, $K$ in the remaining five positions, which can be done in
$$\binom{9}{2}\binom{7}{2}5!$$
ways.
A pair of $M$s are adjacent:  We have nine objects to arrange:  $A, A, A, MM, T, T, E, I, K$.  Choose three of the nine positions for the $A$s, two of the remaining six positions for the $T$s, and arrange the four distinct objects $MM, E, I, K$ in the remaining four positions, which can be done in
$$\binom{9}{3}\binom{6}{2}4!$$
ways.
A pair of $T$s are adjacent:  By symmetry, there are 
$$\binom{9}{3}\binom{6}{2}4!$$
such arrangements.
Arrangements with two pairs of adjacent identical letters:  This can occur in two ways.  Either the pairs are disjoint or they overlap.
Two pairs of $A$s are adjacent:  This can only occur if the three $A$s are consecutive.  Thus, we have eight objects to arrange: $AAA, M, M, T, T, E, I, K$.  Choose two of the eight positions for the $M$s, two of the remaining six positions for the $T$s, and arrange the four distinct objects $AAA, E, I, K$ in the remaining four positions, which can be done in 
$$\binom{8}{2}\binom{6}{2}4!$$
ways.
A pair of $A$s are adjacent and a pair of $M$s are adjacent:  We have eight objects to arrange: $AA, A, MM, T, T, E, I, K$.  Choose two of the eight positions for the $T$s and arrange the six distinct objects $AA, A, MM, E, I, K$ in the remaining six positions in 
$$\binom{8}{2}6!$$
ways.
A pair of $A$s are adjacent and a pair of $T$s are adjacent:  By symmetry, there are
$$\binom{8}{2}6!$$
such arrangements.
A pair of $M$s are adjacent and a pair of $T$s are adjacent:  We have eight objects to arrange: $A, A, A, MM, TT, E, I, K$.  Choose three of the eight positions for the $A$s and then arrange the remaining five distinct objects $MM, TT, E, I, K$ in the remaining five positions in
$$\binom{8}{3}5!$$
ways.
Arrangements with three pairs of adjacent identical letters:  We again consider cases.
Two pairs of $A$s are adjacent and a pair of $M$s are adjacent:  We have seven objects to arrange, $AAA, MM, T, T, E, I, K$.  Choose two of the seven positions for the $T$s and arrange the five distinct objects $AAA, MM, E, I, K$ in the remaining five positions in
$$\binom{7}{2}5!$$
ways.
Two pairs of $A$s are adjacent and a pair of $T$s are adjacent:  By symmetry, there are 
$$\binom{7}{2}5!$$
such arrangements.
A pair of $A$s are adjacent, a pair of $M$s are adjacent, and a pair of $T$s are adjacent:  We have seven objects to arrange: $AA, A, MM, TT, E, I, K$.  Since all the objects are distinct, there are 
$$7!$$ 
such arrangements.  
Arrangements containing four pairs of adjacent identical letters:  We have six objects to arrange: $AAA, MM, TT, E, I, K$.  Since all the objects are distinct, there are 
$$6!$$
such arrangements.
By the Inclusion-Exclusion Principle, there are 
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! - \binom{9}{2}\binom{7}{2}5! - \binom{9}{3}\binom{6}{2}4! - \binom{9}{3}\binom{6}{2}4! + \binom{8}{2}\binom{6}{2}4! + \binom{8}{2}6! + \binom{8}{2}6! + \binom{8}{3}5! - \binom{7}{2}5! - \binom{7}{2}5! - 7! + 6!=47760$$
arrangements in which no two adjacent letters are identical.
