Find the number of permutations of the word MATHEMATICS that satisfy at least one of three restrictions Find the number of permutations of letters of the word ’MATHEMATICS’ where: both letters T
are before both letters A or both letters A are before both letters M or both letters M are before
letter E.
We have 2 Ms 2 As 2 Ts, so the sum of all permutations is $$ \frac{11!}{2!\cdot 2!\cdot 2!} $$
I think i need to do it by the inclusion-exclusion principle, do i need to find that situations
A - Both T before both A $$ \frac {8 \choose2}{2!} \cdot 6!$$
B - Both M before both A  $$ \text{same as } |A| $$
C - Both M before E $$ \frac {9 \choose 2}{2!\cdot2!} \cdot 7! $$
And actually I dont know is it correct, moreover I dont know how to push it forward

Your strategy is correct but not all of your numbers are.  
I will use the same notation that you did. 
Let $A$ be the event that both Ts appear before both As.  
Let $B$ be the event that both Ms appear before both As.
Let $C$ be the event that both Ms appear before E.  
Then, as you observed, the number of permutations of the word MATHEMATICS in which both Ts before both As or both Ms appear before both As or both Ms appear before both E is 
$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
Your correctly observed that the number of distinguishable permutations of the word MATHEMATICS is 
$$\binom{11}{2}\binom{9}{2}\binom{7}{2} \cdot 5! = \frac{11!}{2!2!2!}$$
$|A|$ 
Within a given permutation of the word MATHEMATICS, the letters A, A, T, T can be permuted among themselves in $\binom{4}{2} = 6$ distinguishable ways.  In only one of these arrangements do both Ts appear before both As.  Hence, by symmetry, the number of distinguishable permutations of the word MATHEMATICS in which Ts appear before both As is 
$$\frac{1}{6} \frac{11!}{2!2!2!}$$
$|B|$
Replacing T by M in the preceding argument shows that the number of distinguishable permutations of the word MATHEMATICS in which both Ms appear before both As is 
also 
$$\frac{1}{6} \frac{11!}{2!2!2!}$$
$|C|$
Within a given permutation of the letters of the word MATHEMATICS, the letters E, M, M can be permuted among themselves in $3$ distinguishable ways.  In only one of these arrangements do both Ms appear before E.  Thus, by symmetry, the number of distinguishable permutations of the word MATHEMATICS in which both Ms appear before E is 
$$\frac{1}{3} \frac{11!}{2!2!2!}$$
Thus far, our answers agree.  The reason for this is that the phrase "in only one of these arrangements, ..." applies. For the intersections, this is not true.
$|A \cap B|$
Within a given permutation of the word MATHEMATICS, the letters A, A, M, M, T, T can be permuted among themselves in 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
distinguishable ways.  The requirements that both Ts appear before both As and both Ms appear before both As means that the two As must occupy the last two of the six positions.  The two Ms and two Ts can be arranged in the first four positions in $\binom{4}{2}\binom{2}{2}$ distinguishable ways.  Hence, the fraction of permissible arrangements is 
$$\frac{\dbinom{4}{2}\dbinom{2}{2}}{\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}} = \frac{1}{\dbinom{6}{2}} = \frac{1}{15}$$
Thus, the number of distinguishable permutations of the word MATHEMATICS in which both Ts appear before both As and both Ms appear before both As is 
$$\frac{1}{15} \frac{11!}{2!2!2!}$$
$|A \cap C|$
Within a given permutation of the word MATHEMATICS, the letters A, A, E, M, M, T, T can be permuted among themselves in 
$$\binom{7}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
distinguishable ways.  We require that both Ts appear before both As and both Ms appear before E.  Observe that once we choose four of the seven positions for the two Ts and two As, there is only one permissible arrangement of the letters A, A, E, M, M, T, T since the first two of the four chosen positions must be occupied by Ts, the last two of those positions must be occupied by As, the first two of the remaining three positions must be occupied by Ms, and the final remaining position must be occupied by E.  Hence, the fraction of permissible arrangements is 
$$\frac{\dbinom{7}{4}}{\dbinom{7}{2}\dbinom{5}{1}\dbinom{4}{2}\dbinom{2}{2}} = \frac{35}{630} = \frac{1}{18}$$
Hence, the number of distinguishable permutations of the word MATHEMATICS in which both Ts appear before both As and both Ms appear before E is 
$$\frac{1}{18} \frac{11!}{2!2!2!}$$ 
$|B \cap C|$
Within a given permutation of the word MATHEMATICS, the letters A, A, E, M, M can be permuted among themselves in 
$$\binom{5}{2}\binom{3}{1}\binom{2}{2}$$
distinguishable ways. The requirements that both Ms appear before both As and that both Ms appear before E means the two As must appear in the first two of these five positions. The E and two Ms can be arranged in the last three positions in $\binom{3}{1}$ distinguishable ways.  Hence, the fraction of permissible arrangements is 
$$\frac{\dbinom{3}{1}}{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2}} = \frac{1}{\dbinom{5}{2}} = \frac{1}{10}$$
Hence, the number of distinguishable permutations of the word MATHEMATICS in which both Ms appear before both As and both Ms appear before E is 
$$\frac{1}{10} \frac{11!}{2!2!2!}$$
$|A \cap B \cap C|$
We have already seen that within a given permutation of the word MATHEMATICS, the letters A, A, E, M, M, T, T can be permuted among themselves in 
$$\binom{7}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2} = 630$$
distinguishable ways.  
We require that both Ts appear before both As, both Ms appear before both As, and both Ms appear before E.  Since two Ms and two Ts must appear before the first A, the two As must appear in the last three positions.  Since the two Ms must appear before E, an M cannot appear in the last three positions.  We consider cases.
An E appears in the last three positions:  The last three positions can be filled with two As and an E in $\binom{3}{1}$ ways.  The first four positions can be filled with two Ms and two Ts in $\binom{4}{2}$ ways.  Hence, there are 
$$\binom{4}{2}\binom{3}{1}$$
permissible arrangements of A, A, E, M, M, T, T in this case.
A T appears in the last three positions and E appears in the fourth position:  The last three positions can be filled with two As and a T in $\binom{3}{1}$ ways.  The first three positions can be filled with two Ms and a T in $\binom{3}{1}$ ways.  Hence, there are 
$$\binom{3}{1}\binom{3}{1}$$
permissible arrangements of A, A, E, M, M, T, T in this case.
A T appears in the last three positions and E appears in the third position:  Since both Ms must appear before E, they must occupy the first two positions and a T must occupy the fourth position.  The last three positions can be filled with two As and a T in $\binom{3}{1}$ ways.  Hence, there are
$$\binom{3}{1}$$
permissible arrangements of A, A, E, M, M, T, T in this case.
Thus, there are 
$$\left[\binom{4}{2} + \binom{3}{1} + 1\right]\binom{3}{1} = 30$$
permissible arrangements, so the fraction of permissible arrangements is 
$$\frac{30}{630} = \frac{1}{21}$$
and the number of distinguishable arrangements of the word MATHEMATICS in which both Ts appear before both As and both Ms appear before both As and both Ms appear before both Es is 
$$\frac{1}{21} \frac{11!}{2!2!2!}$$
