Number of ways to form 4 lettered word out of the letters of MATHEMATICS There will be 3 cases to consider for the above problem, out of which the following will be one of them:
2 different and 2 alike letters are chosen. 
Now, we have 3 sets of alike letters, namely ${(M,M)}, (A,A), (T,T)$. So we choose any one of them in $\binom{3}{1}$ ways and out of the remaining 7 different letters, we choose 2 letters in $\binom{7}{2}$ ways and arrange the 4 selected letters in $\frac{4!}{2!}$ ways. Therefore, total ways = $\binom{3}{1} * \binom{7}{2} * \frac{4!}{2!}$.
However, I cannot understand what is wrong if we arrange the alike letters in $\frac{2!}{2!}$ ways and the different letters in $2!$ ways instead of arranging the whole 4 letters? Total no. of ways in this case is $\binom{3}{1} * \frac{2!}{2!} * \binom{7}{2} * 2!$
Which are the cases that are being left if we arranged them individually instead of taking the entire word?
 A: Your second method does not account for the positions of the repeated letter in the four-letter word.
Here is another way of looking at the case of $2$ different and $2$ alike letters:  Choose the repeated letter in three ways.  Choose two of the four positions for that letter.  Choose which two of the other seven letters in the word MATHEMATICS will appear once.  Choose which of them occupies the leftmost open position.
$$\binom{3}{1}\binom{4}{2}\binom{7}{2}\binom{2}{1}$$ 
In your second method, you chose which letter would be repeated, which two letters would appear once, arranged those two letters among themselves, and arranged the repeated letters among themselves.  What you did not do was choose the positions of the repeated letters.  Multiplying your answer in your second attempt by the $\binom{4}{2}$ ways of doing so will fix your answer.  Notice that
$$\binom{3}{1} \cdot \frac{2!}{2!} \color{green}{\cdot \binom{4}{2}}\cdot \binom{7}{2} \cdot 2! = \binom{3}{1}\binom{4}{2}\binom{7}{2}\binom{2}{1}$$
