You need to pick different letters in each sequence, so the alphabet "MATHEMATICS" has "MAT" repeated, so you have $8$ distinct letters.
Now distribute two letters having $8$ letters in total:
$\binom{8}{2} = \frac{8!}{2!6!} = 28$ those are combinations, but you need to permute them in order to include all the pairs, so:
$28.2 = 56$
EDIT: In the case you consider all the letters from "MATHEMATICS" distinct, then you have 11 distinct letters therefore when you pick one letter, the second one cannot be any letter from the left.
MATHEMATICS $= \{1,2,3,4,5,6,7,8,9,10,11\}$
When you pick $11$ there's no possibility since rest of letters are in the left. With $10$ you have 1 possibility, $11$ etc. until you reach $1$ that has $10$ possibilities.
This is summarized as following: $\sum_{i=1}^{10}\binom{i}{1}=55$
with alphabet $\sum = \{1,2,3,4,5,1,2,3,9,10,11\}$
$10=1,9=2,3=5,2=6,1=7,5=3,4=4,3=5,2=6,1=7 \Rightarrow 1+2+5+6+7+3+4+5+6+7 = 46$
Assuming you cannot repeat digits...
Case 1: the first digit is a 2.
There are 2, candidates for the last digit, and 6 ways to fill the remaining digits.
12 arrangements under this condition.
Case 2: the first digit is a 3.
The last digit is a 5. There are still 6 ways to fill the remaining digits.
$6+12 = 18$
Best Answer
Remember that order doesn't matter. According to your reasoning, the order does matter. So your answer is off by a factor of $2! = 2$. Another way to state your answer is: $$ 2 \cdot \binom{250}{2} = 2 \cdot \dfrac{250!}{2!248!} = 2 \cdot \dfrac{250 \cdot 249 \cdot 248!}{2 \cdot 248!} = 250 \cdot 249 $$