How many arrangements of MATHEMATICS are there that have all of the following properties:
(a) TH appears together in the that order
(b) E appears somewhere before C
My attempt at solving this went like this:
Note: we have 2 M's and 2 A's and since TH has to appear in that order, it can be considered a single element instead of two elements. Mathematics has 11 letters but we have 10 spots to fill since TH is again, one element and appears together every time.
so the first thing I considered is the case where C is the second element making E the first element. I came up with the equation
$${ \frac {8!}{2!2!}} $$
since we have 8 elements after choosing E and C and then the $2!$ is there since we have 2 A's and 2 M's.
After this I thought of the next possibility which was C is the third letter and that makes E either the first or second letter, which gave me the following:
$$(2 \times 8) + 7!$$
The left side is that E can be in either position and that the other position will be filled with one of the remaining 8 letters. When I got to the right side is when I realized that I was gonna get stuck.
How do I account for the fact that I might not need to have the $2!$ in the denominator if either A or M is in the left side of the arrangement next to E and the cases where it isn't? My guess is it would be something like
$$\frac{7!}{2!2!} + (2 \times \frac{7!}{2!}) $$
which includes the case where both A and M are on the right side of C and then the case where only A and only M are on that side. As C moves farther down the arrangement and there is less space for elements to be there, I would assume I would have to take into consideration the fact that the left side of C could and will have two A's and/or two M's.
maybe I am over complicating this problem but this is what I could think of. If I'm doing something wrong here please let me know and point me in the right direction.
For part $b)$, instead of considering every case one by one, you can directly say in half of the arrangements, $E$ appears before $C$ and in half of them $E$ appears after $C$. Your approach in part $a)$ seems correct to me so it is just a simplification.
In other words, we can simply say sticking $TH$, we have $\frac{10!}{2! \cdot 2!}$ different arrangements as you suggested and in $\frac{1}{2} \cdot\frac{10!}{2! \cdot 2!}$ of them, $E$ appears somewhere before $C$.
NOTE: The answer I gave is valid since there is only one $E$ and one $C$ in the word. If number of appearances of $C$ and $E$ were different, this propotion ($1:1$ in this case) would have changed of course.