Counting letters 
Arrange the letters IMAGINATORIUM where a 3 I's are separate and NAT appears as a subsequence.

My thinking:
Start by arranging the NAT. There are three spots in the string, so we need three letters. There is one N in the word, so to chose which N and which spot 1C1. Then for the A, there are 3 A's and 1 spot, so 3 C 1, and then the same case for the T, so 1C1. All of those multiply together and we get three possible ways to make NAT.
Then we are going to build figure out what number we are at now, so remove the 3 I's, remove the 3 characters that go into NAT, and then add one character "NAT" because that string can go anywhere in the end result. So we get something like 
$ {(13 - 3 + 1 - 3) ! } \over {2! \cdot 2! \cdot 1! ... } $ = $ 8! \over 2! \cdot 2!$
Now, there are 9 possible spots for the three I's to go, so multiply the result by 9 C 3 to get the answer. 
While this makes sense to me, I usually get these questions wrong, so let me know where my logic breaks please. Thank you!
 A: We have a total of $13$ real letters. However, it is useful to tie NAT together and think of it as a new "letter."
So there are $11$ "letters." I count $3$ I's and $2$ M's and the rest distinct.
First deal with the non-I's. There are $8$  of these, which can be arranged in $8!$ ways, if we paint the two M's different colours. But if we want to think of the two M's as identical, we only have $\frac{8!}{2!}$ words.
Now we need to insert the $I$'s.  Given an $8$-letter word, that determines $7+2$ "gaps." Of these, $7$ are real gaps between letters, and $2$ are the "end gaps." We need to choose $3$ of these to insert $I$ into. That gives a total count of
$$\binom{9}{3}\frac{8!}{2!}.$$
Remark: The above is essentially your analysis, except that you have an additional division by $2$ that I think is not correct. Once we have made a new letter out of NAT, there is only one free  A left.
To see this, let's solve the same problem with the much simpler base word NAAT. There would be $2$ words that met our condition, ANAT and NATA. Your procedure would yield $\frac{2!}{2!}=1$. 
