In this small problem the easy way is to note first that there are $4\cdot3\cdot2=24$ ways to pick three different letters and permute them. Then observe that there are $3$ ways to pair LL with one of the other letters, and $3$ ways to place that odd letter relative to the two L’s, so there are $3\cdot3=9$ more $3$-letter arrangements that include both L’s, for a total of $24+9=33$ arrangements.
I will give a basic counting solution. If you want an answer using generating functions, look at the answer to this question.
If all five letters are different (which happens when you use one of each), you get $5!=120$ words.
If only two of the letters are equal (two 'S', two 'T', or two 'I', and $\binom{4}{3}$ ways to choose the remaining three), you have $\frac{5!}{2!}=60$ words for each of {'S', 'T', 'I'}. In total $720$ words.
If you have "two pairs" (there are three ways to get the pairs, and three ways to get the last letter in each case), there are $\frac{5!}{2!2!}=30$ words. In total for all these cases, there are $270$ words.
If you have "full house" (there are two ways to get three equal, and two ways for each of those to get the last pair), there are $\frac{5!}{3!2!}=10$ words. In total for these cases, there are $40$ words.
If you have three equal and two different (the triple can be had in two ways, and the remaining two can be had in $\binom{4}{2}=6$ ways for each triple), there are $\frac{5!}{3!}=20$ words. For all these cases, there are $240$ words.
Summing all of these cases gives $1390$ words.
Best Answer
To clarify the problem with $7P3$:
The reason one might think that dividing by $2$ is a good idea is that there is a symmetry, swapping the two $O's$ That is perfectly correct but for words like $MAG$ the symmetry has no bearing.
How can we use the symmetry to count correctly? Let's start by numbering the $O's$, so we have $O_1,O_2$. Now of course the answer is exactly $7P3=7\times6\times5=210$. These split into two types: those containing neither one of the $O's$ and the rest. The first type contains $5P3=5\times4\times3=60$ words. The second type contains all those words with at least one $O$, and it has $210-60=150$ elements. Now, if we identify $O_1$ and $O_2$ we see that we have counted every word of the second type twice. For example $MOG$ is counted as $MO_1G$ and $MO_2G$ while $MOO$ is counted as $MO_1O_2$ and $MO_2O_1$. Thus, to solve your problem, we need to divide that group by $2$. Thus the final answer is $$60+\frac {210-60}2=60+75=135$$ as desired.