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.
We have to choose an element, then to choose a permutation of the remaining elements without fixed points (a derangement). Since there are $9$ permutations without fixed points in $S_4$ ($3$ elements with the cycle structure of $(1\,2)(3\,4)$ and $6$ elements with the cycle structure of $(1\,2\,3\,4)$), the answer is $5\cdot 9=45$.
Best Answer
The keyword here is derangements. The formula for the number of derangements of $n$ things is a bit messy:
$$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;.$$
You’ll find some other formulas, less easy to prove but more usable, at the link; perhaps the nicest is
$$d_n=\left\lfloor\frac{n!}e+\frac12\right\rfloor\;,$$
for $n\ge 1$.