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.
M=1
I=2
S=4
A=2
U=1
G=1
Case 1: All 4 letters are same
There is only one arrangement for this.
Case 2: 3 letters are similar
The three letters must be S. The remaining letter can be chosen in ${5 \choose 1}$
ways and these can be arranged in $\frac {4!}{3!}$ ways. This makes a total of 20 arrangements.
Case 3: 2 pairs are similar
The two repeating letters can be chosen in ${3 \choose 2}$ ways and can be arranged in $\frac {4!}{{2!}{2!}}$. This equals 18 arrangements.
Case 4: 2 are similar
The repeating letters can be chosen in ${3 \choose 1}$ ways and the remaining two in ${5 \choose 2}$ ways. These can be arranged in $\frac {4!}{2!}$ ways. This equals 360 arrangements.
Case 5: All 4 are different
The 4 letters can be chosen in ${6 \choose 4}$ ways and can be arranged in $4!$ ways. This equals 360 ways.
Hence the total number of arrangements is $1+20+18+360+360=759$.
Best Answer
We have two possibilities: one letter E or two.
In the case of one E, there are $4$ possible positions for this first letter, and then 6 possibilities for filling the next letter (picking from the remaining letters, and excluding the other E), and then 5 possibilities, and then 4. So we have $4 \cdot 6 \cdot 5 \cdot 4 = 480$.
In the case of two Es, there are $6$ possible positions for these first two letters. We can calculate this as ${4 \choose 2} = 6$ or just by intuitively observing that there are $4$ possibilities for the first E and then 3 for the second, but dividing by two since we don’t care about the order. Then $6$ more possibilities for the third letter, and then $5$. So we have $6 \cdot 6 \cdot 5 = 180$.
So our final total is $480 + 180 = 660$ possible arrangements.