Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI How many ways can a sequence of three letters be chosen from the letters of MISSISSIPPI? 
I'm just a little confused how to go about this since so many letters repeat 
I=4
S=4
P=2
So ultimately there are 4 different letters. 

For a generating function solution, use exponential generating functions.  
Let's take an example that's a bit simpler than yours.  How many three-letter words can be made from the letters in AABBB?  If you were simply forming a multiset of three letters, but not arranging them in order, you could let $r$ be the number of As and $3-r$ be the number of Bs.  Then $r$ could be any of $0,$ $1,$ or $2.$  So the number of such multisets is $3.$  Notice that $3$ is the coefficient in front of $x^3$ in the expansion of $(1+x+x^2)(1+x+x^2+x^3).$  The first factor reflects the fact that you can take $0,$ $1,$ or $2$ As, and the second that you can take $0,$ $1,$ $2,$ or $3$ Bs.  Since As are not distinguishable and Bs are not distinguishable, there is only one way to take a particular number of As or Bs.  This is why the coefficients in front of the powers of $x$ are all $1.$
Now let's worry about how many ways there are to arrange the letters in a multiset.  If our multiset has $r$ As and $n-r$ Bs, there are $\frac{n!}{r!(n-r)!}={}_nC_r$ ways to arrange the letters.  We can account for this by using exponential generating functions instead of ordinary generating functions.  The way this works is that we expand
$$
\left(\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}\right)\left(\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}\right).
$$
Again we find the coefficient of $x^3$ in the expanded function, but this time we multiply the coefficient by $3!.$  An equivalent way to say this is that we find the coefficient of $\frac{x^3}{3!}$ in the expansion.  This coefficient is guaranteed to be a whole number since it is effectively introducing binomial coefficients into the expansion.  In the unordered example above, $x^3$ arose as the sum $1\cdot x^3+x\cdot x^2+x^2\cdot x.$  In the ordered case, we get
$$
\frac{1}{0!\,3!}1\cdot x^3+\frac{1}{1!\,2!}x\cdot x^2+\frac{1}{2!\,1!}x^2\cdot x=\frac{1}{3!}\left({}_3C_0\cdot 1\cdot x^3+{}_3C_1\cdot x\cdot x^2+{}_3C_2\cdot x^2\cdot x\right),
$$
which is exactly what we want.  We find that the number of words is $1+3+3=7.$  This is correct: BBB, ABB, BAB, BBA, AAB, ABA, BAA.
You should be able to generalize this idea to your problem.
