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. 

Our desired $3$-tuples, will consist of $x$  M's, $y$  I's, $z$  S's and $w$  P's.
These should be picked among the letters of the word MISSISSIPI and must sum to $3$: 
\begin{equation}
x+y+z+w=3
\end{equation}


*

*$x$ -> # of M's, thus: $x=0,1$

*$y$ -> # of I's, thus: $y=0,1,2,3,4$ (note that $y=4$, is obviously excluded)

*$z$ -> # of S's, thus: $z=0,1,2,3,4$ (note that $z=4$, is obviously excluded)

*$w$ -> # of P's, thus: $w=0,1$


The possible solutions to the equation $$x+y+z+w=3$$ under the above restrictions on the values of $x,y,z,w$, are displayed in the following table: 
$$
\begin{array}{c|c|c|c}
   x & y & z & w \\ \hline
   0 & 1 & 2 & 0 \\
   0 & 2 & 1 & 0 \\
   0 & 3 & 0 & 0 \\
   0 & 0 & 3 & 0 \\
   1 & 2 & 0 & 0 \\
   1 & 0 & 2 & 0 \\
   1 & 1 & 1 & 0 \\
   0 & 2 & 0 & 1 \\
   0 & 0 & 2 & 1 \\
   0 & 1 & 1 & 1 \\
   1 & 1 & 0 & 1 \\
   1 & 0 & 1 & 1 
\end{array}
$$
What remains to do, is to perform the following summation:
$$\sum_{x+y+z+w=3}\frac{3!}{x!y!z!w!}$$
on the rows of the table above. Consequently we get:
$$\frac{3!}{1!2!}+\frac{3!}{2!1!}+\frac{3!}{3!}+\frac{3!}{3!}+\frac{3!}{1!2!}+\frac{3!}{1!2!}+\frac{3!}{1!1!1!}+\frac{3!}{2!1!}+\frac{3!}{2!1!}+\frac{3!}{1!1!1!}+\frac{3!}{1!1!1!}+\frac{3!}{1!1!1!}= \\ =3+3+1+1+3+3+6+3+3+6+6+6=44$$
