How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga? How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga?
I'm thinking we need permutations? I already tried finding each letter and still don't get it... any help? 
So far I've broken it off to:-
$M=1$
$I=2$
$S=4$
$A=2$
$U=1$
$G=1$
However, I don't really know what to do next.
 A: 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 $1+20+18+360+360=759$.
A: Here's a hint as to one way to do it. Break it up into mutually-distinct cases:


*

*All four letters are different.

*One letter appears twice; two letters appear once.

*Two letters appear twice.

*One letter appears three times; one letter appears once.

*One letter appears four times.


Can you take it from here?
A: The word Mississauga has the following multiplicity: $M(1\times), I(2 \times ), S(4\times), A(2 \times), G(1\times), U( 1 \times)$
You want a word with $4$ letters, so lets take for example $(M,I,S,A,G,U)$.
For a $4$-letter word you have a lot of combinations. Now you just have to write them down.
I'm going to do a few and let you do the rest.
For example if you have a $4$ letter word with one $M$ , one $I$, one $A$ and one $U$. You have $(1,1,0,1,0,1)$. For this you have $\frac{11!}{1!1!1!1!}$
If you take one $M$, two $I$ and one $S$  , you have $(1,2,1,0,0,0)$. For this you have $\frac{11!}{1!2!1!}$
You have to write all of combinations possible of $(M,I,S,A,G,U)$ with the sum value of $4$, where $M,I,S,A,G,U$ can vary up to their multiplicity respectively. In the end you sum up everything and get your result. However the word Missisauga has alot of combinations, so it might take some time.
A: Considering  letters in the order $M,I,S,A,G,U,$ the approach given by MathScientist  can be encapsulated in the formula
Coefficient of $x^4$ in $4!(1+x)(1+x+\frac{x^2}{2!})(1+x+\frac{x^2}2 + \frac{x^3}{3!} +\frac{x^4}{4!})(1+x+\frac{x^2}{2!})(1+x)(1+x)$
which, of course, can be condensed and ordered systematically  to
Coefficient of $x^4$ in $4!(1+x)^3(1+x + \frac{x^2}{2!})^2  (1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac {x^4}{4!})$
