How many $8$ letter words can be formed from the letters $PAAAARRTTM$? How many $8$ letter words can be formed from the letters $PAAAARRTTM$?  
This is what I thought:
Having $8$ letter words is basically just counting the ways of arranging the letters, after we choose to remove $2$ of the letters from the $10$ given.
Thus, I will count the ways of arranging the letters, after I remove some two letters.  
The letters I will remove come in $13$ cases and are:
$$\{P,A\},\{P,R\},\{P,T\},\{P,M\},\{A,R\},\{A,T\},\{A,M\},\{A,A\},\{R,T\},\{R,M\},\{R,R\},\{T,M\},\{T,T\}.$$  
Removing $P$ and some other letter after:
$$= \frac{8!}{3!2!2!} + \frac{8!}{4!2!} + \frac{8!}{4!2!} + \frac{8!}{4!2!2!} = 3780.$$ 
Similarly for $A$:
$$= \frac{8!}{3!1!2!} + \frac{8!}{3!2!1!} + \frac{8!}{3!2!2!}+ \frac{8!}{2!2!2!} = 13440$$
Similarly for $R$:
$$=\frac{8!}{4!2!} + \frac{8!}{4!2!2!} + \frac{8!}{4!2!} = 2100$$
Similarly for $T$:
$$=\frac{8!}{4!2!2!} + \frac{8!}{4!2!}= 1260$$  
So total number of ways is $20580$.
However, the answer is $22260$, with a difference of $1680 = \frac{8!}{4!}$...  
Did I go wrong somewhere?
Also, for these type of questions, (given $n$ letters where some repeat and we have to form $k < n$ words from them), is this the systematic way to approach it?

you made two errors:
for $R$:
$$=\frac{8!}{4!2!} + \frac{8!}{4!} + \frac{8!}{4!2!} $$
for $T$:
$$=\frac{8!}{4!2!} + \frac{8!}{4!2!}$$
$$\text{difference}=(-\frac{8!}{4!2!2!}+\frac{8!}{4!})+(-\frac{8!}{4!2!2!}+\frac{8!}{4!2!})=\frac{8!}{4!}$$
