Words from $\text{ACCOMODATION}$ 
How many 4 letter words can be formed by the letters of the word $ACCOMODATION$?

I thought of first taking the letters $A$,$C$ and $O$ together as separate individual units, but couldn't get any farther.
PS. I know that accommodation has a double m, but the question had spelt accommodation with a single $m$.
 A: Here $\bf{ACCOMODATION}$ contain $\left\{2A,2C,3O,M,D,T,I,N\right\}$
Now we have to form $4$ letter words, So we will form different cases.
$\bullet\; $ If all letters are different, Then we will select $4$ letters out of total $8$ distinct letters
So this can be done by $\displaystyle \binom{8}{4}\times 4! = $
$\bullet\; $ If $2$ letter are same and other $2$ are distinct, Then we will select $1$ same letter pair 
from $3$ pairs and other $2$ distinct letter from $7$ distinct letter
so this can be done by $\displaystyle \binom{3}{1}\times \binom{7}{2}\times \frac{4!}{2!} = $
$\bullet\; $ If $2$ letter are of same kind and other $2$ letter are of same kind, Then
we will select $2$ pairs from $3$ pairs
so this can be done by $\displaystyle = \binom{3}{2}\times \frac{4!}{2!\times 2!}=$
$\bullet\; $ Id $3$ letter are of same kind and $1$ is different, 
This can be done by $\displaystyle \binom{1}{1}\times \frac{7}{1}\times \frac{4!}{3!} = $
Now total no. of arrangements $ = $ Sum of all above cases.
A: Another way is to use a generating function. There is a single occurrence of 5 letters, a double occurrence of two letters, and a triple occurrence of one letter, so just find the coefficient of $x^4$ in $$4!(1+x)^5(1+x+x^2/2!)^2(1+x+x^2/2+x^3/3!)\;\;\text{which works out to}\;\; 2482$$
