Permutation of multiset 
How many 8-permutation are there of the letters of the word
  'ADDRESSES'?

My textbook suggests that we should divide the situation into cases where a different letter is removed. In other words, for the multiset $\{1 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$, we count the number of permutation of the follow set:


*

*$\{0 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$

*$\{1 \cdot A, 1 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$

*$\{1 \cdot A, 2 \cdot D, 0 \cdot R, 2 \cdot E, 3 \cdot S\}$

*$\{1 \cdot A, 2 \cdot D, 1 \cdot R, 1 \cdot E, 3 \cdot S\}$

*$\{1 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 2 \cdot S\}$


It is easy to show that the total number of 8-permutation is $15120$. I am not happy with this case-dividing computation and I want to have a direct computation of the result. Accidentally, I find that $$C_8^9\frac{8!}{2!2!3!}=15120.$$ I further test this formula, e.g. 3-permutation of a multiset with 4 elements, and it actually works. So I think there should be a nice explanation why the formula works. Can anyone explain that for me?

There are nine letters in ADDRESSES, but we have eight positions to fill.  Thus, we must omit one of the letters.  If we omit the A, we can fill two of the eight positions in $\binom{8}{2}$ ways.  We can then fill one of the reamining six positions with an R in $\binom{6}{1}$ ways.  We can fill two of the remaining positions with E's in $\binom{5}{2}$ ways.  Finally, we fill the remaining three positions with S's in $\binom{3}{3}$ ways.  Hence, there are $$\binom{8}{2}\binom{6}{1}\binom{5}{2}\binom{3}{3} = \frac{8!}{2!6!} \cdot \frac{6!}{1!5!} \cdot \frac{5!}{2!3!} \cdot \frac{3!}{0!3!} = \frac{8!}{2!1!2!0!3!} = \frac{8!}{2!2!3!}$$ 
$8$ permutations of the letters of ADDRESSES in which A is omitted.  Since there is only one A and one R in ADDRESSES, there are the same number of $8$ permutations in which an R is omitted (interchange the roles of A and R in the above argument).  
By similar argument, the number of $8$ permutations in which a D is omitted is 
$$\binom{8}{1}\binom{7}{1}\binom{6}{1}\binom{5}{2}\binom{3}{3} = \frac{8!}{2!3!}$$
Since there are two D's and two E's in addresses, there are the same number of permutations in which an E is omitted.  
The number of $8$ permutations in which an S is omitted is 
$$\binom{8}{1}\binom{7}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2} = \frac{8!}{2!2!2!}$$
Adding the cases above yields
\begin{align*}
\frac{8!}{2!2!3!} + \frac{8!}{2!2!3!} + \frac{8!}{2!3!} + \frac{8!}{2!3!} + \frac{8!}{2!2!2!} & = \frac{8!}{2!2!3!}(1 + 1 + 2 + 2 + 3)\\
& = 9 \cdot \frac{8!}{2!2!3!}\\ 
& = \frac{9!}{2!2!3!}
\end{align*}
