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?

An algebraic solution
The permutations of all 9 letters taken together = $\dfrac{9!}{1!2!1!2!3!}=\dfrac{9!}{a!b!c!d!e!}$, say
Leaving 1 letter at a time, we get $8!\left[\dfrac{1}{(a-1)!b!c!d!e!} + \dfrac{1}{a!(b-1)!c!d!e!} . . . +\dfrac{1}{a!b!c!d!(e-1)!}\right]$ 
Putting under a common denominator, we get $8!\left[\dfrac{a+b+c+d+e}{a!b!c!d!e!}
\right]$ 
But $a+b+c+d+e = 9$, so ...... you should be able to see the end.
To simplify, we can leave the multiplication factor as 9 rather than ${9\choose 8}$ 
