TITLE: Permutation of multiset
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: As you observed (in a comment on a different answer), the number of
8-permutations of this multiset is equal to the number of 9-permutations.
As noted in another answer, we can show that this is true
by observing that every 9-permutation can be mapped to an 8-permutation by
dropping the last letter; and the inverse of this mapping exists
(simply add back the letter that was dropped; there is only one possible
way to do this), so it is one-to-one and onto and the two sets of
combinations have equal size.
This is a general observation that is true for any 9-letter word.
The reason not to ignore the method of the book is that the
"one-case" method works in general only for $(n-1)$-permutations
(and of course $n$-permutations) of an $n$-letter word.
If you were looking for 7-permutations of 'ADDRESSES'
then you might need to consider the different cases depending on which
letters were removed.
There will certainly not be a one-to-one mapping to the 9-permutations.