Method 1: We use symmetry.
For the moment, let's consider the number of distinguishable ways we can permute the ten letters in BOOKKEEPER. We can place the three E's in three of the ten locations in $\binom{10}{3}$ ways. We can place the two O's in two of the remaining seven places in $\binom{7}{2}$ ways. We can place the two K's in two of the remaining five places in $\binom{5}{2}$ ways. There are then $3!$ ways of arranging the B, P, and R in the three remaining places. Hence, the number of distinguishable arrangements of BOOKKEEPER is
$$\binom{10}{3}\binom{7}{2}\binom{5}{2} \cdot 3! = \frac{10!}{7!3!} \cdot \frac{7!}{5!2!} \cdot \frac{5!}{3!2!} \cdot 3! = \frac{10!}{3!2!2!}$$
Now, let's restrict our attention to arrangements of the five vowels in BOOKKEEPER. Since there are three E's and two O's, a given permutation of EEEOO is determined by in which three of the five positions the E's are placed. There are $\binom{5}{3} = 10$ ways to do this, of which just one is in alphabetical order.
Hence, the number of permutations of the letters of BOOKKEEPER in which the vowels appear in alphabetical order is
$$\frac{1}{10} \cdot \frac{10!}{3!2!2!} = \frac{9!}{3!2!2!}$$
Method 2: We place the consonants first.
There are $\binom{10}{2}$ ways of choosing the positions of the two K's, eight ways to place the B, seven ways to place the P, and six ways to place the R. Once the consonants have been placed, there is only way to fill the five remaining positions with the vowels in alphabetical order. Hence, the number of distinguishable arrangements of the letters of BOOKKEEPER in which the vowels appear in alphabetical order is $$\binom{10}{2} \cdot 8 \cdot 7 \cdot 6$$
Method 3: We place the vowels first.
There are five vowels in BOOKKEEPER, which has ten letters. We can select positions for the five vowels in $\binom{10}{5}$ ways. There is only one way to arrange the vowels in those positions in alphabetical order. There are $\binom{5}{2}$ ways to place the K's in two of the remaining five positions. There are $3!$ ways to arrange the B, P, and R in the remaining three positions. Hence, the number of distinguishable arrangements of BOOKKEEPER in which the vowels appear in alphabetical order is $$\binom{10}{5}\binom{5}{2} \cdot 3!$$
Suppose you have a $4$ letter string composed of, say, $1$ distinct and $3$ identical letters
There would be $\frac{4!}{1!3!}$ permutations, also expressible as a multinomial coefficient, $\binom{4}{1,3}$
Similarly, for $2$ distinct, $2$ identical, and $3$ distinct, $1$ identical,
it would be $\binom{4}{2,2}\;$ and $\binom{4}{3,1}$ respectively.
In the polynomial expression $4!(1+x/1!)^3(1+x+x^2/2!+x^3/3!)$,
the 4! corresponds to the numerator, whatever the combination; the first term in $( )$ corresponds to choosing one or more from $R,M,N$; and the other term corresponds to choosing $1,2,$ or $3 A's$
It will become evident why this approach works if we expand the first term in ( ), and compare serially with your case approach by just using the appropriate coefficients to get terms in $x^4$
$4!(1 + 3x + 3x^2 + x^3)(1 + x + x^2/2! + x^3/3!)$
To find the coefficient of $x^4$, consider the three cases that produce $x^4$
One from $R,M,N, 3A's : 4!\cdot3\cdot\frac{1}{3!} = 12$
Two from $R,M,N, 2A's : 4!\cdot3\cdot\frac1{2!} = 36$
Three from $R,M,N, 1A : 4!\cdot1\cdot 1 = 24$
Coefficient of $x^4 = 12+36+24 = 72$
We can now clearly see why the coefficient of $x^4$ in the expression automatically gives all possible permutations of $4$ letters
Best Answer
Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.
Then the number of arrangements is $7!$. Imagine that you write down all of them.
Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.