We can encode the five nonempty customer groups in the following way as a $01$-sequence:
$$0\ \ldots\ |0\ \ldots\ |0\ \ldots\ |0\ \ldots\ |0\ \ldots\quad.$$
Here each $0$ denotes a person, each $\ldots$ a nonnegative number of zeros, and each $|$ a separator between groups. Since there have to be $15$ zeros in all we have to distribute $10$ remaining zeros onto the five $\ldots\ $. This can be encoded as a word containing $10$ zeros and $4$ separators (denoted by $\|$ this time). There are ${14\choose 4}$ words of this kind, which implies there are ${14\choose4}$ ways to form $5$ nonempty queues of empty chairs. For each of these configurations there are $15!$ ways to assign each of the $15$ actual persons a place on one of the chairs. Therefore the end result is $15!\cdot{14\choose4}$.
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 $$\binom{10}{5}\binom{5}{2} \cdot 3!$$
Best Answer
Your work is mostly fine.
However, for case $1$ (no letters the same), it should be $^{10}P_4 = \ ^{10}C_4 \cdot 4! = \ ^{10}C_\color{red}{6} \cdot 4!$. It would be better to write this number as a permutation, as which order you place the numbers matters.