Presuming you mean just that the A has to be before the I and the I before the O in the permutation, I don't think the answer is 9*8*7.
Think about it this way: for every "good" permutation, how many "bad" ones are there where the vowels are in the same positions but our of order? For instance the good permutation ALGIROTHM has a matching bad permutation ALGORITHM.
Now if you know how many permutations there are altogether, and how many bad permutations there are for every good one, you can get the answer.
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
So the vowels in "GARDEN" are 'A' and 'E'. Total permutations of the word GARDEN are $6! = 720.$ In half of them, A will occur before E and in the other half of the permutaions, E will occur before A. This is obvious as "A before E" and "E before A" are equally likely and exclusive(nothing other than these two can happen) events.
So both must have probability $1/2$. Hence in half of the words, vowels are in alphabetical order. Hence, answer is $720/2 = 360$.