In how many ways can the letters of the word ${\sf DIRECTOR}$ be arranged so that the three vowels are never together?
I arranged the consonants in $5!/2!$ Then the number of gaps created around these consonants are $6$ so I choose $3$ gaps from these $6$ gaps and arranged the $3$ vowels in $3!$ ways. So my answer is $5!/2! \cdot {6 \choose 3} \cdot 3!$, but the answer given is $1800$ while my answer is coming out to be $7200$.
Best Answer
As you already determined, there are $5!/2!=60$ ways to arrange the sequence of consonants (ignoring vowels) and $3!=6$ ways to arrange the sequence of vowels (ignoring consonants). So just by considering all of these possible combinations, we have $60\times 6 = 360$ ways to sequence vowels and consonants independently.
To get a complete arrangement of all eight letters, we just need to select a pattern for interleaving vowels and consonants. The answer $1800$ implies that there are exactly $5$ such patterns that have no two vowels together. Here are some such patterns with $V$ for any vowel and $c$ for any consonant:
$$ VcVcVccc \\ VcVccVcc \\ VcVcccVc \\ VcVccccV \\ VccVcVcc \\ VccVccVc $$
That's already $6$ ways to interleave the vowels and consonants, and we clearly are nowhere near the end of this list of interleavings. So the answer $1800$ is clearly wrong.
Your answer is well-reasoned and seems correct to me.
Update: As observed in comments and another answer, thanks to the ambiguity of the English language there is another interpretation of the question that arrives at the answer $18000,$ so it is possible (even likely) that the source of this problem intended the other interpretation but mistakenly omitted a zero from the answer. (I'm still not convinced of the other interpretation; there are many ways the problem could easily have been disambiguated in that direction, for example "without three consecutive vowels.")
For the interpretation you chose, we can also work out the excluded arrangements independently of your counting method. We can start by choosing two vowels to place consecutively, for example EO. There are $6$ ways to do this, including the order in which the two vowels are written. Now, taking the double vowel as one "letter," we have $7!/2! = 2520$ arrangements from each pair of letters, for a provisional total of $6 \times 2520 = 15120$ arrangements.
But the $15120$ arrangements count every arrangement with three consecutive vowels twice. For example, arrangements with the sequence IEO are counted once when we take the pair IE and put O after it, and again when we take the pair EO and put I in front of it. There are $6$ ways to arrange the three vowels consecutively; treating the triplet of vowels as one "letter", we have $6!/2! = 360$ arrangements for each triplet, for $6 \times 360 = 2160$ arrangements altogether. Subtract $2160$ from $15120$ to cancel the double-counting of these triplet arrangements, and we have $15120 - 2160 = 12960$ arrangements with vowels together. (These arrangements include $10800$ with only two vowels together and $2160$ with all three vowels together.)
Subtract the arrangements with vowels together from the total of $8!/2! = 20160$ arrangements of DIRECTOR, and we have $20160 - 12960 - 7200$ remaining arrangements, which are all the arrangements in which vowels do not appear together.
Yet another method for counting arrangements with exactly two vowels together. Arrange the consonants in $5!/2! = 60$ ways. Take an ordered pair of vowels in any of $6$ ways and place it in any of the $6$ "gaps" between or next to consonants. Then take the remaining vowel and put it in any of the $5$ remaining "gaps." This can all be done in $60\times 6 \times 6\times 5 = 10800$ ways.