How many ways are there to do this so that no officer picks $3$ students from the same high school? There are $18$ students, three (distinct) students each from $6$ different high schools. 
There are $6$ admissions officers, one from each of $6$ colleges. Each of the officers successively picks a subset of 
three of the $18$ students to go to their college (once a student is chosen, another college cannot choose him later).
How many ways are there to do this so that no officer picks $3$ students from the same high school?
Answer is: 
$\dfrac{18!}{(3!)^6} - 6\dfrac{6 \cdot 15!}{(3!)^5} + {6 \choose 2} \dfrac{6 \cdot 5 \cdot 12!}{(3!)^4} \dots $
Can someone please explain? 
 A: Suppose the colleges are college A, B, C, D, E, F (we can properly order them alphabetically).
By inclusion-exclusion the number of ways no college violates the "don't pick all students from the same high school" rule is:
$$N_0 = N_{\geq 0} - N_{\geq1} + N_{\geq2} - N_{\geq3} + N_{\geq4} - N_{\geq5} + N_{\geq6}$$
Where $N_{\geq k}$ represents the number of ways at least $k$ colleges violate the rule.
$N_{\geq0}$ is simply the number of ways that the colleges can pick the students without regard to the "don't take from a single highschool" rule.  By multiplication principle college A picks three students out of the 18 total, college B picks three students out of the remaining 15, etc... for a total number of $\binom{18}{3}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} = \frac{18!}{(3!)^6}$ number of ways.
For $N_{\geq 1}$, first pick the college that is guaranteed to be in violation of the rule (there are possibly others, hence why we are using inclusion-exclusion): there are six colleges to choose from.  Now, take the offending college and have them pick a highschool to take all three students from: there are six highschools to choose from.  The remainder of the colleges can take randomly from the remaining students for a total of: $\binom{6}{1}\cdot 6\cdot \binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} = 6\cdot 6\cdot \frac{15!}{(3!)^5}$
For $N_{\geq 2}$, first pick two colleges that are guaranteed to be in violation of the rule: there are $\binom{6}{2}$ choices.  From here, let the offending colleges in alphabetical order choose which highschool they want to take from.  The first offending college has 6 choices, the second offending college has 5 choices.  The remaining colleges each take randomly from the remaining students for a total of $\binom{6}{2}\cdot 6\cdot 5\cdot \frac{12!}{(3!)^4}$.
The remaining cases are similar for a final total of:
$$N_0 = \frac{18!}{(3!)^6} - \binom{6}{1} \cdot 6 \frac{15!}{(3!)^5} + \binom{6}{2}\cdot 6\cdot 5\frac{12!}{(3!)^4} - \binom{6}{3}\cdot 6\cdot 5\cdot 4\frac{9!}{(3!)^3} + \binom{6}{4}\cdot 6\cdot 5\cdot 4\cdot 3\frac{6!}{(3!)^2} - \binom{6}{5}\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \frac{3!}{(3!)^1} + \binom{6}{6}\cdot 6!\cdot \frac{0!}{(3!)^0}$$
(note: it is impossible for 7 or more colleges to violate the rule since there are only six colleges that we are worried about)

In case it wasn't made perfectly clear, in response to your comment "but I don't understand why that should be multiplied by an additional 6, or why the next term is multiplied by 30", these numbers appear because in our multiplication principle setup they represent the number of ways of having the violating colleges choose which highschools they will take students from.  Note that since the colleges are distinct, order here matters so we use a permutation instead of a combination.
