<s>[INST] A committee of 4 is to be formed out of 6 superheroes, 7 supervillains, and 4 citizens. How many ways to form a committee? I am, once again, working on a problem which I have a disagreement with the given solution and I was wondering which solution is correct.
A committee of four people is to be formed from a set of 6 superheroes, 7 supervillains,
and 4 ordinary citizens (17 people in total). How many ways are there to form the committee if:
$\textbf{(d)}$ At least one superhero and one supervillain must be on the committee, but Mr. Fantastic (a superhero)
and Dr. Doom (a supervillain) cannot be on the committee together.
$\textbf{Given Solution:}$
$\binom{6}{1}\binom{7}{1}\binom{15}{2}-\binom{15}{2}$

(Here we count the committees with at least one hero and at least one supervillain,
then subtract the committees with Mr Fantastic and Dr Doom on them).
QED
I feel like the solution above is problematic in this way: The term $\binom{6}{1}\binom{7}{1}\binom{15}{2}$ calculates several cases twice. To illustrate this, let us separate the 3 binomial expressions and show that the term overcounts at least one case more than once.
Case 1: in $\binom{6}{1}$ we choose Mr. Fantastic, in $\binom{7}{1}$ we choose Dr. Doom and in $\binom{15}{2}$, we choose Spongebob and Patrick, then in our Comittee we have {Mr. Fantastic, Dr. Doom, Spongebob, Patrick}. 
Case 2: Let's say Spongebob is a superhero and Patrick is a supervillain, then in $\binom{6}{1}, \binom{1}{7}, \binom{15}{2}$, we choose Spongebob, Patrick, Mr. Fantastic and Dr. Doom respectively. This forms the committee {Spongebob, Patrick, Mr. Fantastic, Dr. Doom}, which is the same committee as previously but is counted twice.
$\textbf{My Solution:}$
Consider the case where both Dr. Doom and Mr. Fantastic are both in the committee, we have $1*1*\binom{15}{2}$.
Case 1: Consider the case where there are neither superheroes nor supervillains. $\binom{4}{4}$.
Case 2: Consider the case where there are no superheroes, $\binom{11}{4}$.
Case 3: Consider the case where there are no supervillains, $\binom{10}{4}$.
Our final answer is given by $\binom{17}{4} - \binom{15}{2} - \binom{11}{4} - \binom{10}{4} + \binom{4}{4}$
Note that Case 1 is included in both Case 3 and Case 2, so we add $\binom{4}{4}$ to compensate for the exclusion of Case 1 twice (by the exclusion of case 3 and case 2)
QED
 [/INST] Your solution looks good. 
You can cross-check against the less-elegant method of assembling valid committees and then excluding the $\binom{15}{2}$ of the Fantastic/Doom constraint.
$\binom{6}{1}\binom{7}{1}\binom{4}{2} + \binom{6}{2}\binom{7}{1}\binom{4}{1} + \binom{6}{1}\binom{7}{2}\binom{4}{1} + \binom{6}{3}\binom{7}{1} + \binom{6}{2}\binom{7}{2} + \binom{6}{1}\binom{7}{3} - \binom{15}{2} \\
= 252 + 420 + 504 + 140 + 315 + 210 - 105 = 1736$
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