Problem related to probability from Sheldon Ross' book The stumbled upon following problem from Sheldon Ross's book:

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue and 18 green balls. What is probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn?

In this pdf, the solution is given as $p=\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.
But I guess it should be $\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\color{red}{2\times}\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.
Am I right? (The book has not given answer.)

I'd interpret the question as that exactly one of the two events (3 red or 3 blue) occurs but not both ("either  or" strongly hints at that). This amounts to $P(3 \text{ red}) + P(3 \text{blue}) - 2P(3 \text{ red and } 3 \text{ blue })$ which is
$${{\binom{12}{3}\binom{16+18}{4}}\over{\binom{12+16+18}{7}}} + {{\binom{16}{3}\binom{12+18}{4}}\over {\binom{12+16+18}{7}}} - 2{{\binom{12}{3}\binom{16}{3}\binom{18}{1}} \over {\binom{12+16+18}{7}}}$$
In short, I concur.
