So the judges are going to pick 3 pigs from the 14 to give a ribbon. Since the ribbons are different, the order does matter. So this is a permutation. This can be set up like this:
[tex] \frac{14!}{(14-3)!} [/tex]
Basically, this means we'll be multiplying 14*13*12, which is 2184. This makes A the correct answer :)
