A team squad combination and probability problem A team of 11 is chosen randomly from a squad of 18.
Two of the squad are goal keepers and one of them must be chosen. If neither of the goalkeepers is captain or vice captain, what now is the probability that both the captains and vice captains are selected?
The working I came up with was:
$$\frac{^{14}C_{8}}{^{16}C_{10}}=\frac{3}{8}$$
The answer is correct as I checked the solutions.
What I am wondering was why $^{14}C_{8}$ was used in the working to give the correct answer. (Why use the ways to select 8 from 14?)

Apparently, exactly one goalkeeper is to be selected, and the selection has no bearing on the probability being asked, (other than to exclude both from further computations.)
If both captain and vice-captain are to be included, only $8$ more are needed from $14$ left which explains the numerator.
