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?)

The squad of $18$ has two goalkeepers. Out of these, (exactly) one must be selected. But as none of them is the captain or the vice captain, the result of this selection has no consequence on the probability that has to be found.
Out of the eighteen, the goalkeeper can be selected in $\binom{2}{1}$ ways. Out of the remaining sixteen, ten players can be chosen in $\binom{16}{10}$ ways. The total number of ways is then $\binom{16}{10}\binom{2}{1}$.
In the favourable case, the number of ways of choosing the goalkeeper remains $\binom{2}{1}$. Now, the captain and the vice captain must be chosen, which can be done in $\binom{1}{1}$ ways each. Out of the remaining fourteen players, eight have to be chosen which can be done in $\binom{14}{8}$ ways.
The probability then becomes,
$$\frac{\binom{14}{8}\binom{1}{1}\binom{1}{1}\binom{2}{1}}{\binom{16}{10}\binom{2}{1}}=\frac{\binom{14}{8}}{\binom{16}{10}}$$
