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?)
 A: Apparently we’re supposed to understand that exactly one of the goalkeepers is chosen. Once that choice has been made, we must choose the other $10$ members of the team from the $16$ members of the squad who are not goalkeepers; this can be done in $\binom{16}{10}$ different ways. How many of these $\binom{16}{10}$ possible teams include both the captain and the vice captain? If we’ve selected both of them for the team, there are still $8$ players to be picked, and there are $14$ players left from whom they can be chosen, namely, the $14$ who are neither the captain, the vice captain, nor either of the goalkeepers. Thus, the team can be completed in $\binom{14}8$ ways.
In short, once we’ve chosen our goalkeeper, there are $\binom{14}8$ teams that include that goalkeeper, the captain, and the vice captain out of a total of $\binom{16}{10}$ teams that include that goalkeeper. Assuming that all $\binom{16}{10}$ of those teams are equally likely to be picked, the probability of getting one of the $\binom{14}8$ with the captain and vice captain is
$$\frac{\binom{14}8}{\binom{16}{10}}=\frac{10\cdot9}{16\cdot15}=\frac38\;.$$
A: 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,
$$\frac{\binom{14}{8}\binom{1}{1}\binom{1}{1}\binom{2}{1}}{\binom{16}{10}\binom{2}{1}}=\frac{\binom{14}{8}}{\binom{16}{10}}$$
A: 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.
A: We have the squad $$S=\{G_1, G_2, C, VC, 1, \cdots , 14\},$$ with $|S|=18.$
If we need at least one goalkeeper, we have all the teams settings as $$T = 2 \times {10 \choose 16} = 2 \times \frac{16!}{10! \times 6!}.$$ And the teams with $G_1$ or $G_2$ and $(C, VC)$ are simple $$A = 2 {8 \choose 14} = 2 \times \frac{14!}{8! \times 6!}.$$
Then, we have $$\mathbb{P}(\{G_1 \, || \, G_2 , C, VC, \{1,\cdots, 8\}) = \frac{A}{T} = \frac{3}{8}.$$
In fact, in the answer it was suppressed the goalkeeper permutation directly. But that's where the terms from your answer come from.
