Probability that either one, but not both is chosen A basketball team of 5 is to be selected from 12 players. Find the probability that either one, but not both of the captain or vice-captain are selected. So I thought the answer is: $$2\frac{{10}\choose{4}}{12\choose{5}}$$ Because if the captain is to be in the team, then you select the other 4 from the remaining 10, not 11, because there can't be both captain and vice-captain in the team at the same time. The same goes if the vice-captain is to be included, hence I multiply by two. According to the textbook, however, the answer is $\frac{5}{6}$. Why?

Your answer is correct, it is essentially $\frac{\binom{2}{1}\binom{10}{4}}{\binom{12}{5}} = \frac{35}{66}$, where you choose one of the captain/vice-captain group and 4 from the others.

To verify this we can compute the other probabilities as well:

both are chosen with probability $\frac{\binom{2}{2}\binom{10}{3}}{\binom{12}{5}} = \frac{5}{33} = \frac{10}{66}$, whereas none are chosen with probability $\frac{\binom{2}{0}\binom{10}{5}}{\binom{12}{5}} = \frac{7}{22} = \frac{21}{66}$ and all these nicely sum up to 1.