I attempted the following question
How many ways can a team of 24 hockey players choose a captain and two
alternate captains?
The correct answer in the end turned out to be 6072, computed by multiplying 24 by the result of the combination $C(23, 2) = 253$:
24 * $ \dfrac{23!}{2!(23-2)!} $
I do not understand why this is the case however. Why make the distinction between the 24 and then the combination of $C(23, 2)$? My initial attempt involved simply using the combination formula with with 24 as the total and 3 as the sample, that being the captain and 2 alternate captains in the problem. This would have given me:
$ \dfrac{24!}{3!(24-3)!} = 2024 $
I am struggling to figure out what I am overlooking in this problem.
Best Answer
After having chosen $3$ special players from the $24$ players (via one of $\displaystyle\binom{24}3$ possible ways),
you next need to elect one of the three to be the actual captain (via one of $3$ possible ways).