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).