TITLE: Do I add or multiply in this question?
QUESTION [5 upvotes]: There are 30 soccer players who need to be divided up into teams.
(a) How many ways can you construct 5 teams of 6 players?
(b) If 5 of the players are goalies (and each team has exactly one goalie), how many ways can you form the teams?
For (a) we have $$C(30,6) \times C(24,6) \times C(18,6) \times C(12,6) \times C(6,6)$$
way to construct $5$ teams of $6$ players right ?
Now for part (b), We will have $25$ players and $5$ goalies to choose from right , so we will have $$C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)$$ ways to choose the players, and then we have $5$ goalies, and we want to count the number of ways that those $5$ goalies can be assigned to each team.
I visualize it this way, How many ways can we rearrange
$1 2 3 4 5$, it is basically $5!$ right so we I use the rule of sum here and I add $5!$ to $C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)$ and I am done right ?
I am confused whether to add or multiply the $5!$, But I kinda feel it should be addition and not multiplication 
Is that correct and is there another way to attack this problem

REPLY [4 votes]: It is agreed that it is the multiplication principle that is to be used.
However, you need to be very careful to see whether the computation is for labelled teams or unlabelled teams
(a)
$$\text{The computation}\;\;{30\choose 6}{24\choose 6}{18\choose 6}{12\choose 6}{6\choose 6}\text{ is for labelled teams}$$
This will be more clear by expressing it as $\dfrac{30\cdot29\cdot28...3\cdot2\cdot1}{6!\cdot6!\cdot6!\cdot6!\cdot6!\cdot6!} = \dfrac{30!}{(6!)^5}$
which means line up the 30 with dividing lines after every 6, permute the 30 in all possible ways, and remove the permutations within each team.
If it is specified that the teams are unlabelled, you need to divide by 5!
(b)
Here the teams automatically become labelled by the 5 goalies,so we can assume that the goalies are arranged as $ABCDE,$ and attach all possible combos of 5 to them, hence
$${25\choose5}{20\choose5}{15\choose5}{10\choose5}{5\choose5}\;\; \text{or if you so prefer,}\;\;\frac{25!}{(5!)^5}$$