Do I add or multiply in this question? 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 
 A: 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
$${25\choose5}{20\choose5}{15\choose5}{10\choose5}{5\choose5}\;\; \text{or if you so prefer,}\;\;\frac{25!}{(5!)^5}$$
A: You add when two events cannot be performed at the same time, which is not the case here.  You multiply when each task can be performed separately.  
We can select one team by selecting six of the $30$ players in $\binom{30}{6}$ ways.  For each such choice, we can select a second team by selecting six of the $24$ remaining players in $\binom{24}{6}$ ways.  Thus, there are $\binom{30}{6}\binom{24}{6}$ ways of selecting two teams.  For each way we can select two teams, we can select a third team by selecting six of the remaining $18$ players.  We can then select a fourth team by selecting six of the remaining $12$ players.  The fifth team is then selected by default.  Therefore, there are 
$$\binom{30}{6}\binom{24}{6}\binom{18}{6}\binom{12}{6}\binom{6}{6}$$
ways of selecting five labeled teams.  If the teams were unlabeled, we would have to divide this result by $5!$ to account for the different orders in which the teams could be labeled.
For the second question, there are five goalies. Let's label them A, B, C, D, E.  We can complete the selection of a team by assigning five of the remaining $25$ players to play with goalie A in $\binom{25}{5}$ ways.  We can select a second team by assigning five of the remaining $20$ players to play with goalie B in $\binom{20}{5}$ ways.  We can select a third team by assigning five of the remaining $15$ players to play with goalie C in $\binom{15}{5}$ ways.  We can select a fourth team by assigning five of the remaining $10$ players to play with goalie D in $\binom{10}{5}$ ways.  The fifth team is comprised of goalie E and the five unselected field players.  Since each team is determined by who plays with a particular goalie, the number of ways of selecting the five teams is 
$$\binom{25}{5}\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$$
