<s>[INST] Two combinatorics problems. I'm not 100% confident in my answers These are two problems from my combinatorics assignment that I'm not quite confident in my answer. Am I thinking of these the right way?

Problem 1: On rolling 16 dice. How many of the $6^{16}$ possible (ordered) outcomes would contain exactly three quadruplets? (the same number appearing four times.)
Answer: There are $ \binom{16}{4, 4, 4, 1, 1, 1, 1} $ ways to choose the dice (four quadruplets and four singlets).
There are $\binom{6}{3}$ possible values the quadruplets can have.$^1$
The possible values of the singlets can repeat, yielding $3^4$ in total. But three of those combinations would be
such that they all have the same value; and we can't have another quadruplet. So the number of possible values for the
singlets must be $3^4 - 3$.
Thus, the total number of possible values is
\begin{equation*}
\binom{16}{4,4,4,1,1,1,1}  \binom{6}{3} (3^4 - 3) = 2 \, 361 \, 078 \, 720 \, 000
\end{equation*}



*

*I think this should be changed to $\frac{6!}{3!}$ instead of $\binom{6}{3}$? 



problem: Four friends and 21 boys are to be randomly divided into five teams of 5 players. How many of the $\binom{25}{5,5,5,5,5} \frac{1}{5!}$ possible ways of creating five teams will have each of the four friends play for a different team?
Answer: There are $\frac{4!}{4!}$ ways to distribute the four friends into four teams. There are
$\binom{21}{1}$ ways to choose a leader for the fifth team from the remaining $21$ kids.
There are $\binom{20}{4,4,4,4,4} \frac{1}{5!}$ ways to distribute the remaining $20$ kids among
the five teams.
This yields a total number of possible teams:
\begin{equation*}
\frac{4!}{4!} \binom{21}{1} \binom{20}{4,4,4,4,4} \frac{1}{5!} = 3 \, 547 \, 982 \, 095 \, 257 \, 600
\end{equation*}

 [/INST] Your first answer is right after you make the correction in the footnote: the order of the quadruples does matter.
In the second problem you can’t simply pick a team leader for the fifth team: that counts that team five times, since any member of it could be the leader. I’d analyze it as follows. There are $\binom{21}{5,4,4,4,4}$ ways to split the $21$ boys into a team of $5$ and four teams of $4$, and there are $4!$ ways to assign the friends to the teams of $4$, for a total of $$\binom{21}{5,4,4,4,4} 4! = \frac{21!4!}{5!4!4!4!4!} = \frac{21!}{5!4!4!4!} = 30,798,455,688,000$$ ways of creating the five teams.
Added: I slipped up here. $\binom{21}{5,4,4,4,4}$ actually counts the number of ways to pick the team of $5$ and then to pick the other four teams in a particular order, so we have to divide by $4!$ to get the number of ways to split the $21$ boys into a team of $5$ and four (unordered) teams of $4$ each. Then there are $4!$ ways to pair the friends with the short-handed teams, so the final result should be simply $$\frac{\binom{21}{5,4,4,4,4}}{4!} \cdot 4! = \binom{21}{5,4,4,4,4} = 1,283,268,987,000.$$
Of course as André points out in his solution, we might as well have picked the teams in a particular order to begin with $-$ perhaps in alphabetical order of the friends’ names.
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